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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....

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  • $\begingroup$ Welcome to Math.SE. Please use LaTeX (MathJax) to write math. $\endgroup$
    – MathMajor
    Dec 17, 2014 at 3:17
  • $\begingroup$ 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. $\endgroup$
    – rschwieb
    Dec 17, 2014 at 3:29
  • $\begingroup$ 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. $\endgroup$
    – Kaitlyn
    Dec 17, 2014 at 3:35

2 Answers 2

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

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

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