[Math] G is an abelian group. Prove $G^{(n)}$ is a subgroup of G

abelian-groupsgroup-theory

Let G be an abelian group. Prove that

$$G^{(n)} = \{g \in G | g^n = 1_G \}$$

is a subgroup of G.

How do I go about doing this?

I understand that $G^{(n)}$ is basically the set of all elements whose order divides n. So would I simply need to show that the group axioms hold for $G^{(n)}$ and this would show that it is a subgroup for G?

I already have the identity element at n = 0. Also, the inverse would be any $n < 0$. So now I just have to prove associativity right? How do I go about doing this?

Is this when $g^{(n)^{(m)}} = g^{(m)^{(n)}}$? How would I prove that?

Best Answer

To show that $H$ is a subgroup of $G$, you only need to show that

  1. $1_G \in H$,
  2. If $g, h \in H$, then $gh \in H$, and
  3. If $g \in H$, then $g^{-1} \in H$.

Some people prefer to combine 2 and 3 together and just prove

  • If $g, h \in H$, then $gh^{-1} \in H$.

You can use whichever way you find easier. I'll prove all 3 things for the case $H = G^{(n)}$.

  1. $1_G \in G^{(n)}$ because $1_G^n = 1_G$.
  2. If $g, h \in G^{(n)}$, then $(gh)^n = g^n h^n = 1_G 1_G = 1_G$, so $gh \in G^{(n)}$.
  3. If $g \in G^{(n)}$, then $(g^{-1})^n = (g^n)^{-1} = 1_G^{-1} = 1_G$, so $g^{-1} \in G^{(n)}$.

Therefore, $G^{(n)}$ is a subgroup of $G$. Note that we used the fact that $G$ is abelian in the proof of 2.

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