# Characterization of finite groups generated by one additional element of prime order for every element of prime order

I want to prove or find a counterexample to the following statement.

Let $G$ be a finite group. If for every minimal subgroup $L$ (= cyclic subgroup of prime order), there exists a minimal subgroup $H$ such that $\langle L,H\rangle=G$. Then one of the following cases occurs:

a) $G$ is a Frobenius group whose kernel is an elementary abelian p-group and the complement is a prime order group.

b) $G=Z_{p}\times Z_q$

c) $G$ is a simple group.

Thank you in advance.

• I suppose you mean $nontrivial$ minimal subgroup? So that must be a cyclic group of prime order. – user1688 Aug 14 '16 at 5:25
• (I think you should also allow the case when $G$ has prime order.) Suppose that $G$ has a minimal normal subgroup $N$ which is abelian. Let $L$ be a minimal normal subgroup of $G$ contained in $N$. Let $H$ be as in the hypothesis. If $H\leq N$, then $G$ is abelian and it is easy to see that we have $H=L=G$. Thus, $H\cap N=1$ hence $G=N\rtimes H$ and it is easy to see that we are either in your case a) or b). Hence we may assume that $G$ has no nontrivial normal abelian subgroup. In other words, it has trivial soluble radical. – verret Aug 14 '16 at 5:43
• I'm not sure where to go from here. It's not obvious to me why some almost simple group, say, could not have the required property. – verret Aug 14 '16 at 5:43
• Dear verret. why did you say " Suppose that $G$ has a minimal normal subgroup $N$ which is "abelian"? what will be wrong if we suppose that $G$ has a nonabelian minimal normal subgroup? – H.Shahsavari Aug 14 '16 at 13:27
• Well, that was simply the case I could deal with. After thinking a bit more, I think I can now reduce to the almost simple case. By the previous comment, we may assume that $G$ has trivial soluble radical. Let $N$ be a minimal normal subgroup of $G$. We have $N=T_1\times \cdots \times T_k$ where the $T_i$ are isomorphic nonabelian simple groups. Moreover, they are the only minimal normal subgroups of $N$. Now, pick $L$ in $T_1$ and let $H$ be as in the hypothesis. Say $|H|=p$. – verret Aug 14 '16 at 22:56

## 2 Answers

$\mathrm{P\Sigma L}(2,8)$ is a counter-example.

A corrected version of the statement is that, under the hypothesis, either

a) $G\cong Z_p$ for some prime $p$,

b) $G\cong Z_p\times Z_q$ for some primes $p$ and $q$,

c) $G$ is a Frobenius group with kernel an elementary abelian $p$-group and the complement has prime order, or

d) $G$ is an almost simple group.

Proof:

Let $N$ be a minimal normal subgroup of $G$. We have $N=T_1\times\cdots \times T_k$, where the $T_i$'s are isomorphic simple groups. Take $L\leq T_1$ and let $H$ be as given by the hypothesis.

Case 1) $H\leq N$. In this case, we have $G=N$. If $T$ is abelian, then $G$ is abelian hence $k=1$ and we are in case a). If $T$ is nonabelian then, since $N$ is generated by $T_1$ and $H$ a group of prime order, we find that $k=1$ and $G$ is simple.

Case 2) $H\not\leq N$. Since $H$ has prime order, we have $N\cap H=1$ and thus $G=N\rtimes H$. If $T$ is abelian, then either $H$ is malnormal and $G$ is Frobenius and we have c), or $G$ is abelian and we have b). If $T$ is nonabelian, then the $T_i$'s are the only minimal normal subgroups of $N$ and $H$ must act transitively on $\{T_1,\ldots,T_k\}$. As $H$ has prime order, either $k=1$ (and we are in case d)), or $k=|H|$. In the latter case, $L$ has exactly $k$ conjugates, each of them contained in some $T_i$. On the other hand, these conjugates must generate $N$, which is a contradiction.

Note that this not exactly a characterisation. For example, not all almost simple groups have this property and neither do all groups in c).

• I found this too but you posted first. More generally $L_n(q^f):f$ (i.e. ${\rm PSL}_n({\bf F}_{q^f})$ extended by the Galois automorphism $x \mapsto x^f$ should work for any prime power $q$ and any odd prime $f$. Note that $f=2$ cannot work because if $L$ is a two-element subgroup of $G' := {\rm PSL}_n({\bf F}_{q^f})$ then $H$ would have to be another $2$-element group (else $\langle H,L \rangle \subseteq G'$), but two involutions can only generate a dihedral group. Thus $(n,q,f) = (2,2,3)$ is the first candidate of this kind. – Noam D. Elkies Aug 15 '16 at 4:03
• For your example ${\rm P}\Gamma {\rm L}(2,8)$ (a.k.a. $\Sigma {\rm L}(2,8)$) one can always generate $L$ by a conjugate of the Frobenius map. This may be true in general for ${\rm PSL}_n({\bf F}_{q^f})$. – Noam D. Elkies Aug 15 '16 at 4:16
• I think your comment would be worth be included in your answer (which would not only provide a counterexample but would also provide a correct characterization) – YCor Aug 15 '16 at 11:45
• (correction to my first comment: the Galois automorphism of course takes $x$ to $x^q$, not $x^f$.) – Noam D. Elkies Aug 15 '16 at 15:04
• @YCor Ok, I've done that! – verret Aug 15 '16 at 23:22

For a prime number $p$ the symmetric group $S_p$ is generated by a $p$-cycle and a transposition. These two elements generate individually minimal cyclic subgroups. I think this is case d) of Verret's answer.

• Note that the problem requires that for every minimal subgroup $L$ there exist a minimal subgroup $H$ such that $G=\langle L,H\rangle$. The fact that $S_p$ is generated by a $p$-cycle and a transposition only shows that there exists a minimal subgroup $L$ for which a minimal subgroup $H$ exists with the desired property. For instance, in $S_5$, if $H=\langle (1,2,3)\rangle$, then no $L$ exists: a $3$-cycle and a $3$- or $5$-cycle generate a subgroup of $A_5$; and a $3$-cycle and an element of order $2$ will not give you all of $S_5$. So $S_5$ does not satisfy the hypothesis. – Arturo Magidin Aug 16 '16 at 2:08
• Thanks Arturo Magidin. I did not read the question carefully. I was wondering how a simple one seem to have been missed by many. This explains it! I am the one who missed reading the hypothesis! – P Vanchinathan Aug 16 '16 at 2:12