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. 



Counting cyclic subgroups of order $p^{2}$: $p$ an odd prime vs. $p=2$
The following is true:  

The number of cyclic subgroups of order $4$ of some group is odd iff the Sylow $2$-subgroups are cyclic, dihedral, semi-dihedral or generalized quaternion. 


As in your proof, it suffices to consider $2$-groups $G$ and look at the number mod $4$ of solutions of $x^2=1$ in $G$. Now Theorem 4.9 in Isaacs' book *Character theory of finite groups* tells us that when the number $t$ of involutions in a $2$-group is $t\equiv 1 \mod 4$, then $G$ is cyclic or $\lvert G:G'\rvert =4$. (The theorem is ascribed to Alperin-Feit-Thompson, probably the result meant by Geoff in the comments.) Of course, the only other possibility is $t\equiv -1 \mod 4$. A result of Olga Taussky (Satz III. 11.9 in Huppert's *Endliche Gruppen I*) tells us that the only nonabelian $2$-groups with $\lvert G:G'\rvert = 4$ are the dihedral, semi-dihedral and generalized quaternion groups.  

Conversely, the number of cyclic subgroups of order $4$ in each of these groups is indeed odd, and thus also in a group with such a group as Sylow $2$-subgroup. 


The proof in Isaacs of that theorem of Alperin-Feit-Thompson uses characters, in particular the Frobenius-Schur indicator. 


**Added later:** Trying to find the original source of Alperin-Feit-Thompson, I found instead a [paper](http://dx.doi.org/10.2307/2040939) of Marcel Herzog, *Counting group elements of order $p$ mod $p^2$* ([MR466316](http://www.ams.org/mathscinet-getitem?mr=466316)), which contains (elementary) proofs of the "Alperin-Feit-Thompson"-result and the result for $p$ odd, and references to older proofs in the literature. In his [math review](http://www.ams.org/mathscinet-getitem?mr=466316), Isaacs also mentions a theorem of Kulakoff concerning the case $p$ odd.


