# Subgroup formed by elements of order dividing n

 Importance: Medium ✭✭
 Author(s): Frobenius, Ferdinand G.
 Subject: Group Theory
 Keywords: order, dividing
 Recomm. for undergrads: no
 Posted by: dlh12 on: January 28th, 2008

\begin{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 $G$?

\end{conjecture}

% You may use many features of TeX, such as % arbitrary math (between $...$ and $$...$$) % \begin{theorem}...\end{theorem} environment, also works for question, problem, conjecture, ... % % Our special features: % Links to wikipedia: \Def {mathematics} or \Def[coloring]{Graph_coloring} % General web links: \href [The On-Line Encyclopedia of Integer Sequences]{http://www.research.att.com/~njas/sequences/}

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.

\begin{observation} This conjecture implies the (known) theorem of Frobenius:

\begin{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 $G$. \end{theorem} \end{observation}

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.

## Bibliography

% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries) 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).

### Solved

This conjecture has been proven.