# Show that there exists a positive integer $k$ such that $g^k=1_G$ [duplicate]

This will be my first question at the stack exchange, so please bare with me.

I'm doing a course on Matrix Groups for my Honours year and have really been struggling.

How would I go about solving this?

Let $G$ be a finite group and $g \in G$ is such that $\{g,g^2,g^3,\ldots\}$ is finite. Show that there exists a positive integer $k$ such that $g^k=1_G$

Any help would be greatly appreciated. Thanks!

## marked as duplicate by JP McCarthy, Daniel W. Farlow, Chris Godsil, zz20s, John BApr 8 '16 at 12:43

• $G$ is finite so the elements of $\{g,g^2,\dots\}$ cannot all be distinct. So we must have $g^n=g^m$ for some $n<m$. Now multiply both sides by $g^{-1}$ $n$ times. – almagest Apr 8 '16 at 11:33
$G$ is a finite group, say with $n$ elements. Consider $\{g,g^2, \ldots, g^n\}$. This is $n$ items from $G$, so you either have something repeated or they are all distinct. If something is repeated, say, $g^k = g^l$ where $k< l$, then multiplying both sides by $g^{-k}$ gives $g^{l-k} = 1$. If they're all distinct, then one of the items must be $1$ (since you have n distinct items from a set of n items, i.e. you have all of them).