Show that for G be a abelian group, $g \in G, g^m=1$, where $m=|G|$

Question

I am pretty sure there is a name for the following theorem, but unfortunately, I don't known it.

Theorem. Let $G$ be a abelian group with $m=|G|$, than for any $g \in G, g^m=1$.

Currently I don't have ideas how to approach a proof of this theorem, and will appreciate any help.

This is surely a duplicate... – Zev Chonoles Apr 08 '13 at 10:06 — Zev Chonoles, Apr 08 '13 at 10:06
Why do you require $G$ to be abelian? – wj32 Apr 08 '13 at 10:23 — wj32, Apr 08 '13 at 10:23

Andreas Caranti · Accepted Answer · 2013-04-08T10:45:31.773

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There's a well-known elementary version for the case of abelian groups, that does not appeal to Lagrange's theorem.

Consider $$ X = \prod_{x \in G} x. $$ Since $G$ is abelian, the order within the product is immaterial.

Since the map $$ G \to G, \qquad x \mapsto g x $$ is a bijection, we have $$ X = \prod_{x \in G} x = \prod_{x \in G} g x = g^{m} \prod_{x \in G} x = g^{m} X,\tag{X} $$ where we have used again the fact that $G$ is abelian to take the $g$'s out of the product.

From (X) you get $$ g^{m} = 1. $$

edited Apr 08 '13 at 10:45

answered Apr 08 '13 at 10:34

Andreas Caranti

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Define "well-known"... – user1729 Apr 08 '13 at 10:39
@user1729, for the sole purpose of this post, I will define as well-known something that has already been seen elsewhere on MSE, possibly more than once ;-) I have also taught it for many years, but that clearly does not count. – Andreas Caranti Apr 08 '13 at 10:42
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@user1729: It is the usual proof given for Euler's Theorem, and even Fermat's Theorem, in elementary number theory courses. – André Nicolas Apr 08 '13 at 10:58
This is very nice, clear and elementary. +1 – DonAntonio Apr 08 '13 at 11:14
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I wish I knew how to prove that for non-abelian groups without using Lagrange's theorem. – lhf Apr 08 '13 at 11:25

score 1 · Answer 2 · answered Apr 08 '13 at 10:07

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Hints:

Use Lagrange's theorem:

$$g\in G\;,\;\;|\langle g\rangle|=n:=ord(g)\implies n\mid |G|$$

And then $\,g^k=1\;\;\forall\,k=tn\;,\; t\in\Bbb Z\,$

So your claim isn't true unless you either require $\,m\le |G|\,$ or else $\,m=ord(g)\,$

answered Apr 08 '13 at 10:07

DonAntonio

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score 0 · Answer 3 · answered Apr 08 '13 at 10:31

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HINT: Think about what would happen if you had some $g$ with $g^m$ not equal to $1$

answered Apr 08 '13 at 10:31

user27182

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Show that for G be a abelian group, $g \in G, g^m=1$, where $m=|G|$

3 Answers3