G is an abelian group. Prove $G^{(n)}$ is a subgroup of G 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?
 A: To show that $H$ is a subgroup of $G$, you only need to show that


*

*$1_G \in H$,

*If $g, h \in H$, then $gh \in H$, and

*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 $H = G^{(n)}$.


*

*$1_G \in G^{(n)}$ because $1_G^n = 1_G$.

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

*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.
A: $G^{(n)}$ is a subset of $G$, that is, its elements are elements of $G$ and so associativity is automatically satisfied. The identity $1_G$ is in $G^{(n)}$ because $1_G^n=1_G$, and if $g \in G^{(n)}$, that is, $g^n = 1_G$, then $(g^{-1})^n = (g^n)^{-1} = 1_G^{-1} = 1_G$, so that $g^{-1}$ is also in $G^{(n)}$. Finally, we need to show $G^{(n)}$ is closed under composition.  Now, if $g,h \in G^{(n)}$, then $(gh)^n = g^nh^n = 1_G$, where the first equality comes from the assumption that $G$ is abelian, and the second comes from the assumption $g,h \in G^{(n)}$. Thus, all requirements for $G^{(n)}$ to be a subgroup of $G$ are satisfied.
A: Another way to go:
$G$ is an abelian group. What can we say about the map $f: G \rightarrow G$ defined by $f(g)=g^n$? What would be the kernel of this map?
