Subgroup formed by elements of order dividing n

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Posted	by:	dlh12
	on:	January 28th, 2008

Conjecture

Suppose $G$ is a finite group, and $n$ is a positive integer dividing $|G|$ . Suppose that $G$ has exactly $n$ solutions to $x^{n} = 1$ . Does it follow that these solutions form a subgroup of $G$ ?

If these solutions form a subgroup, they form a characteristic (and therefore normal) subgroup of $G$ . This easily follows from the First Sylow Theorem if $n$ is the highest power of a prime $p$ dividing $|G|$ .

In a 1980 article, Feit commented that the case where $(n, \frac{|G|}{n}) = 1$ (i.e., $n$ 'exactly divides' $|G|$ ) had been reduced to considering $G$ simple. Thus it should be resolvable using the classification of finite simple groups.

It is known that if $n$ divides $|G|$ , the number of solutions of $x^{n} = 1$ in $G$ is a multiple of $n$ . A generalization of this theorem, replacing $x^{n} = 1$ by $x^{n} \in C$ for a conjugacy class $C$ of $G$ , can be found in Marshall Hall Jr.'s book.

Observation This conjecture implies the (known) theorem of Frobenius:

Theorem If $G$ is a finite transitive permutation group in which only the identity has more than one fixed point, then the derangements of $G$ , together with the identity, form a subgroup of $G$ .

To prove Frobenius' Theorem from this (say $G$ is such a permutation group on $k$ points), use the standard elementary counting arguments to show that the solutions to $x^{k} = 1$ are only the derangements and the identity and that there are exactly $k$ of these. This theorem of Frobenius has not yet been proven in general without the use of group characters.