1. Introduction
Let G be a finite group. We say that a subgroup H is isolated in G if for each
$x\in G$
either
$x\in H$
or
$\langle x\rangle \cap H={1}$
. In 2016, Janko [Reference Janko4] first studied isolated subgroups of finite p-groups and obtained many interesting results. Among them, he proved the following result for finite p-groups with isolated minimal nonabelian subgroups.
Theorem 1.1. [Reference Janko4, Theorem 4]
Let G be a nonabelian p-group of exponent greater than p which is not minimal nonabelian and all of whose minimal nonabelian subgroups are isolated in their normalisers. Then we have the following possibilities:
-
(a)
$p>2$
and G has an abelian maximal subgroup A of exponent
$>p$
with
$A =\mathrm{H}_p(G)$
and all minimal nonabelian subgroups are isomorphic to
$\mathrm{S}(p^3)$
(a nonabelian group of order
$p^3$
and exponent p) and, if S is any one of them, then
$N_G(S)$
is of exponent p; -
(b)
$p =3$
and G has a nonmetacyclic minimal nonabelian subgroup H of exponent greater than three and index three with
$H=\mathrm{H}_3(G)$
, where with
$$ \begin{align*}H = \langle a, b \mid a^{3^m} = b^{3^n}= c^3 = 1, [a, b] = c, [c, a] = [c, b] = 1\rangle\end{align*} $$
$m\geqslant 2$
,
$n\geqslant 1$
and either
$m=n$
or
$m =n +1$
, and each minimal nonabelian subgroup S distinct from H is isomorphic to
$\mathrm{S}(27)$
and
$N_G(S) ={\textit {SZ}}(G)$
, where
$Z(G)\cong {\mathrm{E}}_9$
and
${\textit {SZ}}(G) =S'={\mathrm{C}}_3$
.
We will prove that
$p\geqslant 5$
in the possibility (a) in Theorem 1.1, and the possibility (b) in Theorem 1.1 does not occur. Theorem 1.1 can therefore be refined as follows.
Theorem 1.2. Let G be a nonabelian p-group of exponent greater than p which is not minimal nonabelian. Then all minimal nonabelian subgroups of G are isolated in their normalisers if and only if all minimal nonabelian subgroups of G are isomorphic to
${\mathrm{S}}{\mathrm{(}}p^3{\mathrm{)}}$
, where
$p\geqslant 5$
.
We also determine the structure of finite p-groups which have an isolated metacyclic subgroup.
Let G be a finite p-group. We use
$p^{e(G)}$
to denote the exponent of G. For any integer s with
$0\le s\le e=e(G)$
, we define
$$ \begin{align*}\Omega_s(G)=\langle a\in G\mid a^{p^s}=1\rangle \quad \mbox{and} \quad \mho_s(G)=\langle a^{p^s}\mid a\in G\rangle.\end{align*} $$
We use
${\mathrm{E}}_{p^k}$
to denote an elementary abelian group of order
$p^k$
,
${\mathrm{S}}(p^3)\ (p>2)$
a nonabelian group of order
$p^3$
and exponent p and
${\mathrm{D}}_{2^n} \ (n\geqslant 3)$
a dihedral group of order
$2^n$
. The Hughes subgroup
${\mathrm{H}}_p(G)$
is the subgroup of G generated by all elements of order not equal to p. We write ‘
$H\ \mathrm{char}\ G$
’ to indicate that H is a characteristic subgroup of G. The other notation and terminology are standard, as in [Reference Berkovich2].
2. Finite p-groups with isolated minimal nonabelian subgroups
In this section, we prove Theorem 1.2. The first proposition is straightforward.
Proposition 2.1. Let G be a finite p-group.
-
(1) Assume that
$K\leq G$
and
$H\leq G$
. If K is isolated in G, then
$H\cap K$
is isolated in H. In particular, if
$K\leq H$
, then K is isolated in H. -
(2) Assume that
$N\leq H\leq G$
and
$N\trianglelefteq G$
. If H is isolated in G, then
$H/N$
is isolated in
$G/N$
.
We also need the following lemmas.
Lemma 2.2. [Reference Xu, An and Zhang6, Lemma 2.2]
Suppose that G is a finite nonabelian p-group. Then the following conditions are equivalent:
-
(1) G is minimal nonabelian;
-
(2)
$d(G)=2$
and
$|G'|=p;$
-
(3)
$d(G)=2$
and
$\Phi (G)=Z(G)$
.
Lemma 2.3. Let G be a finite
$2$
-group. If G has a proper subgroup H which is isolated in G, then H is abelian.
Proof. Let
$T\leq G$
be such that
$|T:H|=2$
. Since H is isolated in G, H is isolated in T and so
$o(x)=2$
for all
$x\in T-H$
. Let
$y\in H$
. Then
$o(xy)=2$
. So,
$y^x=y^{-1}$
. Hence,
$(ab)^x=(ab)^{-1}$
for all
$a,b\in H$
. On the other hand,
$(ab)^x=a^xb^x=a^{-1}b^{-1}$
. So,
$ab=ba$
, that is, H is abelian.
Lemma 2.4. Suppose that G is a finite
$3$
-group, M is maximal in G and
$g^3=1$
for all
$g\in G-M$
. If
$[a,b]=1$
for some
$a,b\in M$
, then
$[a^x,b]=1$
for all
$x\in G-M$
.
Proof. Let
$x\in G-M$
and
$y\in M$
. Then
$(xy)^3=1$
. That is,
$yxy=x^{-1}y^{-1}x^{-1}$
. Now
$$ \begin{align*} [a^x,b]&=x^{-1}a^{-1}xb^{-1}x^{-1}axb\\ &=x^{-1}a^{-1}(xb^{-1}x)(xax)b\\ &=x^{-1}a^{-1}(bx^{-1}b)(a^{-1}x^{-1}a^{-1})b\\ &=x^{-1}a^{-1}bx^{-1}a^{-1}bx^{-1}a^{-1}b\\ &=(x^{-1}a^{-1}b)^3=1.\\[-32pt] \end{align*} $$
Lemma 2.5. Suppose that G is a finite p-group and H is a proper minimal nonabelian subgroup of G. If H is isolated in G, then H is isomorphic to
${\mathrm{S(}}p^3{\mathrm{)}}$
.
Proof. By Lemma 2.3, we have
$p>2$
. Let
$T\leq G$
be such that
$|T:H|=p$
. Since H is isolated in G, H is isolated in T and so all elements in
$T-H$
are of order p.
If
$p\geqslant 5$
, then, by [Reference Qu, Yang, Xu and An5, Corollary
$3.4$
], T has at least two minimal nonabelian subgroups of index p. Moreover, by [Reference An, Li, Qu and Zhang1, Theorem
$2.7$
], we have
$\mathrm{cl}(G)\leqslant 3$
. It follows from [Reference Berkovich2, Theorem
$7.1$
] that T is regular. Notice that
$T=\langle T-H\rangle $
and all elements in
$T-H$
are of order p. We have
$\exp (T)=p$
. Moreover,
$\exp (H)=p$
. Hence,
$H\cong {\mathrm{S}}{\mathrm{(}}p^3{\mathrm{)}}$
.
In the following, assume that
$p=3$
.
First, we claim that
$T'\le \mathrm{Z}(H)=\Phi (H)$
. In fact, since H is minimal nonabelian,
$\mathrm{Z}(H)=\Phi (H)$
and
$|H:\mathrm{Z}(H)|=3^2$
by Lemma 2.2. Let
$h\in H-\mathrm{Z}(H)$
. It follows from
$|H:\mathrm{Z}(H)|=3^2$
that
$\mathrm{C}_H(h)=\langle h\rangle \mathrm{Z}(H)$
and
$\mathrm{C}_H(h)$
is maximal in H. Notice that
$o(t)=3$
for all
$t\in T-H$
. By Lemma 2.4,
$[h^t,h]=1$
. So,
$h^t\in \mathrm{C}_H(h)$
. It follows that
$\mathrm{C}_H(h)\trianglelefteq T$
. Since
$|T/ \mathrm{C}_H(h)|=3^2$
, we have
$T'\leq \mathrm{C}_H(h)$
. Since h is arbitrary,
$T'\leq \Phi (H)=\mathrm{Z}(H)$
.
Next, we claim that
$H\cong {\mathrm{S}}{\mathrm{(}}27{\mathrm{)}}$
. Otherwise,
$\exp (H)>\exp (\mathrm{Z}(H))$
. Hence,
$a\not \in \mathrm{Z}(H)$
for all
$a\in H$
with
$o(a)=\exp (H)$
. Let
$x\in T-H$
. Then
$o(x)=o(xa^{-1})=3$
. Since
$T'\leq \mathrm{Z}(H)$
,
$$ \begin{align*}1=(xa^{-1})^3=x^3[x,a]^{{3}\choose{2}}[x,a,x][x,a,a]a^{-3}=[x,a]^3[x,a,x]a^{-3}.\end{align*} $$
Now, we consider
$[x,a,x]$
. Without loss of generality, we may assume that
$H=\langle a,b\rangle $
and
$H'=\langle c\rangle $
. Notice that
$[x,a]\in T'\leq \mathrm{Z}(H)=\Phi (H)=\langle a^3,b^3,c\rangle $
. We may assume that
$[x,a]=a^{3i}b^{3j}c^k$
. Since
$|H'|=3$
, we have
$H'\leq \mathrm{Z}(T)$
. So,
$[c^k,x]=1$
. Hence,
$$ \begin{align*}[x,a,x]=[a^{3i}b^{3j}c^k,x]=[a^{3i}b^{3j},x]=[a^{3i},x][b^{3j},x]=[a^i,x]^3[b^j,x]^3.\end{align*} $$
Since
$o(a)=\exp (H)>\exp (\mathrm{Z}(H))$
,
$o([x,a]^3)<o(a^{-3})$
and
$o([x,b]^3)<o(a^{-3})$
. So,
$o([x,a,x])<o(a^{-3})$
. It follows that
$[x,a]^3[x,a,x]a^{-3}\neq 1$
. This is a contradiction.
Lemma 2.6. [Reference Zhang7, Theorem 4.5]
Assume that G is a finite nonabelian p-group and p an odd prime. Then all minimal nonabelian subgroups of G are isomorphic to
${\mathrm{S}}{\mathrm{(}}p^3{\mathrm{)}}$
if and only if
$\exp (G)=p$
or
$G=\mathrm{H}_p(G)\rtimes \langle a\rangle $
, a semidirect product of
$\mathrm{H}_p(G)$
by
$\langle a\rangle $
, where the Hughes subgroup
$\mathrm{H}_p(G)=B_1\times B_2\times \cdots \times B_{n}$
is an abelian subgroup of index p and
$o(a)=p$
. Moreover, for all i with
$1\le i\le n$
,
$B_i\langle a\rangle $
is a group of maximal class with an abelian subgroup
$B_i$
of index p, and all elements of
$B_i\langle a\rangle - B_i$
are of order p.
Lemma 2.7. Let G be a nonabelian p-group of exponent greater than p, where
$p\geqslant 5$
. If all minimal nonabelian subgroups of G are isomorphic to
${\mathrm{S}}{\mathrm{(}}p^3{\mathrm{)}}$
, then
$\mathrm{N}_G(S)$
is of exponent p for any minimal nonabelian subgroup S of G.
Proof. Assume the contrary and
$\exp (\mathrm{N}_G(S))>p$
for some minimal nonabelian subgroup S of G. Then there exists
$a\in \mathrm{N}_G(S)- S$
such that
$o(a)=p^2$
. Let
$L=S\langle a\rangle $
. Then
$|L|\leqslant p^5$
and hence
$\mathrm{cl}(L)\leqslant 4$
. So, L is regular. By [Reference Berkovich2, Proposition 10.28], L is generated by minimal nonabelian subgroups. So,
$\exp (L)=p$
, which is a contradiction.
Proof of Theorem 1.2
$(\Leftarrow :)$
By Lemma 2.7, S is isolated in
$\mathrm{N}_G(S)$
for any minimal nonabelian subgroup S of G.
$(\Rightarrow :)$
By Lemma 2.3,
$p>2$
. By Lemma 2.5, all minimal nonabelian subgroups of G are isomorphic to
${\mathrm{S}}{\mathrm{(}}p^3{\mathrm{)}}$
. By Lemma 2.6,
$G=\mathrm{H}_p(G)\rtimes \langle a\rangle $
has the structure described as Lemma 2.6. Assume that
$p=3$
. We will prove that this is a contradiction.
If
$\mathrm{cl} (G)=2$
, then G is regular. Notice that all elements of
$G- \mathrm{H}_3(G)$
are of order three and
$G=\langle G-\mathrm{H}_3(G)\rangle $
. We have
$\exp (G)=3$
. This contradicts the hypothesis of
$\exp (G)>3$
.
If
$\mathrm{cl} (G)\geqslant 3$
, then there exists some j such that
$\mathrm{cl}(B_j\langle a\rangle )\geqslant 3$
. Let
$M=\mathrm{Z}_3(B_j\langle a\rangle )\langle a\rangle $
, where
$\mathrm{Z}_3(B_j\langle a\rangle )$
is the third term of the upper central series of
$B_j\langle a\rangle $
. Then M is of maximal class and
$|M|=3^4$
. By checking the list of groups of order
$3^4$
, there is a
$c\in M$
such that
$c^3\ne 1$
. Let S be a minimal nonabelian subgroup of M. Then
$M\le \mathrm{N}_G(S)$
. Since
$c^3\cap S\neq 1$
, S is not isolated in its normaliser. This is a contradiction.
Remark 2.8. If G is the group in Theorem 1.2 and all minimal nonabelian subgroups of G are isomorphic to
${\mathrm{S}}{\mathrm{(}}p^3{\mathrm{)}}$
, then, by Lemma 2.6, G has an abelian maximal subgroup A of exponent greater than p with
$A ={\mathrm{H}}_p(G)$
. By Lemma 2.7,
$\mathrm{N}_G(S)$
is of exponent p for any minimal nonabelian subgroup S of G. This means that Theorem 1.2 contains all conclusions in Theorem 1.1.
Theorem 2.9. Let G be a nonabelian p-group which is not minimal nonabelian. Then all minimal nonabelian subgroups of G are isolated in G if and only if
$\exp (G)=p$
.
Proof.
$(\Leftarrow :)$
It is obvious.
$(\Rightarrow :)$
Since all minimal nonabelian subgroups of G are isolated in G, all minimal nonabelian subgroups of G are isolated in their normalisers. By Lemma 2.5, all minimal nonabelian subgroups of G are isomorphic to
${\mathrm{S}}{\mathrm{(}}p^3{\mathrm{)}}$
.
Assume that
$\exp (G)>p$
. Then, by Lemma 2.6,
$G=\mathrm{H}_p(G)\rtimes \langle a\rangle $
has the structure described in Lemma 2.6. Since
$\exp (G)>p$
, there exists some j such that
${\exp (B_j\langle a\rangle )>p}$
. So,
$\exp (B_j)>p$
and hence
$\mho _1(B_j)\neq 1$
. Since
$B_j\langle a\rangle $
is a group of maximal class,
$|\mathrm{Z}(B_j\langle a\rangle )|=p$
. Hence,
$\mathrm{Z}(B_j\langle a\rangle )\leq \mho _1(B_j)$
. Notice that
$B_j$
is abelian. Then there exists
$b\in B_j$
such that
$\mathrm{Z}(B_j\langle a\rangle )=\langle b^p\rangle $
. Let
$D=\mathrm{Z}_2(B_j\langle a\rangle )\langle a\rangle $
, where
$\mathrm{Z}_2(B_j\langle a\rangle )$
is the second term of the upper central series of
$B_j\langle a\rangle $
. Then D is a nonabelian group of order
$p^3$
. So,
$D\cong {\mathrm{S}}{\mathrm{(}}p^3{\mathrm{)}}$
. Since
$1\neq \langle b^p\rangle =\mathrm{Z}(B_j\langle a\rangle )\leq D$
, D is not isolated in G. This is a contradiction.
Remark 2.10. By Theorem 2.9, the p-groups satisfying the hypothesis of [Reference Janko4, Corollary 5] do not exist.
3. Finite p-groups with an isolated metacyclic subgroup
Firstly, we prove that if H is an isolated metacyclic subgroup, then H is abelian.
Theorem 3.1. Let G be a nonabelian p-group and H a proper metacyclic subgroup of G. If H is isolated in
$N_G(H)$
, then H is abelian. In particular, if H is isolated in G, then H is abelian.
Proof. Assume the contrary and that H is nonabelian. By Lemma 2.3,
$p>2$
. Let
$N=N_G(H)$
and
$K=\Phi (H')$
. Since K is a characteristic subgroup of H,
$K\unlhd N$
. By Proposition 2.1,
$H/K$
is isolated in
$N/K$
. By Lemma 2.2, it is easy to see that
$H/K$
is minimal nonabelian. By Lemma 2.5,
$H/K$
is isomorphic to
${\mathrm{S(}}p^3{\mathrm{)}}$
, which contradicts the assumption that H is metacyclic. So, H is abelian.
In the following, we consider finite p-groups with an isolated abelian subgroup.
Lemma 3.2. [Reference Hogan and Kappe3]
Let G be a finite p-group with
${\mathrm{H}}_p(G)\neq 1$
. Then
$|G:{\mathrm{H}}_{p}(G)|\leqslant p$
if G satisfies one of the following conditions:
-
(a)
$G_i/G_{i+1}$
is cyclic for all
$i\geqslant 2$
; -
(b) G is metabelian.
Theorem 3.3. Let G be a nonabelian p-group and H an abelian subgroup of G with
$\exp (H)>p$
. If H is isolated in
$N_G(H)$
, then the following assertions hold:
-
(1)
$N_G(H)/H$
has no subgroup isomorphic to
${\mathrm{E}}_{p^2}$
; -
(2)
$N_G(H)=G$
; -
(3)
$C_G(H)=H.$
Proof. (1) Assume the contrary and let
$L/H\leq N_G(H)/H$
be such that
$L/H\cong {\mathrm{E}}_{p^2}$
. Notice that H is isolated in
$N_G(H)$
. So, all elements in
$L-H$
are of order p. Hence,
$\mathrm{H}_p(L)\leq H$
. Since
$\exp (H)>p$
, we have
$H\leq \mathrm{H}_p(L)$
. So,
$H=\mathrm{H}_p(L)$
. It follows that
$|L:\mathrm{H}_p(L)|=p^2$
. On the other hand, since
$L/H\cong {\mathrm{E}}_{p^2}$
, we have
$L'\leq H$
. Hence, L is metabelian. So, by Lemma 3.2,
$|L:\mathrm{H}_p(L)|\leqslant p$
, which is a contradiction.
(2) By (1),
$N_G(H)/H$
has a unique subgroup of order p. Set
$M/H\leq N_G(H)/H$
be such that
$|M/H|=p$
. Then
$M\ \mathrm{char} \ N_G(H)$
. Notice that all elements in
$M-H$
are of order p. We have
$\mathrm{H}_p(M)=H$
. It follows that
$H\ \mathrm{char}\ M$
and hence
$H \ \mathrm{char} \ N_G(H)$
. So,
$N_G(N_G(H))=N_G(H)$
. Thus,
$N_G(H)=G$
.
(3) Assume the contrary and let
$A\le C_G(H)$
be such that
$|A:H|=p$
. Since H is isolated in
$N_G(H)$
, H is isolated in A. Hence, all elements in
$A-H$
are of order p. Since A is abelian and
$A=\langle A-H\rangle $
, we have
$\exp (A)=p$
, contradicting
$\exp (H)>p$
.
Theorem 3.4. Let G be a nonabelian p-group and H an abelian subgroup of G with
$\exp (H)>p$
. If H is isolated in G, then
$p\leqslant d(H)+1$
.
Proof. If
$p=2$
, then the conclusion is obvious. In the following, we may therefore assume that
$p>2$
. Let
$T\leq G$
be such that
$|T:H|=p$
. Notice that H is isolated in T. So, all elements in
$T-H$
are of order p. Hence,
$\mho _1(T)=\mho _1(H)$
. So,
$|T:\mho _1(T)|=|T:\mho _1(H)|=p|H:\mho _1(H)|=p|H:\Phi (H)|=p^{d(H)+1}$
.
Assume that
$p>d(H)+1$
. Then
$|T/\mho _1(T)|\leqslant p^{p-1}.$
Hence, T is absolutely regular. So, by [Reference Berkovich2, Theorem 9.8], T is regular. Since
$T=\langle T-H\rangle $
, we have
$\exp (T)=p$
, contradicting
$\exp (H)>p$
. Hence,
$p\leqslant d(H)+1$
.
Corollary 3.5. Let G be a nonabelian p-group and H a proper metacyclic subgroup of G with
$\exp (H)>p$
. If H is isolated in G, then
$p=2$
or
$3$
.
Corollary 3.6. Let G be a nonabelian p-group and H a cyclic subgroup of G with
$\exp (H)>p$
. If H is isolated in G, then
$G\cong {\mathrm{D}}_{2^{e+1}}$
, where
$e\geqslant 2$
.
Proof. Let
$H=\langle a\rangle $
and
$|H|=p^e$
, where
$e\geqslant 2$
. By Theorem 3.4,
$p=2$
. By Theorem 3.3,
$H\unlhd G$
and there is a unique subgroup M such that
$|M/H|=2$
. Let
$M=\langle a,b\rangle $
. Since H is isolated in G, we have
$b^2=(ba)^2=1$
. Hence,
$a^b=a^{-1}$
. So,
$M\cong {\mathrm{D}}_{2^{e+1}}$
. We claim that
$M=G$
. Otherwise, there is a
$c\in G$
such that
$c^2=b$
. Let
$a^c=a^i$
. Then
$a^b=a^{c^2}=a^{i^2}$
. It follows that
$i^2 \equiv -1 \ (\mathrm{mod }\ 2^e)$
, which is a contradiction. Hence,
$G=M\cong {\mathrm{D}}_{2^{e+1}}$
.
Corollary 3.7. Let G be a nonabelian p-group and H an isolated abelian subgroup of G. Assume that H is noncyclic with type invariants
$$ \begin{align*}(e_1,e_2,\ldots,e_{d}),\quad \mbox{where } e_1\geqslant e_2\geqslant\cdots \geqslant e_{d} \mbox{ and } e_1\geqslant 2.\end{align*} $$
If
$p>2$
or
$p=2$
and
$|G:H|\geqslant 4$
, then
$e_1-e_2\leqslant 1.$
Proof. Assume that
$e_1-e_2\geqslant 2$
. By Theorem 3.3,
$H\unlhd G$
. Hence,
$\Omega _{e_2}(H)\unlhd G$
. Consider the quotient group
$\overline {G}=G/\Omega _{e_2}(H)$
. Obviously,
$\overline {H}\cong {\mathrm{C}}_{p^{e_1-e_2}}$
. By Corollary 3.6,
$\overline {G}\cong {\mathrm{D}}_{2^{e_1-e_2+1}}$
. Hence,
$|\overline {G}:\overline {H}|=2$
, contradicting
$|G:H|\geqslant 4$
. Thus,
$e_1-e_2\leqslant 1$
.
Proposition 3.8. Suppose that
$G\cong {\mathrm{C}}_{p^2}\times {\mathrm{C}}_p$
and
$P\in {\mathrm{Syl}}_p({\mathrm{Aut}}(G))$
. If
$p=2$
, then
$P\cong {\mathrm{D}}_8.$
If
$p>2$
, then
$P\cong {\mathrm{S}}(p^3)$
.
Proof. Let
$G=\langle x\rangle \times \langle y\rangle \cong {\mathrm{C}}_{p^2}\times {\mathrm{C}}_p,\ \mathrm{where}\ o(x)=p^2, o(y)=p.$
By calculation,
$|{\mathrm{Aut}}(G)|= p^3(p-1)^2$
. Hence,
$|P|=p^3$
. Let
$$ \begin{align*} \alpha:\ & x\rightarrow x, y\rightarrow yx^p, \\ \beta:\ & x\rightarrow xy, y\rightarrow y, \\ \gamma:\ & x\rightarrow x^{1-p}, y\rightarrow y. \end{align*} $$
By straightforward calculation,
$\alpha ,\beta , \gamma \in {\mathrm{Aut}}(G)$
and
$$ \begin{align*}o(\alpha)=o(\beta)=o(\gamma)=p, \quad [\alpha, \beta]=\gamma, \quad [\gamma,\alpha]=[\gamma,\beta]=1.\end{align*} $$
Hence,
$\langle \alpha ,\beta ,\gamma \rangle \in {\mathrm{Syl}}_p({\mathrm{Aut}}(G))$
and the conclusions in the theorem hold.
In the following result, we determine finite p-groups with an isolated metacyclic subgroup.
Theorem 3.9. Let G be a nonabelian p-group. Then G has a metacyclic subgroup with exponent greater than p which is isolated in its normaliser if and only if G is isomorphic to one of the following nonisomorphic groups:
-
(I)
${\mathrm{D}}_{2^{e+1}}$
, where
$e\geqslant 2;$
-
(II)
$\langle a,b,t\mid a^{2^{e}}=b^{2^{e'}}=t^2=1,[a,b]=1,a^t=a^{-1},b^t=b^{-1}\rangle $
, where
$e\geqslant e'$
and
$e\geqslant 2;$
-
(III)
$\langle a,x,b\mid a^{2^{e}}=b^{2^{e'}}=x^4=1,[a,b]=1,a^x=ab,b^x=b^{-1}a^{-2}\rangle $
, where
$e\geqslant e'\geqslant e-1\geqslant 1;$
-
(IV)
$\langle a,t,b\mid a^{3^{e}}=b^{3^{e'}}=t^3=1,[a,b]=1,[a,t]=b,[b,t]=a^{-3}b^{-3}\rangle $
, where
$e\geqslant e'\geqslant e-1\geqslant 1.$
Proof. (
$\Rightarrow :$
) Let H be a metacyclic subgroup of G with exponent greater than p and H be isolated in
$N_G(H)$
. By Theorem 3.1, H is abelian. Moreover, by Theorem 3.3,
$N_G(H)=G$
. So, H is isolated in G and
$H\unlhd G$
.
If H is cyclic, then, by Corollary 3.6,
$G\cong {\mathrm{D}}_{2^{e+1}}$
, giving the group (I) in the theorem.
In the following, we assume that
$d(H)=2$
. By Corollary 3.5,
$p=2$
or
$3$
.
Case 1:
$p=2$
. Assume that
$H=\langle a\rangle \times \langle b\rangle \cong {\mathrm{C}}_{2^{e}}\times {\mathrm{C}}_{2^{e'}},$
where
$e\geqslant e'$
,
$e\geqslant 2$
. By Theorem 3.3,
$H\unlhd G$
and there is a unique subgroup T such that
$|T/H|=2$
. Since H is isolated in G, H is isolated in T and so all elements in
$T-H$
are of order two.
Let
$t\in T- H$
. Then
$(ht)^2=1$
for all
$h\in H$
. Hence,
$t^2=1$
and
$h^t=h^{-1}$
for all
$h\in H$
. If
$T=G$
, then we get the group (II) in the theorem.
Next, assume that
$T<G$
. By Corollary 3.7,
$e\geqslant e'\geqslant e-1\geqslant 1$
. Let
$M\leq G$
be such that
$|M:T|=2$
. By Theorem 3.3,
$M/H\cong {\mathrm{C}}_4$
. Let
$x\in M- T$
. Since
$H\unlhd M$
, we may assume that
$a^x=a^ib^j, b^x=a^kb^l$
, where
$2\nmid il-jk$
. Notice that
$x^2\in T-H$
.
If
$2\mid i$
, then
$e=e'$
and
$2\nmid j$
since
$o(a^x)=o(a)$
. Replacing b with
$a^{i-1}b^j$
, we have
$a^x=ab$
.
If
$2\nmid i$
and
$2\nmid j$
, then, replacing b with
$a^{i-1}b^j$
, we have
$a^x=ab$
.
If
$2\nmid i$
and
$2\mid j$
, then
$2\nmid l$
since
$2\nmid il-jk$
. We claim that
$2\nmid k$
. Otherwise,
${4\mid kj}$
. By calculation,
$a^{-1}=a^{x^2}=a^{i^2+kj}b^{ij+lj}$
. It follows that
$-1\equiv i^2\ (\mathrm{mod }\ 4)$
, which is a contradiction. Hence,
$2\nmid k$
. Moreover,
$e=e'$
. Replacing a and b with b and
$a^{k}b^{l-1}$
, we also have
$a^x=ab$
.
Hence, we may assume that
$a^x=ab$
. By calculation,
$a^{-1}=a^{x^2}=(ab)^{x}=abb^x$
. So,
$b^x=b^{-1}a^{-2}$
.
We claim that
$M=G$
. If not, we have
$|G/H|\geqslant 8$
. Without loss of generality, we may assume that
$|G/H|=8$
. Notice that
$\mho _2(H)\ {\mathrm{char}}\ H$
and
$H\trianglelefteq G$
. We have
$\mho _2(H)\trianglelefteq G$
. Let
$\overline G=G/\mho _2(H)$
and
$\overline H=H/\mho _2(H)$
. By Proposition 2.1,
$\overline H$
is isolated in
$\overline G$
. Obviously,
$|\overline G:\overline H|=8$
and
$\exp (\overline H)=4$
. So, we always assume that
$\exp (H)=4$
. Notice
$d(H)=2$
and H is abelian. We have
$H\cong {\mathrm{C}}_4\times {\mathrm{C}}_2$
or
$H\cong {\mathrm{C}}_4^2$
. For the latter, let
$H_1\leq H$
be such that
$|H_1|=2$
and
$H_1\trianglelefteq G$
. Consider the quotient group
$G/H_1$
. It is easy to see that
$H/H_1$
is isolated in
$G/H_1$
,
$|G/H_1:H/H_1|=8$
and
$H/H_1\cong {\mathrm{C}}_4\times {\mathrm{C}}_2$
. So, we always assume that
$H\cong {\mathrm{C}}_4\times {\mathrm{C}}_2$
. Notice that H is isolated in G. By Theorem 3.3,
$C_G(H)=H$
. Hence,
$$ \begin{align*}G/H=N_G(H)/C_G(H)\lesssim {\textrm{Aut}}(H).\end{align*} $$
Notice that
$|G/H|=8$
. By Proposition 3.8,
$G/H\cong {\mathrm{D}}_8$
. So,
$G/H$
has a subgroup isomorphic to
${\mathrm{E}}_{4}$
. This contradicts Theorem 3.3.
Summarising, we have
$M=G$
. So, we get the group (III) in the theorem.
Case 2:
$p=3$
. Assume that
$H\cong {\mathrm{C}}_{3^{e}}\times {\mathrm{C}}_{3^{e'}}$
, where
$e\geqslant e'$
,
$e\geqslant 2$
. By Corollary 3.7,
$e\geqslant e'\geqslant e-1\geqslant 1$
. By Theorem 3.3,
$H\unlhd G$
and there is a unique subgroup T such that
$|T/H|=3$
. Since H is isolated in G, H is isolated in T and so all elements in
$T-H$
are of order three.
Let
$M<H$
be such that
$|M/\mho _1(H)|=3$
and
$M\unlhd G$
. Take
$t\in T- H$
,
$a\in H-M$
and
$b\in M-\mho _1(H)$
. Then
$t^3=1$
and
$H=\langle a\rangle \times \langle b\rangle $
. Notice that
$\langle b^3,a^9\rangle =\mho _1(M)$
is a normal subgroup of T and
$|\mho _1(H)/\mho _1(M)|=3$
.
We claim that
$[a,t]\in M-\mho _1(H)$
. Otherwise,
$[a,t]\in \mho _1(H)$
. It follows that
$[a,t,t]\in \mho _1(M)=\langle b^3,a^9\rangle $
. By calculation,
$1=(at^{-1})^3=a^3[a,t]^{{3}\choose {2}}[a,t,t]$
. Hence,
$a^3=[a,t]^{-3}[a,t,t]^{-1}\in \langle b^3,a^9\rangle $
, which is a contradiction. So,
$[a,t]\in M-\mho _1(H)$
. Replacing b with
$[a,t]$
, we have
$[a,t]=b$
and
$[b,t]=a^{-3}b^{-3}$
.
Now, we claim that
$T=G$
. Otherwise, we have
$|G/H|\geqslant 9$
. Without loss of generality, we may assume that
$|G/H|=9$
. By Theorem 3.3,
$G/H\cong {\mathrm{C}}_9$
. By a similar argument to that for
$p=2$
, we always assume that
$H\cong {\mathrm{C}}_9\times {\mathrm{C}}_3$
. By Theorem 3.3, we have
$C_G(H)=H$
. Hence,
$$ \begin{align*}G/H=N_G(H)/C_G(H)\lesssim {\textrm{Aut}}(H).\end{align*} $$
By Proposition 3.8,
$G/H\lesssim {\mathrm{S}}(27)$
. Hence,
$\exp (G/H)=3$
. This contradicts
$G/H\cong {\mathrm{C}}_9$
. Hence,
$T=G$
and we get the group (IV) in the theorem.
(
$\Leftarrow :$
) For the group (I), it is obvious that the unique cyclic maximal subgroup of G is isolated in G.
For the group (II), it is easy to see that
$\langle b,t\rangle $
is a dihedral group. Hence,
$o(b^jt)=2$
. Since
$\langle a, b^jt\rangle $
is also a dihedral group,
$o(a^ib^jt)=2$
. Hence,
$H=\langle a,b\rangle $
is isolated in G.
For the group (III), since
$\langle a,b,x^2\rangle $
is isomorphic to the group (II), we have
$o(a^ib^ix^2)=2$
and
$(a^ib^jx^{2k}x)^2\not \in H$
. Hence,
$H=\langle a,b\rangle $
is isolated in G.
The group (IV) is a 3-group of maximal class. Its uniform elements are all of order three. Hence,
$H=\langle a,b\rangle $
is isolated in G.
The following corollary is a refinement of [Reference Janko4, Theorem 13].
Corollary 3.10. Let G be a nonmetacyclic p-group of exponent greater than p. Then all maximal metacyclic subgroups are isolated in their normalisers in G if and only if
$G=\langle a,t,b\mid a^{3^{e}}=b^{3^{e'}}=t^3=1,[a,b]=1,[a,t]=b,[b,t]=a^{-3}b^{-3}\rangle $
, where
$e\geqslant e'\geqslant e-1\geqslant 1$
,
$H=\mathrm{H}_3(G)=\langle a\rangle \times \langle b \rangle $
is noncyclic abelian, all other maximal metacyclic subgroups are isomorphic to
${\mathrm{E}}_9$
and G is of maximal class.
Proof. By Theorem 3.1, metacyclic subgroups of G are of exponent p or abelian. Hence, for
$p=2$
, there is no nonabelian metacyclic subgroup in G. By Theorem 3.9, G is isomorphic to one of the groups (II)–(IV). For the group (II),
$\langle a^{2^{e-2}},t\rangle $
is a nonabelian metacyclic subgroup. For the group (III),
$\langle a^{2^{e-2}},x^2\rangle $
is a nonabelian metacyclic subgroup. Hence, G is isomorphic to the group (IV).
Now, it is easy to see that
$H=\mathrm{H}_3(G)=\langle a\rangle \times \langle b \rangle $
is noncyclic abelian and G is of maximal class. Let
$H_1$
be a maximal metacyclic subgroup of G such that
$H_1\neq H$
. Since
$G=HH_1$
, we have
$|H_1:H_1\cap H|=3$
and
$H\cap H_1\le Z(G)$
. Hence,
$H_1=\langle x, Z(G)\rangle $
, where
$x\in G-H$
. Since
$x^3=1$
, we have
$H_1\cong {\mathrm{E}}_9$
. It is easy to see that
$N_G(H_1)=\langle x, Z_2(G)\rangle $
, where
$Z_2(G)=\Omega _1(H)$
is the unique normal subgroup of order nine in G. Hence,
$N_G(H_1)\cong \mathrm{S}(27)$
and
$H_1$
is isolated in
$N_G(H_1)$
. Summarising, all maximal metacyclic subgroups are isolated in their normalisers in G.
