A theorem of Frobenius states that if $n$ divides the order of a finite group $G$, then the number of solutions to $x^n = 1$ in $G$ is a multiple of $n$. Frobenius conjectured that if the number of solutions is exactly $n$, then the set of solutions form a characteristic subgroup of $G$. The conjecture was eventually proved in the 90's, and the full proof uses the classification of finite simple groups.

The theorem feels a bit isolated for me.. I'm not sure how the conjecture fits into a wider context, either. What is their importance, if any? Are there any good examples of applications of the theorem or the conjecture? If I'm interested in finite groups, why should I care about the theorem or the conjecture, other than that they are kind of neat?

One example I know is that if $G$ has every Sylow subgroup cyclic, then with Frobenius theorem we can show that the Sylow subgroup corresponding to the largest prime divisor of $G$ is normal. Also (this one is too easy, but I like it) for any prime $p$, the number of elements satisfying $x^p = 1$ in the symmetric group $S_p$ is $(p-1)! + 1$, so Frobenius theorem implies $(p-1)! \equiv -1 \mod p$.

  • $\begingroup$ Perhaps it is worth noting that both your "applications" follow easily without Frobenius's theorem. $\endgroup$
    – Steve D
    Oct 7, 2012 at 0:32
  • 2
    $\begingroup$ Nice proof of Wilson's Theorem. I suppose we can look at other conjugacy classes in $S_p$ and get similar congruences. $\endgroup$ Oct 7, 2012 at 14:45

7 Answers 7


Danny Gorenstein, in his book Finite Simple Groups, An Introduction to their Classification discusses a very significant application of Frobenius' theorem (see p. 95). He is interested in the following central problem in the Classification of Finite Simple Groups:

Let $G$ be a simple group in which the structure of the centralizer of an involution is given. Determine the order of $G$.

This problem naturally splits into different cases, the most difficult of which concerns the case where $G$ has a single conjugacy class of involutions. One proceeds by studying the $2$-local structure of $G$, then the $p$-local structure, for those primes $p$ dividing $|X|$ where $X$ is the given involution centralizer.

Gorenstein remarks that

With this information, one can now obtain a congruence for the order of $G$ with the aid of Sylow's theorem and a result of Frobenius.

The `result of Frobenius' he refers to is the one in this question. I'm unqualified to write more so I recommend Gorenstein's book for more details (on this and anything else to do with CFSG).

One final remark: Gorenstein notes in a footnote that Hall's book The Theory of Groups contains a strengthening of Frobenius' result (see Theorem 9.1.1, I'm not sure if this is the same generalization mentioned by Anton above):

If $C$ is a conjugacy class of elements of a finite group $G$, then for any positive integer $n$, the number of solutions in $G$ of the equation $x^n=c, c\in C$, is a multiple of ${\rm gcd}(|C|n, |G|)$.

  • 1
    $\begingroup$ @Nick, just to clarify. The theory of groups was written by Marshall Hall, and the paper I cited is authored by Philip Hall. The theorem you mentioned about the number of solution to $x^n=c$ belongs to Frobenius (according to Philip Hall, see the same paper). P.Hall's generalisation is much more complicated. $\endgroup$ Nov 7, 2012 at 23:47
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    $\begingroup$ @Anton, thanks for the clarification. I should keep a count of the number of times in my career I've confused Marshall Hall and Philip Hall. I am certainly in double figures by now! $\endgroup$
    – Nick Gill
    Nov 8, 2012 at 9:31

An application of this theorem to $p$-adic analysis is the $p$-integrality of the coefficients of the Artin-Hasse exponential $$ {\rm AH}_p(X) = e^{X + X^p/p + X^{p^2}/p^2 + \cdots}. $$ That means when this series is expanded as $\sum_{k \geq 0} a_kX^k$, the coefficients $a_k$ don't have their denominator divisible by $p$. This is obvious for $k = 0$ since $a_0 = 1$. We have $a_k = 1/k!$ for $1 \leq k \leq p-1$, so $a_1,\dots,a_{p-1}$ are all $p$-integral. And $a_p = ((p-1)!+1)/p!,$ which is $p$-integral by Wilson's theorem. To handle the general case we'll use the Frobenius theorem.

For $k \geq 1$, a non-obvious combinatorial/probabilistic formula for the coefficients is $$ a_k = \frac{\#\{g \in S_k : g \text{ has } p\text{-power order}\}}{k!}. $$ Using the Frobenius theorem with $n$ being the largest power of $p$ dividing $|S_k|$, the numerator of $a_k$ is divisible by the largest power of $p$ in the denominator. Hence each $a_k$ is $p$-integral.

There are simpler proofs of the $p$-integrality of the coefficients of this series. See the Wikipedia page on the Artin-Hasse exponential.


(Remarks, than a "real" answer): The first theorem is a kind of precise generalization of the Theorem of Cauchy ( that there is an element of prime order $p$ in the finite group $G$ when $|G|$ is divisible by $p$). Frobenius's theorem does imply Cauchy's theorem, and therefore Sylow's theorem. However, one has to take care to avoid circularity. Most published proofs of Frobenius's theorem of which I am aware assume Cauchy's theorem, at least implicitly, but this can be avoided with care. To be precise, a counterexample to Frobenius's theorem with first $|G|$, then $n$ minimal, would reduce to the case that $n$ is prime and $|Z(G)|$ is divisible by $n$. Hence a unified proof of Frobenius's theorem and Cauchy's theorem can be given. Also a unified proof of Cauchy's theorem and Sylow's theorem can be given. Hence Frobenius's theorem, Cauchy's theorem and Sylow's theorem can all be seen as part of the same circle of ideas. The fact that if there are just $n$ solutions (and $n$ divides $|G|$), then they form a subgroup has a satisfactory ring to it, and is a natural question. However, I do not know any immediate applications of this fact myself.

MUCH later edit: I might add that if it were possible to prove this conjecture of Frobenius ( that if there are just $n$ solutions for $n$ a divisor of the order of $G,$ then they form a subgroup), and it could be done without using characters, then it would provide a character free proof of the other famous theorem of Frobenius ( that if each non-identity element of a finite transitive permutation group fixes at most one point, then the elements which do not fix a unique point form a subgroup). There is still no completely character free proof of that theorem, though Terry Tao has a proof which essentially reduces it to a question about characters of finite Abelian groups.

The reason that the conjecture implies the (known) theorem is that if $G$ is the permutation group, and $H$ is a point stabilizer, then $ H \cap H^{g} = 1$ for all $g \in G \backslash H$. Hence there are $[G:H](|H|-1)$ non-identity elements which fix a unique point, and $[G:H]$ elements which do not fix a unique point. Set $n = [G:H].$ Then $n$ is relatively prime to $|H|$ since $[G:H] \equiv 1$ (mod $|H|$), since, eg, the double coset $HgH$ contains $|H|^{2}$ elements for any $g \in G \backslash H.$ Since every element which fixes a unique point is in a conjugate of $H,$ no such element has order dividing $n,$ and $G$ contains exactly $n$ solutions of $x^{n} = 1.$

Later edit: In view of the answers by K.Conrad and J. Moller, I will outline a proof that in the special case that $n= |G|_{p}$ for a prime $p,$ Frobenius's theorem admits a fairly direct inductive proof. This proof is related to, but slightly different from, the proof that appears in the AMM paper that Marty Isaacs and I wrote- the proof bypasses the need to assume Cauchy's Theorem, though in some ways it is more sophisticated : We proceed by induction on $|G|,$ and we are trying to prove that the number of $p$-elements of $G$ (including the identity as a $p$-element) is divisible by $|G|_{p}.$ Suppose first that there is an element $y \neq 1$ of order prime to $p$ with $y \in Z(G).$ We note that there is a bijection between $p$-elements of $G$ and $p$-elements of $G/Y$ where $Y = \langle y \rangle.$ Clearly, the image of a $p$-element of $G$ in $G/Y$ is still a $p$-element. On the other hand, given any $x \in G$ such that $xY$ is a $p$-element in $G/Y,$ we see that $xy^{j}$ is a $p$-element for some value of $j$ with $0 \leq j < |Y|.$ This value of $j$ is unique, since otherwise $y^{k}$ is a $p$-element for some $k$ with $0 < k < |Y|,$ which is not the case. Since we are only interested in the coset $xY,$ we might as well suppose that $x$ itself is a $p$-element. What we have really proved that given a coset $xY$ in $G/Y$ which is a $p$-element, there is a unique coset representative which is a $p$-element. By induction, the number of $p$-elements of $G/Y$ is a multiple of $[G:Y]_{p} = |G|_{p}.$ Hence we may suppose that there is no such central non-identity element $y$ of order prime to $p.$

Now given any element $x \in G$ which is not a $p$-element, we may (uniquely) write $x = yz = zy$ where $y \neq 1$ is an element of order prime to $p$ and $z$ is a $p$-element of $C_{G}(y).$ For any given element $y$ of order prime to $p,$ the number of choices of of $z$ is the number of $p$-elements of $C_{G}(y)$ which, by induction is divisible by $|C_{G}(y)|_{p}$ as $y \not \in Z(G).$ If we count the contribution from the conjugacy class of $y,$ we get a multiple of $[G:C_{G}(y)]|C_{G}(y)|_{p},$ so certainly a multiple of $|G|_{p}.$ Doing this for every conjugacy class of non-identity elements of order prime to $p,$ we see that the number of elements of $G$ which are NOT $p$-elements is divisible by $|G|_{p}.$ Since $|G|$ is certainly divisible by $|G|_{p},$ we see that the number of $p$-elements of $G$ is an integer multiple of $|G|_{p}.$

  • $\begingroup$ There is also a nice direct proof by Brauer (in the paper "On A Theorem of Frobenius" in AMM), which does not require induction at all. Basically if $n = p^\alpha m$, $p$ prime, $(m,p) = 1$, then using a subgroup of order $p^\alpha$, we can define an equivalence relation $\sim$ on $G$ such that each $\sim$-class has order $p^\alpha$, and the set of solutions to $x^n = 1$ is an union of $\sim$-classes. This proof also gives you more general statements about the number of solutions in a fixed double coset $HyH$, for example. $\endgroup$ May 12, 2014 at 17:35
  • $\begingroup$ @Mikko: Yes, thanks. That proof is referenced in my paper with Isaacs $\endgroup$ May 12, 2014 at 17:51

Since nobody gave any examples of applications of Frobenius conjecture, here's a small one I read about recently. We will only consider finite groups in what follows. For a group $G$, denote the set of elements of $G$ satisfying $x^n = 1$ by $a_n(G)$.

If for each positive integer $n$, the groups $G$ and $H$ have the same number of elements of order $n$, we say that $G$ and $H$ have the same order structure. It is not hard to see that $G$ and $H$ have the same order structure if and only if $a_n(G) = a_n(H)$ for every integer $n$.

Question: if $H$ has the same order structure as $G$, how does the structure of $G$ affect the structure of $H$? It is clear that we will have $|G| = |H|$. We cannot say that $G$ and $H$ must be isomorphic: take $G$ and $H$ to be $p$-groups both of same order, both of exponent $p$, $G$ abelian, $H$ nonabelian. So $G$ abelian does not imply $H$ abelian. However, $G$ cyclic does imply that $H$ must be cyclic. We can also show that $G$ nilpotent implies that $H$ is nilpotent. It has been proven (using CFSG) that if $G$ is simple, then $G$ and $H$ must be isomorphic if they have the same order structure.

There is an open problem (due to Thompson, I think) that asks the following question:

Problem: Suppose that $G$ is solvable and that $G$ and $H$ have the same order structure. Does this imply that $H$ is solvable?

Using Frobenius conjecture, we can deduce a partial result:

Theorem: If $G$ is supersolvable and $G$ and $H$ have the same order structure, then $H$ is solvable.

Proof: Suppose $G$ has order $|G| = p_1^{a_1} \ldots p_t^{a_t}$, where $p_1 < p_2 < \ldots < p_t$ are primes. We know that $|G| = |H|$. Furthermore, since $G$ supersolvable, we see that $a_n(G) = n$ for every $n = p_i^{a_i} p_{i+1}^{a_{i+1}} \ldots p_t^{a_t}$. Hence $a_n(H) = n$ for every such $n$ since $G$ and $H$ have the same order structure. By Frobenius conjecture, $H$ has a unique subgroup of order $p_i^{a_i} p_{i+1}^{a_{i+1}} \ldots p_t^{a_t}$ for every $i$ and thus $H$ must be solvable.

  • $\begingroup$ I have a question on your answer. Please let me know why you said that "since G supersolvable, we see that $a_{n}(G)=n$ for every $n$". Thanks in advance. $\endgroup$
    – A R
    Sep 4, 2023 at 12:15
  • $\begingroup$ @AR If a finite group $G$ is supersolvable, it has a normal subgroup $N$ of order $n = p_i^{a_i} p_{i+1}^{a_{i+1}} \ldots p_t^{a_t}$, for all $1 \leq i \leq t$. Here $\gcd(|N|,[G:N]) = 1$, which implies that $a_n(G) = n$ (and $N$ is the unique subgroup of order $n$ in $G$). $\endgroup$ Sep 4, 2023 at 12:25

P. Hall's paper On a theorem of Frobenius contains a generalisation of the Frobenius theorem and some applications of this result. For example, he obtained the following Sylow-like theorem.

THEOREM 4.6. If $p$ is an odd prime, and if the $p$-Sylow subgroups of $G$ are of order $p^l$ and not cyclical, then, for $0 < k < l$, the total number of subgroups of order $p^k$ in $G$ is congruent to $1+p\pmod {p^2}$.


Here is an application of Frobenius theorem of $1907$. Let $G$ be a finite group and $p$ a prime number. Write $G_p$ for the set of elements of $G$ of $p$-power order and $|G|_p$ for the $p$-part of the group order. Then $|G|_p$ divides $|G_p|$ according to Frobenius' theorem.

Let now $\mathcal S_G^*$ denote the Brown poset of nonidentity $p$-subgroups of $G$. Write $\chi(\mathcal S_G^*)$ for the Euler characteristic and $\widetilde{\chi}(\mathcal S_G^*) = \chi(\mathcal S_G^*)-1$ for the reduced Euler characteristic of $\mathcal S_G^*$. Then $$ \sum |H| \widetilde{\chi}(\mathcal S_{N_G(H)/H}^*) = -|G_p| $$ where the sum runs over all $p$-subgroups $H$ of $G$. This identity follows from the analysis of the Euler characteristic of the orbit category from https://arxiv.org/abs/1301.0193. Note that the trivial subgroup contributes $\chi(\mathcal S_G^*)$ to the sum on the left. By induction over group order we see that $|G|_p$ divides $\widetilde{\chi}(\mathcal S_G^*)$. This is Brown's theorem of $1975$.

  • $\begingroup$ Actually, I find the displayed formula more interesting that Brown's Theorem. And it would appear that Brown's theorem implies the case of $n = |G|_{p}$ of Frobenius's theorem as well. In fact, in a paper of Marty Isaacs and myself, we showed that that case essentially is sufficient to do every $n.$ $\endgroup$ May 7, 2014 at 16:06
  • $\begingroup$ @GeoffRobinson: Yes, the displayed formula shows that Brown's theorem and the $n=|G|_p$ case of Frobenius' theorem are equivalent. Thanks to your work with Marty Isaacs we know how to get from this special case to the general case of Frobenius' Theorem. Thus the theorems of Frobenius and Brown are equivalent. Let me mention the general relation $\sum \widetilde{\chi}(\mathcal{S}^*_{N_G(H)/H})=-1$ related to the displayed formula. $\endgroup$ May 7, 2014 at 18:52

An application on Frobenius' theorem in here: Relationship between the number of Sylow subgroups with element orders in finite group


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