1
$\begingroup$

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 $a$.

$\endgroup$
  • $\begingroup$ Such an $n$ is called an exponent for the group. $\endgroup$ – lhf Oct 11 '13 at 2:35
  • $\begingroup$ @lhf My bad. =) $\endgroup$ – Pedro Tamaroff Oct 12 '13 at 2:59
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 . $\endgroup$ – lhf Oct 11 '13 at 2:39
  • $\begingroup$ 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? $\endgroup$ – Stefan Smith Oct 12 '13 at 3:08
4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ So a to any power would be a subgroup within G where the result is the identity? $\endgroup$ – sprgrl11 Oct 10 '13 at 19:09
  • $\begingroup$ @sprgrl11 $a$ to a power is an element, not a subgroup. $\endgroup$ – user61527 Oct 10 '13 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.