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

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 2:35
• @lhf My bad. =) – Pedro Tamaroff Oct 12 '13 at 2:59

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

• Given $a\in G$, there is $n_a\in\mathbb N$ such that $a^{n_a}=e$.

• $a^{kn_a}=e$ for all $k\in\mathbb N$.

• Consider $n=\operatorname{lcm}_{a\in G} n_a$.

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

• 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 2:39
• 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 3:08

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

• 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. – user61527 Oct 10 '13 at 19:55