If G is abelian and $n>1$ is an integer, then $A=\{a^n \mid a \in G\}$ is a subgroup of $G$.

Question

I need to proove:

If $G$ is an abelian group, and we have $n>1 \in \mathbb{Z}$, then $A_n = \{a^n \mid a \in G\} $ is a subgroup for $G$ for all $n$.

To me, it looks like I need to prove by induction, with a base case of $n=2$, but I am unsure of how to prove the statement regardless.

Subgroup proofs typically use closure and inverses, but I am really new to groups and I am unsure of how to proceed. Any help is appreciated!

Answer 1

Hints:

$$\begin{align*} &(1)\;\;1=1^n\\{}\\ &(2)\;\;\forall x,y\in G\;,\;\;(xy)^n=x^ny^n\end{align*}$$

Answer 2

To prove that a set is a subgroup, prove that:

1. The set is closed under the binary operation of $G$.

2. There is an identity element in the set.

3. There is an inverse for every element of the set in the set.

Note that associativity comes for free, since $G$ is a group.