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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!

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ $G$ 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. $\endgroup$ – almagest Apr 8 '16 at 11:33
  • $\begingroup$ en.wikipedia.org/wiki/Pigeonhole_principle $\endgroup$ – JP McCarthy Apr 8 '16 at 11:33
  • $\begingroup$ I don't think its necessarily a duplicate. That does solve the problem, but you don't need that stronger statement. $\endgroup$ – Batman Apr 8 '16 at 11:52
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$G$ 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).

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  • $\begingroup$ This is very clear. Thank you, Batman! $\endgroup$ – Iapetus Apr 8 '16 at 12:55
  • $\begingroup$ There is a second part to the question that I was struggling with. Does it just require relying on the definition? $\endgroup$ – Iapetus Apr 8 '16 at 13:29
  • $\begingroup$ Put that as a different question. $\endgroup$ – Batman Apr 8 '16 at 13:39

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