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

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

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. ...

2. $\displaystyle <a>$ is a cyclic subgroup of $\displaystyle G$; let's say it has order $\displaystyle m$. By Lagrange, $\displaystyle m|n$, which means $\displaystyle bm=\sum_{i=1}^bm=n$ for some integer $\displaystyle b$.

So then $\displaystyle a^n=a^{m+m+...+m}=a^m\cdot a^m\cdot...\cdot a^m=e\cdot e\cdot ... \cdot e = e$.

3. ## 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.

4. 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: $\displaystyle bm=m+m+...+m$.