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


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.

  • $\begingroup$ 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

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


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