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Question: Let $G$ be a group of order $n$, and positive integer $d$ divide $n$. If there is only one subgroup of order $d$ in $G$, can we say that the number of solutions $x^d = 1$ equals $d$?

Motivation: There are two useful conclusions in group theory:

Frobenius's Theorem: Let $G$ be a group of order $n$, and positive integer $d$ divide $n$, then the number of solutions $x^d = 1$ is a multiple of $d$. (An elementary proof without character theory is here).

Frobenius's Conjecture: If, in addition, the number of solutions of $x^d = 1$ is exactly $d$, then these solutions form a (normal, characteristic) subgroup. (This has been proved as a consequence of the classification of finite simple groups in 1991).

My question is the converse proposition of Frobenius's Conjecture. It's difficult to construct an adequate counterexample for me since I am only familiar with the structures of some lower-order groups. But it's easy to see that if there exists a counterexample, then its $\gcd(d,n/d) \neq 1$ by a basic lemma at here.

Whether I miss some trivial facts or construction? I appreciate any help!

M. A.
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Shuning Zhang
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1 Answers1

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The answer is no: consider for example the symmetric group $G = S_n$, $n > 1$.

Then $G$ has a unique subgroup of order $d = n!/2$, namely the alternating group $A_n$.

Assume $n \geq 4$. Then $x^d = 1$ for all $x \in G$, so there are $2d$ solutions to $x^d = 1$.

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