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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 $(p-1)! \equiv -1 \mod p$.

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Perhaps it is worth noting that both your "applications" follow easily without Frobenius's theorem. –  Steve D Oct 7 '12 at 0:32
Nice proof of Wilson's Theorem. I suppose we can look at other conjugacy classes in $S_p$ and get similar congruences. –  Pantelis Damianou Oct 7 '12 at 14:45
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5 Answers

up vote 11 down vote accepted

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 $X$ is a conjugacy class of elements of the 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 ${\rm gcd}(|C|n, |G|)$.

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@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. –  Anton Klyachko Nov 7 '12 at 23:47
@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! –  Nick Gill Nov 8 '12 at 9:31
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(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.

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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 $1+p\pmod {p^2}$.

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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.

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