# If $G$ is abelian, prove that $H = \{g \in G \mid g^2 = e\}$ is a subgroup of $G$. [duplicate]

Let $G$ be an Abelian group. Prove that $H = \{g \in G \mid g^2 = e\}$ is a subgroup of $G$.

I know something similar to this has been asked, but I just want to check my understanding/reasoning:

We want to show: identity inverse closure we know: $G$ is abelian, therefore for all $a,b \in G$, $ab=ba \in G$. Is this reasoning correct: Because $G$ is abelian and $g$ is an element of $G$, $H$ must be closed under multiplication.

We are also given that $gg=e$, and since $H$ is closed under multiplication, $e \in H$.

We must now show that every element of $H$ is its own inverse. <---- This is where I get stuck....

• Welcome to Math.SE. Please use LaTeX (MathJax) to write math. Dec 17, 2014 at 3:17
• Please look around for duplicates next time. At least one or two of the linked duplicates would have shown up in the similar questions tool while you entered the question. Dec 17, 2014 at 3:29
• I have looked at the two duplicates, however they did not answer my question, and this was about verifying my reasoning and understanding, not just giving me an answer. Dec 17, 2014 at 3:35

If $h^2=e$ then $(h^{-1})^2=(h^2)^{-1}=e$ so $H$ is closed under inverse.

• How do we know this? Dec 17, 2014 at 3:17
• Also, is the rest of my reasoning correct? Dec 17, 2014 at 3:17
• You need a proof that $H$ is closed under operation. Your 'proof' has some fault. However, it is not hard to prove: since $(ab)^2=a^2b^2$. Dec 17, 2014 at 3:20

For closure.

Let $g_0,g_1\in H$, then $(g_0*g_1)^2=(g_0g_1)(g_0g_1)=(g_0*g_0)(g_1*g_1)= g_0^2g_1^2= g_0^2*g_1^2=e*e=e.$

This tells us that we have closure.

For inverse. Let $g\in H$, then $g*g=e$. Since $g\in H$, what can we say about its inverse?