While playing with Frobenius' problem (about finite groups $G$ in which, for some positive integer $n \mid |G|$, there are exactly $n$ elements of order dividing $n$), I came up with the following result:

**Theorem.** Let $G$ be a finite group and let $p$ be an odd prime. Then the Sylow $p$-subgroups of $G$ are cyclic, or the number of cyclic subgroups of order $p^{2}$ in $G$ is a multiple of $p$.

I know that problem of Frobenius has been solved as an application of the Classification of Finite Simple Groups; I am not asking about it, but that does not prevent me from playing with it.

My question is this:

Is there an analogous result for the prime 2? If so, how is it stated?

For those who are interested, here is my proof of the theorem:

Step 1. Reducing to the problem for $p$-groups.

Let $G$ be a finite group. We assume $p^{2} \mid |G|$, since the problem is trivial otherwise. Then let $P$ be a Sylow $p$-subgroup of $G$. We let $P$ act by conjugation on the cyclic subgroups of order $p^{2}$ in $G$. Let $C$ be a cyclic subgroup of order $p^{2}$ in $G$. If $P$ normalizes $C$, then $CP$ is a $p$-subgroup of $G$, so $CP = P$ and $C \leq P$. If $P$ does not normalize $C$, then the number of $P$-conjugates of $C$ is a multiple of $p$. So the number of cyclic subgroups of order $p^{2}$ in $G$ is congruent, modulo $p$, to the number of cyclic subgroups of order $p^{2}$ in $P$ and the reduction is proven.

Step 2. It suffices to prove that the number of solutions to $x^{p} = 1$ in $P$ is a multiple of $p^{2}$ (when $|P| \geq p^{2}$ and $P$ is noncyclic).

Proof of Step 2. $|P| \geq p^{2}$ by an assumption justified in Step 1. Then, by a theorem of Frobenius, the number of solutions to $x^{p^{2}} = 1$ is a multiple of $p^{2}$. This set consists of $x$ such that $x^{p} = 1$ and $x$ which have order $p^{2}$. So the size of one set is a multiple of $p^{2}$ if and only if the size of the other set is a multiple of $p^{2}$. The number of elements of order $p^{2}$ is $p^{2}-p$ times the number of cyclic subgroups of order $p^{2}$, so it's a multiple of $p^{2}$ if and only if the number of cyclic subgroups of order $p^{2}$ in $P$ is a multiple of $p$.

Now let $P_{0}$ be a minimal counterexample to the theorem.

Step 3. Narrowing down the structure of $P_{0}$. (Denote $|P_{0}| = p^{n}$.)

Trivial but useful: $P_{0}$ is not elementary abelian, since an elementary abelian group has no cyclic subgroups of order $p^{2}$.

Also: $P_{0}$ has no cyclic subgroup of index $p$: Since $p$ is odd, if $P_{0}$ had a cyclic subgroup of index $p$, we would have $P_{0}$ being cyclic (obviously not a counterexample), $P_{0} \cong C_{p^{n-1}} \times C_{p}$, or $P_{0} \cong C_{p^{n-1}} \rtimes C_{p}$. The latter two have exactly $p^{2}$ solutions to $x^{p} = 1$, so they are not counterexamples either.

Step 4. Counting solutions to $x^{p} = 1$ in $P_{0}$ to get a final contradiction.

Let the maximal subgroups of $P_{0}$ be $Q_{1}, \ldots , Q_{k}$. Then the number of solutions to $x^{p} = 1$ in $P_{0}$ is

$$ \sum_{ R = Q_{a_{1}} \cap \ldots \cap Q_{a_{j}} } -\mu(R,P_{0})| \{x \in R \mid x^{p} = 1 \}| $$

where $\mu$ is the Mobius function for the subgroup lattice of $P_{0}$. We prove this is a multiple of $p^{2}$ by proving every term is a multiple of $p^{2}$.

If $R$ is one of the $Q_{i}$, then $[P_{0}:Q_{i}] = p$ so $Q_{i}$ is noncyclic and $| \{x \in Q_{i} \mid x^{p} = 1 \} |$ is a multiple of $p^{2}$ by induction.

If $R$ is an intersection of multiple $Q_{i}$, then let $d$ be chosen so that $R$ is an intersection of $d$ of them but not an intersection of $d-1$ of them. Then $R$ contains the Frattini subgroup $\Phi(P_{0})$, which is nontrivial because $P_{0}$ is not elementary abelian. Then, since $P_{0}/ \Phi(P_{0})$ is a vector space over $\mathbb{Z}/(p)$, $\mu(R,P_{0}) = (-1)^{d}p^{\frac{d^{2}-d}{2}}$. Since $d \geq 2$, $p \mid \mu(R,P_{0})$. Since $R \geq \Phi(P_{0})$ is nontrivial, Frobenius' theorem implies that $p \mid | \{x \in R \mid x^{p} = 1 \} |$.

In either case, $p^{2}$ divides the term corresponding to $R$, so the sum is a multiple of $p^{2}$ and the proof is complete.

5more comments