# Subgroup formed by elements of order dividing n

**Conjecture**

Suppose is a finite group, and is a positive integer dividing . Suppose that has exactly solutions to . Does it follow that these solutions form a subgroup of ?

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

In a 1980 article, Feit commented that the case where (i.e., 'exactly divides' ) had been reduced to considering simple. Thus it should be resolvable using the classification of finite simple groups.

It is known that if divides , the number of solutions of in is a multiple of . A generalization of this theorem, replacing by for a conjugacy class of , can be found in Marshall Hall Jr.'s book.

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

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

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

## Bibliography

Marshall Hall Jr., Theory of Groups, Macmillan (1959)

Walter Feit, On a Conjecture of Frobenius, Proceedings of the American Mathematical Society, Vol.7, No. 2 (Apr. 1956), 177-187.

* indicates original appearance(s) of problem.

## Solved... again

it can be proved by a isomorphism between the solutions of x^n=1 in G and the solutions of the same eq in C (complex numbers).