Show there exists a fixed positive integer $n$ such that $a^n = e$ for all $a\in G$.

Question

Let $G$ be a finite group. Show there exists a fixed positive integer $n$ such that $a^n = e$ for all $a\in G$.

We know: $n$ is independent of $a$.

Such an $n$ is called an exponent for the group. – lhf Oct 11 '13 at 02:35 — lhf, Oct 11 '13 at 02:35

score 5 · Answer 1 · answered Oct 10 '13 at 19:07

5

Hint: For any element $g \in G$, what can you say about $g^{|G|}$?

answered Oct 10 '13 at 19:07

So a to any power would be a subgroup within G where the result is the identity? – sprgrl11 Oct 10 '13 at 19:09
@sprgrl11 $a$ to a power is an element, not a subgroup. – Oct 10 '13 at 19:55

score 2 · Accepted Answer · answered Oct 11 '13 at 02:25

2

Here is an outline for an elementary argument that avoids Lagrange's theorem:

Of course, that $G$ is finite is essential here.

answered Oct 11 '13 at 02:25

lhf

Lagrange's theorem implies that you can take $n=|G|$ but I wish I knew a proof that does not depend on Lagrange's theorem . – lhf Oct 11 '13 at 02:39
I deleted my answer because it was so similar to yours. I did show show why $n_a$ existed in case the OP didn't know, but I deleted it anyway. The proof of Lagrange's theorem doesn't seem so bad. Why do you want to avoid it? – Stefan Smith Oct 12 '13 at 03:08

2 Answers2