Subgroup formed by elements of order dividing n

Importance: Medium ✭✭
Subject: Group Theory
Keywords: order, dividing
Recomm. for undergrads: no
Posted by: dlh12
on: January 28th, 2008

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.


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).


This conjecture has been proven.

Re: Solved

Could you add a reference please?


Iiyori, Nobuo; Yamaki, Hiroyoshi On a conjecture of Frobenius. Bull. Amer. Math. Soc. (N.S.) 25 (1991), no. 2, 413--416.

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