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.

  • 1
    $\begingroup$ Note that the result you proved for odd primes is false for p = 2, with the quaternion group Q_8 being a counterexample (the Sylow 2-subgroup is Q_8, and there are three cyclic subgroups of order 4 -- one generated by each of i, j and k). So the statement will need a non-trivial modification. $\endgroup$ Oct 21, 2015 at 19:31
  • 2
    $\begingroup$ Please, in future, try to use titles for your questions that give some clue as to what your questions are actually about. (Even better, start now and edit to change the title of this question to be more informative.) Further, when you post a long question, as this one is, try to summarize what you're asking in the initial paragraph, then fill in details. $\endgroup$ Oct 22, 2015 at 0:56
  • $\begingroup$ How does it look now? $\endgroup$ Oct 22, 2015 at 2:02
  • $\begingroup$ Now the title refers to odd $p$, but the question is about $p=2$. $\endgroup$
    – Stefan Kohl
    Oct 22, 2015 at 10:12
  • $\begingroup$ To enforce Joe Silverman's point: You should ask your question directly following on the statement of your Theorem, and then include the proof "for those interested". $\endgroup$ Oct 22, 2015 at 10:54

3 Answers 3


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 of Marcel Herzog, Counting group elements of order $p$ mod $p^2$ (MR466316), 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, Isaacs also mentions a theorem of Kulakoff concerning the case $p$ odd.


All the above mentioned results are well known (see \S 1 in `Groups of Prime Power Order', vol. 1 by Berkovich (there also given the autorship of the presented below results). Moreover, if a p-group G is neither absolutely regular (i.e., $|G/\mho_1(G)|\ge p^p$) nor of maximal class, then the number of solutions of $x^p=1$ in G is a multiple of $p^p$, and the number of cyclic subgroups of given order >p is a multiple of $p^{p-1}$. If the number of cyclic subgroups of order $p^n>p$ in a $p$-group $G$ is not a multiple of $p$, then $G$ is either cyclic or a $2$-group of maximal class (see Theorem 1.17(b) in the mentioned book).

If $G=Q\cdot P$ is a Frobenius group with kernel $P\cong{\rm E}_9$ and complement $Q\cong{\rm Q}_8$, then the number of cyclic subgroups of order $4$ in $G$ is $27$.


Let $G$ be a noncyclic $p$-group. If $p>2$, then the number of cyclic subgroups of order $p^k>p$ in $G$ is a multiple of $p$ (G.A. Miller). If $p=2$ and, in addition, $G$ is not of maximal class, then the number of cyclic subgroups of order $2^k>2$ in $G$ is even. All this one can find in Theorem 1.17 of the book Berkovich, Groups of Prime Power Order, volume 1. In Theorem 13.2 of the same book one can find stronger results.


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