Let $G$ be an abelian group with identity $e$. Let $H=\left\{g\in G\mid g^2=e\right\}$. Show that $H\le G$.

Question

Let $G$ be an abelian group with identity $e$. Let $ H = \left \{ g \in G \mid g^2 =e \right \}$. Show that this is a subgroup of $G$.

My answer:

We have four principles to be valid in order for $H$ be a subgroup of $G$:

(i) is closed

Obvious point, easy to prove

(ii) is associative

This is more trickier and I might need your help here. I thought of the following solution for $ h \in G $ $ g*(g*h)=(g*g)*h$ / multiply both sides by $g$ $g^2 *gh=g^2 *hg $ so $ e*gh=e*hg $ since it's abelian $gh=hg$ Is it a right way of thinking about it?

(iii) Identity, I thought of $e$ but it doesn't work $ g*e \neq g$ So what might it be?

(iv) Inverse of $g$ would be $g$

I formulate questions related mainly to point (iii).

Thank you

Answer 1

You only have to check two properties, given $H$ is not empty (it contains $e$):

(i) If $h, h'\in H$, then $hh'\in H$, i.e. $$\forall h\:\forall h'\enspace(h^2=e)\wedge(h'^2=e)\implies (hh')^2=e,$$ which results from $G$ being abelian.

(ii) If $h\in H$, then $h^{-1}\in H$, i.e. $$\forall h\enspace h^2=e\implies (h^{-1})^2=e,$$ which results from $hh^{-1}=e$, so $$e=e^2=(hh^{-1})^2=h^2(h^{-1})^2=e(h^{-1})^2=(h^{-1})^2.$$

Answer 2

One way to prove $H\le G$ is the one-step subgroup test.

Since $e=ee=e^2$, we have $e\in H$, so $H$ is nonempty.

Let $g, h\in H$. We show that $gh^{-1}\in H$. Indeed,

$$\begin{align} (gh^{-1})^2&=gh^{-1}gh^{-1}\\ &=g^2(h^{-1})^2\\ &=e, \end{align}$$ since $G$ is abelian.

Hence $H\le G$.