[SOLVED] problem 12 - a^n=e for all a in G

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• Apr 27th 2009, 07:02 PM
yvonnehr
[SOLVED] problem 12 - a^n=e for all a in G
Please help.
12) Let G be a group of finite order n. Prove that a^n = e for all a in G.

This is what I have so far. I believe I should use Lagrange's Theorem but I get stuck.

Proof. Let G be a group of finite order n and a be in G. Since a is in G, then a^-1 is also in G. ...
• Apr 27th 2009, 07:50 PM
hatsoff
$$ is a cyclic subgroup of $G$; let's say it has order $m$. By Lagrange, $m|n$, which means $bm=\sum_{i=1}^bm=n$ for some integer $b$.

So then $a^n=a^{m+m+...+m}=a^m\cdot a^m\cdot...\cdot a^m=e\cdot e\cdot ... \cdot e = e$.
• Apr 27th 2009, 08:03 PM
yvonnehr
summation notation?
I haven't used the summation notation in class. Does the summation in this case refer to the definition of cosets? I'm not sure what it means.
• Apr 27th 2009, 08:08 PM
hatsoff
Quote:

Originally Posted by yvonnehr
I haven't used the summation notation in class. Does the summation in this case refer to the definition of cosets? I'm not sure what it means.

Okay, just think of it this way, then: $bm=m+m+...+m$.