7
$\begingroup$

I need to show that elements of finite order in an abelian group form a subgroup of that group. Where do i start ?

$\endgroup$
11
$\begingroup$

Let $U$ be the set of all elements of finite order.

  • The neutral element $1$ has order 1. So $1\in U$.
  • Let $g,h\in U$. Then there are positive integers $n,m \geq 1$ with $g^n = 1$ and $h^m = 1$. So $$(gh)^{nm} \overset{gh = hg}{=} g^{nm} h^{nm} = (g^n)^m (h^m)^n = 1^m 1^n = 1.$$ Hence $\operatorname{ord}(gh) \le nm$ and therefore $gh \in U$.
  • Let $g\in U$. Then there is a positive integer $n$ with $g^n = 1$. Multiplication with $(g^{-1})^n$ yields $$\underbrace{(g^{-1})^n g^n}_{=1} = (g^{-1})^n.$$ So $\operatorname{ord}(g^{-1})\le n$ and $g^{-1}\in U$.

Therefore, $U$ is a subgroup.

$\endgroup$
  • $\begingroup$ where did we need the condition that the group $G$ should be abelian ? $\endgroup$ – Aman Mittal Sep 4 '13 at 20:10
  • 1
    $\begingroup$ For $(gh)^{nm} = g^{nm} h^{nm}$. Look at the explanation $gh = hg$ over the equation sign. $\endgroup$ – azimut Sep 4 '13 at 20:15
  • $\begingroup$ yes, but even without it the proof would hold good. as the LHS will still be equal to 1 $\endgroup$ – Aman Mittal Sep 4 '13 at 20:16
  • $\begingroup$ No, that is not true. For example, without $gh = hg$ you only get $(gh)^2 = ghgh$. $\endgroup$ – azimut Sep 4 '13 at 20:17
  • $\begingroup$ Oh !! Got it . Thanks !! $\endgroup$ – Aman Mittal Sep 4 '13 at 20:19
4
$\begingroup$

The start: let $a^m=b^n=1$. Then $(ab)^{mn}=1$.

$\endgroup$

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.