Finite Nilpotent Groups Whose Cyclic Subgroups are TI-Subgroups

Abstract

A subgroup H of a group G is called a TI-subgroup if $H^g\cap H=1$ or H for all $g\in G$; and H is called quasi TI if $\mathcal {C}_G(x)\le \mathcal {N}_G(H)$ for all non-trivial elements $x\in H$. A group G is called (quasi CTI-group) CTI-group if every cyclic subgroup of G is a (quasi TI-subgroup) TI-subgroup. It is clear that TI subgroups are quasi TI. We first show that finite nilpotent quasi CTI-groups are CTI. In this paper, we classify all finite nilpotent CTI-groups.

1 Introduction and Results

A subgroup H of a finite group G is called a TI-subgroup, if $H \cap H^g=1$ or H for all $g\in G$. A group G is called a TI-group if all of whose subgroups are TI-subgroups. A group G is called an ATI-group if all of whose abelian subgroups are TI-subgroups. Similarly, a group G is called a CTI-group if any cyclic subgroup of G is a TI-subgroup or equivalently $\langle {x}\rangle ^g\cap \langle {x}\rangle =1$ or $\langle {x}\rangle$ for all elements $x,g\in G$.

A subgroup H of a finite group G is called a quasi TI-subgroup or a QTI-subgroup if $\mathcal {C}_{G}(x)\le \mathcal {N}_{G}(H)$ for any $1\ne x\in H$. A group G is called a QTI-group if all of whose subgroups are QTI-subgroups. Similarly, a group G is called a AQTI-group (a quasi CTI-group, respectively) if all of whose abelian subgroups (cyclic subgroups, respectively) are QTI-subgroups.

In [1], Walls classified finite groups all of whose subgroups are TI-subgroups. In [2] and [3], Guo, Li and Flavell classified finite groups whose abelian subgroups are TI-subgroups. Also in [4], Qian and Tang classified finite groups all of whose abelian subgroups are QTI-subgroups.

A subgroup H of G is called n-embedded in G if for every $1\ne K\leqslant H$ we have $\mathcal {N}_G(K)\leqslant \mathcal {N}_G(H)$, and a group G is called n-group if every subgroup H of G is n-embedded in G. If G is nilpotent, then any quasi CTI-subgroup of G is n-embedded in G and quasi CTI-group is n-group. The n-groups have been classified by Kazarin in [5, 6], but those Russian papers are not accessible. In [7] one can find English translation of [5] but without any proofs. So we give here a proof of this particular case independently with full details.

Here, we show that a finite nilpotent group is CTI if and only if it is quasi CTI. The structure of non-nilpotent finite CTI-groups are classified in [8]. In this paper, we complete the classification of finite CTI-groups by classifying nilpotent ones.

Our main results are the following.

Theorem 1.1

Every finite nilpotent group is quasi CTI if and only if it is CTI.

Theorem 1.2

Let G be a finite nilpotent group. Then G is a CTI-group if and only if one of the following holds:

• (1) G is Dedekindian, i.e., all subgroups of G are normal in G;

• (2) G is a p-group of exponent p for some prime p;

• (3) G is a p-group for some prime p such that $G'=\Omega _1(G^p)$ is of order p and $\Phi (G)=G^p$ is a central cyclic subgroup of G; and

• (4) G is a 2-group such that $G=A\langle {x}\rangle$, where A is an abelian subgroup of G and x is an involution in $G\setminus A$ such that $a^x=a^{-1}$.

Our notations are as follows. Let G be a group, $x\in G$ and $H\le G$. Then, $\mathcal {C}_G(x)$ denotes the centralizer of x in G and $\mathcal {N}_G(H)$ denotes the normalizer of H in G. The derived subgroup, the center, and the Frattini subgroup of G are denoted by $G'$, Z(G), and $\Phi (G)$, respectively; and if G is a p-group, $G^p$ denotes $\langle {x^p \;|\; x\in G}\rangle$ and $\Omega _1(G)$ denotes $\langle {x\in G \;|\; x^p=1}\rangle$. The cyclic group of order n is denoted by $\mathbb {Z}_n$ and $Q_8$ denotes the quaternion group of order 8. The exponent of a finite group G is denoted by $\exp (G)$, and the order of an element $x\in G$ is denoted by |x|. For two elements x and y in a group, [x, y] denotes the commutator $x^{-1}y^{-1}xy$.

2 Quasi CTI-Groups

Clearly a TI-subgroup is a QTI-subgroup. So TI-, ATI-, and CTI-groups are QTI-, AQTI-, and quasi CTI-groups, respectively. In [4] Qian and Tang proved that:

Theorem 2.1

([4], Theorem  2.1). For a finite p-group G, the following statements are equivalent:

• (i) G is a TI-group;

• (ii) G is an ATI-group; and

• (iii) G is an AQTI-group.

In Theorem 1.1, we show that a finite p-group is CTI if and only if it is quasi CTI. We need the following proposition in the proof of Theorem 1.1.

Proposition 2.2

Let G be a finite quasi CTI-group and H is a non-normal cyclic subgroup of G.

• (i) $H \cap Z(G)=1$ and in particular no non-trivial subgroup of H is normal in G.

• (ii) If $x\in H$ is of prime order, then $\mathcal {C}_G(x)=\mathcal {N}_G(H)$.

• (iii) If |Z(G)| divides by two distinct primes, then G is Dedekindian.

Proof

• (i) If $1\not =x\in H\cap Z(G)$, then $G=\mathcal {C}_G(x)\leqslant \mathcal {N}_G(H)$, a contradiction.

• (ii) Since $\langle {x}\rangle \trianglelefteq \mathcal {N}_G(H)$, $x\in Z(\mathcal {N}_G(H))$ and so $\mathcal {C}_G(x)=\mathcal {N}_G(H)$.

• (iii) Let $h\in G$ be of prime order. By assumption, Z(G) contains an element x of order coprime to |h|. By (i) $\langle {xh}\rangle \trianglelefteq G$. Since $\langle {h}\rangle \leqslant \langle {xh}\rangle$, $\langle {h}\rangle \trianglelefteq G$ . Therefore, any subgroup of prime order of G is normal and so G is Dedekindian by (i).$\square$

We can now prove Theorem 1.1. Let us restate the statement of Theorem 1.1.

Theorem 1.1

Every finite nilpotent group is quasi CTI if and only if it is CTI.

Proof

Suppose that G is a finite nilpotent non-Dedekindian quasi CTI-group. By Proposition 2.2 (iii), G must be a non-abelian p-group for some prime p. Suppose H is a cyclic subgroup of G and $x\in H$ is of order p. If $H\cap H^g\ne 1$ for some element $g\in G$, then $\langle {x}\rangle =\langle {x}\rangle ^g$. Hence $g\in \mathcal {C}_G(x)=\mathcal {N}_G(H)$ by Proposition 2.2 (ii). Therefore $H=H^g$. This completes the proof. $\square$

3 CTI-p-Groups with Center of Exponent $>p$

The following lemma shows that to classify finite nilpotent CTI-groups it is enough to classify the finite CTI-p-groups. Recall that a group is called Dedekindian (or Hamiltonian) if all of its subgroups are normal.

Lemma 3.1

([8], Corollary 1.3). Let G be a CTI-group with non-trivial center.

• (i) Assume that the order of $1\ne g\in G$ is coprime to the order of an element of Z(G). Then $\langle {g}\rangle \trianglelefteq G$.

• (ii) If two distinct primes p and q divide the order of Z(G), then G is a Dedekindian group.

The preceding lemma implies that a finite non-Dedekindian nilpotent CTI-group is necessarily a non-Dedekindian p-group.

It is easy to see that a finite non-Dedekindian CTI-p-group contains a non-normal (non-central) subgroup of order p. This justifies the hypothesis of the existence of a non-central element of order p in the following lemma:

Lemma 3.2

Let G be a non-Dedekindian finite CTI-p-group such that the exponent of Z(G) is at least $p^2$ and x is a non-central element of order p of G.

• (i) For any $g\in G\backslash \mathcal {C}_G(x)$, [x, g] is an element of $Z(G)^p$ of order p.

• (ii) $\mathcal {C}_G(x)$ is a maximal subgroup of G.

• (iii) $Z(G)\cong \mathbb {Z}_{p^e}\times E$ where $e\ge 2$ and E is an elementary abelian subgroup.

• (iv) If for any non-central element y of order p, $\mathcal {C}_G(x)=\mathcal {C}_G(y)$, then $\mathcal {C}_G(x)$ is abelian.

Proof

• (i) Let $y\in Z(G)$ be of order at least $p^2$. Since $(yx)^p$ is central, $\langle {yx}\rangle \trianglelefteq G$. Hence for any $g\in G$, $yx^g\in \langle {yx}\rangle$, so $yx^g=y^ix^i$ for some i where $(i,p)=1$, then $[x,g]=x^{i-1}y^{1-i}$. Therefore $[x,g]^p=[x^p,g]=1$. Accordingly $y^{p(i-1)}=1$, thus p divides $i-1$ and so $x^g=y^{i-1}x$. This implies that $[x,g]\in Z(G)^p$ and $\langle {y,x}\rangle \trianglelefteq G$. Let $Z\leqslant \langle {y}\rangle$ be of prime order. Then

  $$\begin{aligned} \langle {x^g,x}\rangle =\langle {Z,x}\rangle =\Omega _1(\langle {y,x}\rangle )\trianglelefteq G. \end{aligned}$$

• (ii) By (i), $\langle {x^g,x}\rangle$ contains any conjugate of x, so $|G:\mathcal {N}_G(\langle {x}\rangle )|=p$, now (ii) follows from the equality $\mathcal {C}_G(x)=\mathcal {N}_G(\langle {x}\rangle )$.

• (iii) Let $\langle {y_1}\rangle$ be a central subgroup of order at least $p^2$ such that $\langle {y}\rangle \cap \langle {y_1}\rangle =1$. By (i), we have $\langle {Z,x}\rangle =\langle {x^g,x}\rangle =\langle {Z_1,x}\rangle$, where $Z_1\leqslant \langle {y_1}\rangle$ is of order p. Then we have $Z=Z_1$, which is a contradiction.

• (iv) The hypothesis implies that $\mathcal {C}_G(x)$ is Dedekindian, as it is a CTI-group and any subgroup of prime order is normal in $\mathcal {C}_G(x)$. Since $\mathcal {C}_G(x)$ contains Z(G), so $\exp (Z(\mathcal {C}_G(x)))$ is at least $p^2$. Hence $\mathcal {C}_G(x)$ is an abelian group. $\square$

Theorem 3.3

Let G be a finite non-abelian p-group such that $\exp (Z(G))>p$. If G is a CTI-group then every cyclic subgroup of order at least $p^2$ is normal in G.

Proof

Suppose that G is a non-abelian CTI-p-group having a cyclic subgroup $\langle {a}\rangle$ which is not normal in G and $|a|=p^2$. Thus $a^p\not \in Z(G)$ and $\langle {a}\rangle \cap Z(G)=1$.

Step 1 We show that $\mathcal {C}_G(a)$ is abelian and $\Omega _1(G)\leqslant \mathcal {C}_G(a)$.

Since $a\in Z(\mathcal {C}_G(a))$ and $\langle {a}\rangle \cap Z(G)=1$, it follows from Lemma 3.2 (iii) that $\mathcal {C}_G(a)$ is abelian. Now for any non-central element y of order p, we have $a^p\in \mathcal {C}_G(y)$, since $\mathcal {C}_G(y)$ is a maximal subgroup of G by Lemma 3.2 (ii). Hence $y\in \mathcal {C}_G(a^p)$. Since G is CTI, $\langle {a}\rangle \trianglelefteq \mathcal {C}_G(a^p)$ and so $[a,y]\in \langle {a}\rangle$ because $y\in C_G(a^p)$. Also $[a,y]\in Z(G)$ by Lemma 3.2 (i). Hence $[a,y]=1$ (otherwise $\langle {a}\rangle \trianglelefteq G$) and so $\mathcal {C}_G(a)$ contains any element of order p of G.

Step 2 $M=\mathcal {C}_G(a^p)$ is abelian.

We show that for any non-central element y of order p, $\mathcal {C}_G(y)=M$. Suppose, for a contradiction, that $y\in G\backslash Z(G)$ is of order p such that $\mathcal {C}_G(y)\ne M$. By Step 1, $C=\mathcal {C}_G(a)$ is abelian and $y\in C$. Thus $C\leqslant \mathcal {C}_G(y)$, so $C\ne M$ and by Normalizer–Centralizer theorem C is maximal in M. Now we have $C=M\cap \mathcal {C}_G(y)$. Let $g\in M\backslash C$, then $|g|\ge p^2$ as by the previous part $\Omega _1(G)\le C$. Since $M=\langle {g}\rangle C$, we have $\langle {g}\rangle \trianglelefteq M$ (because $g^p\in Z(M)$). Therefore

$$\begin{aligned} 1\ne [a,g]\in \langle {g}\rangle \cap \langle {a}\rangle =\langle {a^p}\rangle , \end{aligned}$$

and so $\langle {g}\rangle \ntrianglelefteq G$, because $\langle {x}\rangle \ntrianglelefteq G$. Now since $g\not \in \mathcal {C}_G(y)$ and $\langle {g}\rangle \trianglelefteq M$, then $1\ne [g,y]\in \langle {g}\rangle$ is a central element of prime order, a contradiction. Hence, $M=\mathcal {C}_G(y)$ for any non-central element y of prime order and it follows from Lemma 3.2 (iv) that M is abelian.

Let $x\in G\backslash M$. Then $|x|\ge p^2$. Since $G=M\langle {x}\rangle$ and M is abelian, $\langle {x^p}\rangle \leqslant M\cap \langle {x}\rangle \leqslant Z(G)$. Therefore $\langle {x}\rangle \trianglelefteq G$. Thus $[a,x]\in M\cap \langle {x}\rangle \le Z(G)$. Hence $[a^p,x]=[a,x^p]=1$ and so $a^p\in Z(G)$, a contradiction. $\square$

In Theorem 4.3 (see Sect. 4), we shall prove that all finite CTI-p-groups G satisfy the property mentioned in the conclusion of Theorem 3.3, i.e., every cyclic subgroup of order at least $p^2$ is normal in G. In the proof of Theorem 4.3 we use Theorem 3.3.

Theorem 3.4

Let G be a finite non-Dedekindian p-group such that $\exp (Z(G))>p$. If G is a CTI-group then the followings hold:

• (i) $G'$ is of order p and $G'=\langle {x}\rangle$ for all $x\in G$ such that $|x|\ge p^2$;

• (ii) $G^p$ is a cyclic central subgroup of G and $G'=\Omega _1(G^p)$. In particular, $\Phi (G)=G^p$ is a cyclic central subgroup of G.

Proof

• (i) Let H be a non-normal subgroup of G and $h\in H$ such that $h^g\not \in H$ for some $g\in G$. Let $p^e=\exp (Z(G))$. Since $\langle {h}\rangle \ntrianglelefteq G$, Theorem 3.3 implies that $|h|=p$ and it follows from Lemma 3.2 (iii) that $[g,h]\in Z$, where $Z=Z(G)^{p^{e-1}}$. Hence $\langle {hZ}\rangle \trianglelefteq G/Z$. Therefore G / Z is Dedekindian. If $p>2$ or $p=2$ and $\exp (Z(G))>4$, then G / Z is abelian and so $G'=Z$ is of order p.

Now let $p=2$ and $\exp (Z(G))=4$ and assume that $G/Z\cong Q_8 \times E$ for some elementary abelian 2-group E. It follows that there exists a normal subgroup K of G such that $K/Z\cong Q_8$. By [9, 3.2.10], $Z\lvertneqq Z(K)$. Thus $K/Z(K)\cong \mathbb {Z}_2\times \mathbb {Z}_2$ which implies that $|K'|=2$. Hence $K'\leqslant Z(G)$ since $K\trianglelefteq G$. It follows that $K'\cap Z=1$. Now we have $K\cong Q_8\times Z$ and so G has a subgroup H isomorphic to $\mathbb {Z}_4\times Q_8$ which is not CTI, since $\Omega _1(H)\leqslant Z(H)$, a contradiction. Therefore, in this case again G / Z is abelian and $|G'|=2$.

Since G is not Dedekindian, it follows from Lemma 3.2 (i) that $G'=\Omega _1(Z(G)^p)$. Let x be an element of G of order greater than p. By Theorem 3.3, $\langle {x}\rangle \trianglelefteq G$. If $x\not \in Z(G)$, then $[x,g]\not =1$ for some $g\in G$. Since $|G'|=p$, $G'=\langle {[x,g]}\rangle =\langle {x^{\frac{|x|}{p}}}\rangle$ as $\langle {x}\rangle \trianglelefteq G$. Now assume that $x\in Z(G)$. Then it follows from Lemma 3.2 (iii) that $Z(G)^p \cap \langle {x}\rangle \not =1$ and so $G'=\langle {x^{\frac{|x|}{p}}}\rangle$.

• (ii) Note that $G^p=\langle a^p\;:\; |a|\ge p^2 \rangle$. Since $G^p\le Z(G)$, $G^p$ is abelian and so

  $$\begin{aligned} \Omega _1(G^p)=\langle { a^{\frac{|a|}{p}}\;:\; |a|\ge p^2 }\rangle . \end{aligned}$$

Now by part (i), $G'=\langle {a^{\frac{|a|}{p}}}\rangle$ for all $a\in G$ such that $|a|\ge p^2$. Therefore $G'=\Omega _1(G^p)$. $\square$

4 CTI-p-Groups with Center of Exponent $=p$

In this section, we classify CTI-p-groups whose centers are elementary abelian.

The following lemma holds in any CTI-group.

Lemma 4.1

Let G be a finite CTI-group and g be any element of G. Then $\langle {g^x,g}\rangle$ is abelian for all $x\in \mathcal {N}_G(\mathcal {N}_G(\langle {g}\rangle ))$.

Proof

Since $x\in \mathcal {N}_G(\mathcal {N}_G(\langle {g}\rangle ))$, $\langle {g}\rangle ^x\le \mathcal {N}_G(\langle {g}\rangle )$ and $\langle {g}\rangle ^{x^{-1}}\le \mathcal {N}_G(\langle {g}\rangle )$. It follows that $\langle {g}\rangle ^x$ and $\langle {g}\rangle$ are normal subgroups of $\langle {g,g^x}\rangle$. Thus $\langle {g,g^x}\rangle '\le \langle {g}\rangle ^x\cap \langle {g}\rangle$. Suppose, if possible, that $\langle {g,g^x}\rangle '\not =1$. Since G is a CTI-group, $\langle {g}\rangle ^x=\langle {g}\rangle$ and so $[g^x,g]=1$. This completes the proof. $\square$

The following Lemma is a consequence of Theorem 3.3.

Lemma 4.2

Let G be a finite non-abelian CTI-p-group and let H be any normal non-abelian subgroup of G such that Z(H) is of exponent at least $p^2$. Then any cyclic subgroup of order at least $p^2$ of H is normal in G.

Proof

By Theorem 3.4, $H'\le \langle {x}\rangle$ for all $x\in H$ of order at least $p^2$. Now Theorem 3.4 implies that $|H'|=p$ and since H is normal in G, $H'\le Z(G)$. It follows that $\langle {x}\rangle$ is normal in G. $\square$

As we promised in Sect. 2, we now prove that the conclusion of Theorem 3.4 holds for all finite CTI-p-groups.

Theorem 4.3

Let G be a finite CTI-p-group and g be any element of order $p^2$ of G. Then $\langle {g}\rangle$ is normal in G.

Proof

Suppose, for a contradiction, that $\langle {g}\rangle$ is not normal in G. Thus $\mathcal {N}_G(\langle {g}\rangle )\not =G$ and so $\mathcal {N}_G(\mathcal {N}_G(\langle {g}\rangle ))\setminus \mathcal {N}_G(\langle {g}\rangle )\not =\varnothing$ since G is a nilpotent group. Let $x\in \mathcal {N}_G(\mathcal {N}_G(\langle {g}\rangle ))\setminus \mathcal {N}_G(\langle {g}\rangle )$. Let $H=\langle {g,x}\rangle$. Note that $\langle {g}\rangle$ is not also normal in H.

• (i) It follows from Lemma 4.1 that the normal closure $\langle {g}\rangle ^H$ is abelian.

• (ii) The centralizer $\mathcal {C}_H(g)$ is normal in H, since $H'\le \langle {g}\rangle ^H\le \mathcal {C}_H(g)$. By Lemma 4.2, the centralizer $\mathcal {C}_H(g)$ is abelian, otherwise $\langle {g}\rangle$ is normal.

• (iii) Let k be the largest positive integer such that $|[g,_k x]|=p^2$ and $|[g,_{k+1} x]|\le p$. Note that such k exists; for G is nilpotent and so $[g,_ c x]=1$, where c is the nilpotency class of H and $|[g,x]|= p^2$, otherwise $[g,x]^p=1$, implies then $(g^{-1}g^x)^p=g^{-p}(g^p)^x=1$ and so $g^p=(g^p)^x$. It follows that $\langle {g}\rangle \cap \langle {g}\rangle ^x\not =1$ and so $\langle {g}\rangle =\langle {g}\rangle ^x$, a contradiction (note that $[g,x]^{p^2}=[g^{p^2},x]=1$). Since $[g,_{k+1} x]^p=1$, $[g,_k x]^p=([g,_k x]^p)^x$ and so $[g,_k x]^x=[g,_k x]^i$ for some integer i such that $1\le i\le p^2$ and $\gcd (i,p)=1$. Therefore, $\langle {[g,_k x]}\rangle \trianglelefteq H$ and so $\mathcal {C}_G([g,_kx])$ is a maximal subgroup of H.

Suppose, for a contradiction, that $\mathcal {C}_H([g,_k x])$ is not abelian. Since $\langle {g}\rangle ^H\le \mathcal {C}_H([g,_k x])$, $H'\le \mathcal {C}_H([g,_k x])$ and so $\mathcal {C}_H([g,_k x])$ is a normal non-abelian subgroup of H such that the exponent of its center is greater than p (as $[g,_k x]\in Z(\mathcal {C}_H([g,_k x]))$). Now Lemma 4.2 implies that $\langle {g}\rangle$ is normal in H, a contradiction. Hence $\mathcal {C}_H([g,_k x])$ is abelian. By (ii), $C_H(g)$ is abelian and so $\mathcal {C}_H(g)=\mathcal {C}_H([g,_k x])$, because $[g, [g,_k x]]=1$. Hence $\mathcal {C}_H(g)$ is a maximal subgroup of H. Therefore, $x^p \in \mathcal {C}_H(g)$ and so $[g,x^p]=1$. Thus

$$\begin{aligned} 1=[g,_{k-1} x, x^p]=[g,_k x]^{x^{p-1}+\cdots +x+1}=[g,_k x]^{i^{p-1}+\cdots +i+1}. \end{aligned}$$

If $i=1$, then it follows that $[g,_k x]^p=1$, a contradiction. Thus $i>1$, and so

$$\begin{aligned} i^{p-1}+\cdots +1=\frac{i^p-1}{i-1}. \end{aligned}$$

Since $[g,_k x]^p=([g,_k x]^p)^x$, $i=pk+1$ for some integer. Therefore, $\frac{i^p-1}{i-1}=p(1+p\ell )$ for some integer $\ell$ and so $[g,_k x]^p=1$, a contradiction. This last contradiction completes the proof. $\square$

Lemma 4.4

Let G be a finite CTI-p-group and x and y be elements of G of orders at least $p^2$. Then either $\langle {x}\rangle \cap \langle {y}\rangle \not =1$ or $\mathcal {C}_G(x)=\mathcal {C}_G(y)$ is abelian.

Proof

Suppose that $\langle {x}\rangle \cap \langle {y}\rangle =1$ and let $H=\mathcal {C}_G(x)$. We show that H is abelian and $H=\mathcal {C}_G(y)$. By Theorem 4.3, $\langle {x}\rangle$ and $\langle {y}\rangle$ are normal subgroups of G and so $y\in H$. Suppose, for a contradiction, that H is not abelian. Since $x\in Z(H)$, it follows from Theorem 3.4 (ii) that $[g,y]\in \langle {x^p}\rangle$ for all $g\in H$. Since $\langle {y}\rangle \lhd G$, $[g,y]\in \langle {x^p}\rangle \cap \langle {y}\rangle$ and so $y\in Z(H)$. This means that $\langle {x}\rangle \times \langle {y}\rangle \le Z(H)$ contrary to Lemma 3.2. Hence H is abelian and by the symmetry between x and y, $\mathcal {C}_G(y)$ is also abelian. Since x commutes with y and both $\mathcal {C}_G(x)$ and $\mathcal {C}_G(x)$ are abelian, it follows that $\mathcal {C}_G(x)=\mathcal {C}_G(y)$. This completes the proof. $\square$

Theorem 4.5

Let G be a non-abelian CTI-p-group such that $\exp (Z(G))=p$ and $\exp (G)>p$. If $p>2$, then $\exp (G)=p^2$ and $G'=G^p$ is of order p. In particular, $\Phi (G)=G'=G^p$ is a central subgroup of G of order p.

Proof

We first prove that $\exp (G)=p^2$. Suppose, for a contradiction, that $g\in G$ is of order $p^3$. By Theorem 4.3, $\langle {g}\rangle \trianglelefteq G$ and so $\langle {g^p}\rangle \trianglelefteq G$. Since $\exp (Z(G))=p$, $g^p\not \in Z(G)$ and so $\mathcal {C}_G(g^p)$ is a maximal subgroup of G. Assume that x is an element of G. Suppose that $|x|=p$. Since $p>2$, $\langle {g}\rangle \trianglelefteq G$, and $|g|=p^3$, $g^x=g^{1+kp^2}$ for some $k\in \{1,\dots ,p\}$. It follows that $[g^p,x]=1$. Now assume that $|x|\ge p^2$. By Theorem 4.3 $\langle {x}\rangle \trianglelefteq G$ and so $[g,x]\in \langle {g}\rangle \cap \langle {x}\rangle$. Since $\langle {g}\rangle \cap \langle {x}\rangle \in \{\langle {g^p}\rangle ,\langle {x}\rangle , \langle {g^{p^2}}\rangle , 1\}$. It follows that $|[g,x]|\le p$ and so $[g^p,x]=[g,x]^p=1$. Therefore, $g^p\in Z(G)$ which is a contradiction as $\exp (Z(G))=p$. Hence $\exp (G)=p^2$.

Now let $x,y\in G$ be of the same order $p^2$. We show that $\langle {x^p}\rangle =\langle {y^p}\rangle$. The latter is equivalent to say $\langle {x}\rangle \cap \langle {y}\rangle \not =1$. It follows from Lemma 4.4 that $\mathcal {C}_G(x)=\mathcal {C}_G(y)$ is abelian. We can find an h element of order $p^2$ in $G\setminus C_G(x)$: for if $g\not \in \mathcal {C}_G(x)$ and $|g|=p$, then $|xg|=p^2$ and $xg\not \in \mathcal {C}_G(x)$. Since $1\ne [x,h]\in \langle {x}\rangle \cap \langle {h}\rangle$, $\langle {x^p}\rangle =\langle {g^p}\rangle$, similarly $\langle {y^p}\rangle =\langle {g^p}\rangle$. Therefore $\langle {x^p}\rangle =\langle {y^p}\rangle$.

Hence $G^p$ is cyclic of order p, since $\exp (G)=p^2$.

Now we show that $G'=G^p$. We show that $[x,y]\in G^p$ which follows that $G'=G^p$. Suppose a is an element of G of order $p^2$. Since $\mathcal {C}_G(a)$ is maximal, $G=\mathcal {C}_G(a)\langle {b}\rangle$ for some $b\in G$. It follows that $G'=\langle [m,b], [m,m'] \;|\; m,m'\in \mathcal {C}_G(a)\rangle ^G$.

Let x and y be elements of G. If |x| or |y| is greater than p, then by Theorem 4.3, $[x,y]\in \langle {x^p}\rangle$ or $\langle {y^p}\rangle$ and so $[x,y]\in G^p$.

If $\mathcal {C}_G(a)$ is not abelian, it follows from Theorem 3.4 (ii) that $\mathcal {C}_G(a)'\le G^p$. The latter is clearly valid if $\mathcal {C}_G(a)$ is abelian. Thus $[m,m']\in G^p$ for all $m,m'\in \mathcal {C}_G(a)$. By the previous paragraph, it remains to show that $[m,b]\in G^p$ for $m\in \mathcal {C}_G(a)$ whenever $|m|=|b|=p$. Since $|ab|=p^2$ and $[m,ab]=[m,b]$, it follows from the previous paragraph that $[m,b]\in G^p$. This shows that $G'\le G^p$. $\square$

The following result is well known and easy to prove. We need it in the sequel. We give it for the reader’s convenience.

Lemma 4.6

Let G be a finite CTI-2-group. Let x and $y\in G$ such that $|x|=2$ and $|y|=2^n\ge 4$. Then $y^x\in \{y,y^{-1},y^{1+2^{n-1}}\}$.

Proof

Suppose that $y^x\not =y$. By Theorem 4.3, $\langle {y}\rangle \trianglelefteq G$ and so $y^x=y^r$ for some odd integer $r\in \{2,\dots ,2^n-1\}$. Thus $2^n$ divides $r^2-1$. If $n=2$, then $y^x=x^{-1}$ easily follows. Suppose that $n>2$. Since $2^n$ divides $(r-1)(r+1)$ and $\gcd (r-1,r+1)=2$, $2^{n-1}$ divides exactly one of $r-1$ or $r+1$. Therefore $r=1+2^{n-1}$ or $r=-1+2^{n-1}$ as $1<r<2^n$. We now show that the latter case does not happen. Since $|xy|=4$ and $(xy)^x=yx$ and G is CTI, $\langle {xy}\rangle =\langle {yx}\rangle$. Thus $xy=(yx)^{-1}$ and so $x^2=y^{-2}$, a contradiction. This completes the proof. $\square$

Theorem 4.7

Let G be a non-abelian CTI-2-group of exponent $2^e$. If Z(G) is of exponent 2, then

• (i) any 2-generated non-abelian subgroup of G is either a dihedral group or $Q_8$;

• (ii) $G'=G^2=\Phi (G)$. In addition, if $\exp (G)=4$ and G does not contain any subgroup isomorphic to $\mathbb {Z}_4 \times \mathbb {Z}_4$, then $G^2$ is of order 2.

Proof

If G is Dedekindian, then (i) and (ii) obviously occur. Hence we consider non-Dedekindian case.

• (i) First we note that for any two elements g, x such that $[g,x]\ne 1$, if $|g|\le 4$ and $|x|=2$, then the subgroup $\langle {g,x}\rangle$ is a dihedral group.

Let $g\in G$ be of order $2^n\ge 4$ and $Z\leqslant \langle {g}\rangle$ be of order 4. By Theorem 4.3, $\mathcal {C}_G(Z)$ is a maximal subgroup of G. Assume that $x\in G\backslash \mathcal {C}_G(Z)$, then $G=\mathcal {C}_G(Z)\langle {x}\rangle$.

First suppose that $|x|=2$, by Lemma 4.6, $g^x=g^{-1}$ or $g^{1+2^{n-1}}$. In the latter case, since $[g,x]=g^{2^{n-1}}\in Z(G)$ and $g^2\in Z(\mathcal {C}_G(Z))$ so $g^2\in Z(G)$ which implies $n=2$. Therefore $\langle {g,x}\rangle \cong D_{2^{n+1}}$.

If $|x|\ne 2$, then $[g,x]\in \langle {g^2}\rangle \cap \langle {x^2}\rangle \leqslant Z(G)$, as $g^2\in Z(\mathcal {C}_G(Z))$ by Theorem 3.4. Therefore $[g^2,x]=1$ hence $g^2\in Z(G)$ and so $|g|=4$ which implies $\langle {g}\rangle =Z$. Since $x^4\in Z(\mathcal {C}_G(Z))$, $x^4\in Z(G)$ is of order at most 2, so $|x|\le 8$ and $|x|=4$ if $\mathcal {C}_G(Z)$ is abelian.

Assume that $|x|=8$. Then $\mathcal {C}_G(Z)$ is non-abelian. Since $Z(\mathcal {C}_G(Z))=\langle {g}\rangle \times E$, where E is elementary abelian, it follows from Theorem 3.4 (ii) that $(\mathcal {C}_G(Z))^2\leqslant \langle {g}\rangle$. Now let y be a non-involution element of $\mathcal {C}_G(Z)$. Then

$$\begin{aligned}{}[y,x]\leqslant \langle {y^2}\rangle \cap \langle {x^2}\rangle \leqslant \langle {g}\rangle \cap \langle {x^2}\rangle =\langle {g^2}\rangle \leqslant Z(G). \end{aligned}$$

So $[y,x^2]=1$. If y is a non-central involution of $\mathcal {C}_G(Z)$, since $|gy|=4$, it follows that $[x^2,y]=[x^2,gy]=1$. Therefore $x^2\in Z(G)$ which is impossible. Hence $|x|=4$ and $\langle {g,x}\rangle \cong Q_8$.

• (ii) By the previous part, for any $g\in G$ of order $2^n\ge 4$, if there exists $x\in G\backslash \mathcal {C}_G(g)$ of order 2, then $\langle {g,x}\rangle \cong D_{2^{n+1}}$. So $g^2=[x,g]\in G'$. Otherwise $\langle {g,x}\rangle \cong Q_8$ and so $g^2=[g,x]$. Therefore $G^2\leqslant G'$.

Let $x,y\in G$ be of the same order 4. By Theorem 4.3, $\langle {x}\rangle$ and $\langle {y}\rangle$ are normal in G. If $\langle {x}\rangle \cap \langle {y}\rangle =1$, then $\langle {x}\rangle \times \langle {y}\rangle \cong \mathbb {Z}_4 \times \mathbb {Z}_4$, a contradiction. Thus $\langle {x}\rangle \cap \langle {y}\rangle \not =1$ and so $\langle {x^2}\rangle =\langle {y^2}\rangle$. It follows that $G^2=\langle {a^2}\rangle$ for any element a of order 4 of G. This completes the proof. $\square$

Lemma 4.8

Let G be a CTI-2-group such that $\exp (Z(G))=2$ and $\exp (G)=2^e\ge 8$. Then $\mathcal {C}_G\left( g^{2^{e-2}}\right)$ is abelian for any element g of order $2^e$.

Proof

Let $a=g^{2^{e-2}}$ so that $|a|=4$. Suppose, for a contradiction, that $\mathcal {C}_G(a)$ is non-abelian. Since $g\in \mathcal {C}_G(a)$, it follows from Theorem 3.4 that $g^2\in Z(\mathcal {C}_G(a))$ and so $\mathcal {C}_G(a)\le \mathcal {C}_G(g^2)$. By Theorem 4.3, $\mathcal {C}_G(a)$ is maximal in G. It follows that $\mathcal {C}_G(a)=\mathcal {C}_G(g^2)$, since $g^2\not \in Z(G)$. Now suppose that there exists an element $h\in G\setminus \mathcal {C}_G(g^2)$ of order greater than 2. Then $G=\mathcal {C}_G(g^2)\langle {h}\rangle$. By Theorem 4.3, $\langle {g}\rangle$ and $\langle {h}\rangle$ are both normal subgroups of G and so $[g,h]\le \langle {g^2}\rangle \cap \langle {h^2}\rangle$. Since $\langle {g^2}\rangle \cap \langle {h^2}\rangle \le Z(G)$ and $\exp (Z(G))=2$, $[g,h]^2=1$ and so $[g^2,h]=1$ which implies that $g^2\in Z(G)$, a contradiction. Therefore, all elements in $G\setminus \mathcal {C}_G(a)$ are of order 2. Now fix an element h of $G\setminus \mathcal {C}_G(a)$ and let $b\in \mathcal {C}_G(a)$. Then $|bh|=|h|=2$ and so $b^h=b^{-1}$ for all $b\in \mathcal {C}_G(a)$. This implies that $\mathcal {C}_G(a)$ is abelian. This completes the proof. $\square$

Theorem 4.9

Let G be a CTI-2-group such that $\exp (Z(G))=2$. Then one of the following holds:

• (1) G is Dedekindian,

• (2) $G'=G^2$ is of order 2, and

• (3) $G=A\langle {x}\rangle$ for some abelian subgroup A and involution $x\in G\setminus A$ such that $a^x=a^{-1}$ for all $a\in A$.

Proof

Suppose that (1) and (2) do not hold. We show that G has the structure described in (3).

• We first show that there exists an element y of order 4 such that $A=\mathcal {C}_G(y)$ is abelian.

Since G is not abelian and we are assuming that (2) does not hold, it follows from Theorem 4.7 that either $\exp (G)\ge 8$ or G contains a subgroup isomorphic to $\mathbb {Z}_4 \times \mathbb {Z}_4$. If the former happens, then such an element y exists according to Lemma 4.8 and if the latter happens, Lemma 4.4 guaranties the existence of an element y of order 4 with abelian centralizer in G.

• Now we prove that there exists an involution $x\in G\setminus A$.

By Theorem 4.3, A is a maximal subgroup of G. Note that if $g\in A\setminus Z(G)$, then $A=\mathcal {C}_G(g)$ and if $h\in A\setminus Z(G)$ and $[t,h]=1$ for some $t\in G$, then $t\in A$. We use the latter note in the sequel of the proof.

Since G is non-Dedekindian, it follows from Theorem 4.3 that there exists a non-central element z of order 2. If $z\not \in A$, we take $x=z$. Now assume that $z\in A$ and there exists an element $t\in G\setminus A$ of order $2^n\ge 4$. If $t^z=t$, then by the above note $t\in A$, which is impossible. Thus $t^z\not =t$. It follows from Lemma 4.6 that $t^z=t^{-1}$ or $t^z=t^{1+2^{n-1}}$. If the latter happens, then $[t^2,z]=1$ and so $t^2\in A$. If $t^2\not \in Z(G)$, then $t\in A$, a contradiction. Thus $t^2\in Z(G)$ and so $|t|=4$. Hence $t^z=t^{3}=t^{-1}$. Now take $x=tz$ which is of order 2 and $x\not \in A$.

• Now we prove that $a^x=a^{-1}$ for all $a\in A=\mathcal {C}_G(y)$.

By Lemma 4.6, $y^x=y^{-1}$ since $|y|=4$. Let $a\in A$ be of order 2. Then $|ay|=4$ and so by Lemma 4.6, $(ay)^x=(ay)^{-1}$ or $(ay)^x=ay$. The latter does not happen, otherwise $x\in A$ by the above note. Therefore $a^x=a^{-1}=a$. If $a\in A$ is of order 4, then by Lemma 4.6 $a^x=a^{-1}$ or $a^x=a$. The latter does not happen according to the above note. Now assume that $a\in A$ of order $2^e\ge 8$. Then by Lemma 4.6 and the above note, $a^x=a^{-1}$ or $a^x=a^{1+2^{e-1}}$. If the latter happens, then $[a^2,x]=1$ and so by the above note $a^2\in Z(G)$ which is not possible as $\exp (Z(G))=2$. This completes the proof. $\square$

5 Proof of the Main Theorem

In this section, we give the proof of Theorem 1.2. Let us restate the statement of Theorem 1.2.

Theorem 1.2

Let G be a finite nilpotent group. Then G is a CTI-group if and only if one of the following holds:

• (1) G is Dedekindian;

• (2) G is a p-group of exponent p for some prime p;

• (3) G is a p-group for some prime p such that $G'=\Omega _1(G^p)$ is of order p and $\Phi (G)=G^p$ is a central cyclic subgroup of G; and

• (4) G is a 2-group such that $G=A\langle {x}\rangle$, where A is an abelian subgroup of G and x is an involution in $G\setminus A$ such that $a^x=a^{-1}$.

Proof

Suppose G satisfies one (1), (2), (3), or (4).

If G is a p-group of exponent p, every non-trivial cyclic C subgroup is of order p and so $C^x\cap C$ is clearly equal to C or 1. Therefore G is CTI.

If G is a Dedekindian group, then every subgroup is normal in G and so G is CTI.

Let G be a p-group such that $G'=\Omega _1(G^p)$ is of order p. If $a\in G$ is of order at least $p^2$, then it follows that $G'\le \langle {a}\rangle$ and so $\langle {a}\rangle \trianglelefteq G$. Elements of order p obviously satisfy the condition of being CTI. This implies that G is a CTI-group.

Suppose G is of the form described in (3). Then every element in $G\setminus A$ is of order 2 and so if $|g|\ge 4$, $g\in A$. Since $g^h=g$ or $g^{-1}$ for all $g\in G$, $\langle {g}\rangle \trianglelefteq G$. Hence G is CTI.

Now assume that G is a nilpotent CTI-group which is not Dedekindian. By Lemma 3.1, G is a non-Dedekindian p-group for some prime p. Suppose further that G is not of exponent p. If $\exp (Z(G))>p$, then it follows from Theorem 3.4 that G satisfies (3). If $\exp (Z(G))=p>2$, then it follows from Theorem 4.5 that $G'=G^p$ of order p. Therefore G satisfies (3), since $\Omega _1(G^p)=G^p$ in this case. If $\exp (Z(G))=2$, then it follows from Theorem 4.9 that $G'=G^2$ is of order 2 or G is of the form described in (4). This completes the proof. $\square$

We note that any extra-special group is CTI, since the central factor of such a group is elementary abelian. Let G be a finite p-group such that $\exp (Z(G))\ne p$. In this case, $\Phi (G)$ is cyclic by Theorem 3.4. If $\Phi (G)$ is of order p, then G has the following structure.

Theorem 5.1

Let G be a non-Dedekindian CTI-p-group such that $\exp (G)\ge p^2$ and $\exp (Z(G))>p$. If $|\Phi (G)|=p$, then $G= AZ(G)$, where $A=A_1*\cdots *A_s$ is an extra-special p-group and $|A_1|=\cdots =|A_s|=p^3$.

Proof

Since in this case $|G'|=p$ and $G/G'$ is elementary abelian, then [10, Lemma 4.2] completes the proof. $\square$

Theorem 5.2

Let G be a non-Dedekindian CTI-p-group of exponent $p^e$ such that$\exp (Z(G))>p$. If $|\Phi (G)|>p$ and $Z(G)=\langle {z}\rangle \times E$, where E is elementary abelian. Then $G=K\times E$, where K is a non-abelian CTI-group with cyclic center. Also $K\cong M\rtimes \langle {x}\rangle$ for some maximal subgroup M of K and some prime order element $x\not \in \Phi (G)$.

• (i) If M is abelian, then either $M\cong \langle {z}\rangle \times \langle {y}\rangle$ for some prime order element $y\in M$ or M is maximal cyclic subgroup of G such that $Z(K)=M^p$. The subgroup K has one of the following presentations:

  • (1) $\langle {z,y,x}\,|\,{z^{p^e}=y^p=x^{p^e}=[z,y]=[z,x]=1, y^x=yz^{p^{e-1}}}\rangle$;

  • (2) $M_{p^{e+1}}=\langle {z,x}\,|\,{z^{p^e}=x^p=1, z^x=z^{1+p^{e-1}}}\rangle$.

• (ii) If M is not abelian then $K=AZ=(A_1*\cdots *A_s)Z=Z(A)TZ$ where Z is a maximal cyclic subgroup of K containing $\Phi (G)$ and $A_i$ is minimal non-abelian for all i and T is extra-special.

Proof

By Theorem 3.4, $\Phi (G)\cap E=1$. It follows from [11, Hilfssatz 4.4] that $G=H\times E$ and $\Phi (G)=\Phi (H)\leqslant \langle {z}\rangle$. Since H is non-abelian, it contains a non-normal cyclic subgroup $\langle {x}\rangle$ of order p. Obviously, $x\not \in \Phi (H)$ and so H has a maximal subgroup M such that $x\not \in M$. Therefore $H\cong M\rtimes \langle {x}\rangle$.

• (i) Since $Z(K)=\langle {z}\rangle$ and $|z|>p$, it follows from Corollary 3.4 that $\Phi (K)\leqslant Z(K)$. Now let x be a non-central element of order p. By Lemma 3.2 (ii), $C=\mathcal {C}_K(x)$ is maximal and $M\cap C$ is an abelian maximal subgroup of C. Since $x\in Z(C)$ and $x\not \in M\cap C$, $C=(M\cap C)\langle {x}\rangle$ is abelian. Therefore, $Z(K)=M\cap C$ is a maximal cyclic subgroup of M. So either $M\cong \langle {z}\rangle \times \langle {y}\rangle$ for some $y\in M$ of order p or M is cyclic and $Z(K)=M^p$.

• (ii) In this case, $G'=K'$ and $\Phi (G)=\Phi (K)\leqslant Z(K)\leqslant M$. Let $Z \leqslant K$ be cyclic of maximal order such that $\Phi (G)\leqslant Z$. Then $|Z|=p|\Phi (G)|$. Assume that H is a subgroup of K such that $|H:Z|=p$. Thus $H=Z\langle {y}\rangle$ for some prime order $y\in K$ so $T(H)=\langle {G', y}\rangle \trianglelefteq G$. Now for any non-central element y of K, $\langle {Z, y}\rangle$ has such properties and $T(\langle {Z, y}\rangle )\trianglelefteq G$. Let A be generated by T(H) where $|H: Z|=p$. So A contains any non-central element of prime order. If A is abelian, then for any two non-central elements $y_1$ and $y_2$, $\mathcal {C}_G(y_1)=\mathcal {C}_G(y_2)$ is abelian by Lemma 3.2 (ii), (iv). Since M is non-abelian, we can choose a non-central element y of M so that $\mathcal {C}_M(y)=\mathcal {C}_M(x)$ is a maximal in M. Now let $g\in M\backslash \mathcal {C}_M(x)$, so $[g,y]=[g,x]$ is central. Thus $[g, xy^i]=1$ for some i and so $xy^i$ is central, since $|xy^i|=p$. Then $xy^i\in Z(K)\leqslant M$, which is impossible as $x\not \in M$. Therefore, A is non-abelian and H satisfies the hypothesis of [10, Theorem 4.4]. This completes the proof. $\square$

Remark 5.3

Let G be a non-Dedekindian CTI-p-group of exponent $p^e$ such that$\exp (Z(G))=p$. If $e\ge 3$ or G contains an abelian subgroup of type (4, 4), then $p=2$ and G is of type (4) in the Theorem 1.2. Otherwise $\exp (G)\le p^2$ and G does not contain any abelian subgroup of type (4, 4). In this case, we have the following theorem:

Theorem 5.4

Let G be a non-Dedekindian CTI-p-group of exponent $p^2$ such that$\exp (Z(G))=p$. If G does not contain any abelian subgroup isomorphic to $\mathbb {Z}_4 \times \mathbb {Z}_4$, then $G\cong E\times H$, where H is an extra-special p-group and $Z(G)=E\times \Phi (G)$ for some elementary abelian subgroup E.

Proof

By Theorems 4.5 and 4.7, $|\Phi (G)|=p$. So we may write $Z(G)=\Phi (G)\times E$, where E is an elementary abelian subgroup of G. By [11, Hilfssatz 4.4], $G\cong E\times H$. Hence $\Phi (G)=\Phi (H)$. This completes the proof. $\square$

6 Some Applications in the Structure of Solvable CTI-Groups with Trivial Center

Let G be a finite solvable CTI-group with trivial center which is not isomorphic to the symmetric group of degree 4. Then by [8, Proposition 3.2 and Theorem 3.4], $G= F(G)H$ is a Frobenius group whose kernel is the Fitting subgroup F(G) and H is the complement. By [8, Theorem 3.5], F(G) is abelian if F(G) is not of prime power order or |H| is even. Assume that $n=|H|$ is odd. Then H is cyclic and F(G) is of prime power order. Thus F(G) is the Sylow p-subgroup of G for some prime p such that $p\not \mid n$.

Corollary 6.1

Under the above assumptions and notations, suppose further that F(G) is not abelian. Now

• (i) if p is odd and $\exp (F(G))\not =p$, then n divides $p-1$;

• (ii) if $p=2$ then $F(G)=A\langle {x}\rangle$, where A is an abelian subgroup and $x\in F(G)\setminus A$ of order 2 such that $a^x=a^{-1}$ for all $a\in A$.

Proof

• (i) If $p>2$ and $\exp (F(G))\ne p$, then it follows from Theorem 1.2 (3) that F(G) contains a characteristic subgroup of order p. Thus H is embedded in the automorphism group of $\mathbb {Z}_p$. Hence $n \mid p-1$.

• (ii) If $\exp (Z(F(G)))\ne 2$, then by Theorem 3.4 $F(G)'$ is of order 2 and so $F(G)'\le Z(G)$, a contradiction. Thus $\exp (Z(F(G))=2$. Now Theorem 4.9 implies that F(G) satisfies one of the cases (1), (2), or (3). If F(G) satisfies (2), then $Z(G)\not =1$ which is not possible. Then F(G) satisfies (1) or (3). If F(G) is Dedekindian, $F(G)\cong Q_8 \times E$ for some elementary abelian 2-group E. It follows that $F(G)'$ is of order 2 and it is contained in Z(G), a contradiction. This completes the proof. $\square$