Is there a finite group $G$ and a divisor $d$ of $|G|$ so that $G$ contains exactly two subgroups of order $d$?

The motivation for this question is an old qual problem (see http://www.math.wisc.edu/~passman/qualjan01.pdf, problem 1). If $G$ is such a group and $A$ and $B$ are the two subgroups of order $d$, then it is easy to see that $A \unlhd \langle A, B \rangle$ and $B \unlhd \langle A, B \rangle$ and so $$ \langle A,B \rangle / (A \cap B) \cong A/(A \cap B) \times B/(A \cap B) $$ is another group (with possibly smaller order) with this property. Assume therefore that $G \cong A \times B$ and the only two subgroups of $G$ with order $d$ are $A \times 1$ and $1 \times B$. Since $(\mathbb{Z}/p\mathbb{Z})^{2}$ has $p+1$ subgroups of order $p$, we see that $d$ is not prime.

Moreover, if $A$ has a subgroup $C$ of index $p^{k}$, then $B$ has a subgroup of order $p^{k}$, then $C \times D$ is another subgroup of $G$ that also has order $d$. This implies that neither $A$ nor $B$ can have any subgroups of prime power index, and this implies that they are both perfect.

I searched Magma's database of perfect groups for pairs $(A,B)$ with the same order. For each pair $(A,B)$, there is a subgroup of $A \times B$ of the form $C \times D$, where $C$ is a non-trivial subgroup of $A$ and $D$ is a non-trivial subgroup of $B$.

Note: I remember asking Isaacs this question. His recollection was that this was once discussed on the Group Pub Forum, and that the answer was yes, but he didn't recall details of the construction.

  • $\begingroup$ Is it a good idea to consider the semi direct product rather than direct product? $\endgroup$ Mar 28, 2014 at 13:20
  • $\begingroup$ @Jeremy: Have you looked at groups of order 24? If I recall correctly, for example, SL(2,3) has just two Sylow 3-subgroups. Anyway, there are lots of interesting nonabelian groups of that order. $\endgroup$ Mar 28, 2014 at 14:31
  • $\begingroup$ @JimHumphreys The order cannot be the order of a Sylow subgroup, as $2$ is not congruent to $1$ mod any prime. $\endgroup$ Mar 28, 2014 at 14:34
  • $\begingroup$ @JimHumphreys It appears that SL(2,3) has 4 Sylow 3-subgroups. The argument I sketch above also implies that (among many other things) any such $G$ cannot be solvable, so groups of order $24$ won't provide examples. $\endgroup$ Mar 28, 2014 at 15:01
  • $\begingroup$ @Jeremy, Tobias: Sorry for the too-quick comment. $\endgroup$ Mar 28, 2014 at 15:12

1 Answer 1


I asked Bob Guralnick this question in class this morning. I expected he would have something to contribute. I did not expect (although perhaps I should have) that the existence of such a group $G$ and such a divisor $d$ of $|G|$ is his own result dating back to 2002.

The title of the article he sent me is "Groups with exactly two subgroups of a given order," and it appears in Communications in Algebra, Vol. 30, No. 9, pp. 4401-4406.

I'll restate the main theorem of the paper here for convenience.

Let $m$ be a positive integer. There exist a finite group $G$ and a positive integer $n$ such that $G$ has precisely $m$ subgroups of order $n$. Moreover, this can be done in such a way that all subgroups of order $n$ in $G$ are normal.

The question of this post is resolved by the theorem in the case $m=2$. Guralnick constructs two groups $A_1$ and $A_2$ which are both nonsplit extensions of $E_8(q)$ having the same order. Then $G=A_1\times A_2$ is the desired group, with $A_1\times 1$ and $1\times A_2$ the desired subgroups. Details can be found in the reference. The paper does not claim that this construction yields the smallest such example (in fact it probably does not), and leaves open the problem of determining the smallest such $G$.

Of note, and also included in the paper, is that the theorem is quickly resolved for $m>2$. This can be seen for $m\ne 6$ as $S_m$ has precisely $m$ subgroups of order $(m-1)!$, which are the stabilizers of points. For $m=6$, the group $(\mathbb{Z}/5\mathbb{Z})^2$ has $6$ subgroups of order $5$.


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