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Let $G$ be a finite group (not necessarily abelian). Let $d \mid |G|$, where $d \in \mathbb{N}$. Does there exist at least $d$ solutions to the equation $x^d = 1_G$ for $x \in G$?

I can prove it for abelian groups (Not hard).

Is it true for all groups? If not please give counterexample.

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See also mathoverflow.net/questions/109027/… –  j.p. Nov 7 '12 at 21:33
    
Thanks for the link! –  Shubhodip Mondal Nov 7 '12 at 22:03
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1 Answer 1

up vote 7 down vote accepted

This is Frobenius theorem.

Suppose $G$ is a finite group such that $d$ divides $|G|$. Then the number of solutions to $x^d = 1$ is a multiple of $d$.

Since the identity is always a solution, there exist at least $d$ solutions. This is true for all finite groups, but it is not very easy to prove.

A proof can be found in the following Monthly article:

I. M. Isaacs, G. R. Robinson, On a Theorem of Frobenius: Solutions of $x^n = 1$ in finite groups, Amer. Math. Monthly, Vol. 99, No. 4, 352-354, (1992).

The theorem can be generalized in many different ways. A proof of the following generalization is given in Marshall Hall's group theory book:

Let $G$ be a finite group and $C$ a conjugacy class of $G$. The number of elements $x$ such that $x^n \in C$ is a multiple of $\gcd(n|C|, |G|)$.

The original theorem follows from the case $C = \{1\}$.

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