Suppose $G$ is an abelian group and $k$ is a natural number.
Prove $H = \{ g \in G : g^k = 1 \}$ is a subgroup of G.
I know I need to show that $1_G \in H$, existence of inverse element in group, and closure, but how?
Suppose $G$ is an abelian group and $k$ is a natural number.
Prove $H = \{ g \in G : g^k = 1 \}$ is a subgroup of G.
I know I need to show that $1_G \in H$, existence of inverse element in group, and closure, but how?
$1 \in H$ since $1^k = 1$.
Suppose $g \in H$. That is, $g^k = 1$. Then $(g^{-1})^k = (g^{k})^{-1} = 1$. So $g^{-1} \in H$.
If $x,y \in H$, this means $x^k = 1$ and $y^k = 1$. Then $(xy)^k = x^ky^k = 1$ since $G$ is an abelian group. So $xy \in H$.
$H$ is a subgroup.
It's probably overkill but $x \mapsto x^k$ is a homomorphism $G\to G$ when $G$ is abelian. The set $H$ is the kernel of that homomorphism and hence is a subgroup.