# If $G$ is abelian, then the set of all $g \in G$ such that $g = g^{-1}$ is a subgroup of $G$

Prove that if $G$ is abelian then the set $H$ of all elements of $G$ that are their own inverses is a subgroup of $G$.

Naturally in an abelian group, $ab = ba$ for $a, b \in G$, however I'm not sure how to show the set elements that are their own inverses is a subgroup of $G$ using arbitrary elements.

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You must show the following: Let $G$ an abelian group, that means $ab = ba$ for all (not some $a,b \in G$). Then $H = \{g \in G\mid g = g^{-1}\}$ is a subgroup of $G$ (that means it is not empty and $g,h \in H$ implies $gh^{-1}\in H$. – martini May 18 '12 at 23:55

A different way to phrase the same argument everyone gave:

The map $a\in G\mapsto a^2\in G$ is a group homomorphism and your subset $H$ is its kernel: it is therefore a subgroup of $G$.

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+1 Conceptualize before calculating! – Bill Dubuque May 19 '12 at 1:17

$\newcommand{\N}{\Bbb N}$ Let $G$ an abelian group, let $e$ denote its identity element. For each $m\in\N$ define $$G(m):=\{g\in G: g^m=e\}.$$ $G(m)$ is a subgroup of $G$. Indeed, you can see that $e\in G(m)$. If $g,h\in G$, since $G$ is abelian we have $$(gh^{-1})^m=g^m(h^{-1})^m=e(h^m)^{-1}=e^{-1}=e.$$ Therefore $G(m)\leq G$ as claimed.

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@Goober, if $G$ is finite, you might want to know that $G(m) = G$ if and only if $gcd(m,|G|)=1$. So in your original case, iff the abelian group $G$ is of odd order. – Nicky Hekster May 20 '12 at 20:14

You'll need to show only closure under multiplication (that is, that $ab\in H$ for all $a,b\in H$), since the identity is trivially its own inverse, so is in $H$, and since every element of $H$ is its own inverse, you don't need to check inverses, either. The fact that $G$ (so also $H$) is abelian makes checking closure fairly trivial.

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Generally the one-step subgroup test is faster but in this case you can just check the group axioms: the only non-trivial one is closure. If $a^2=b^2=e$, can you see that $ab$ is its own inverse, given the group is Abelian?

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Write H = {x in G: x*x = e}, where e is the identity element.

1. Show e is in H: since e*e = e, e is in H and so H is nonempty.

2. Consider x in H. Then x*x=e. Since the inverse of x is x and x is in H, H is closed under inverses.

3. Now consider x,y in H. Then x*x = e and y*y = e. So x*x*y*y = e. Since G is Abelian, so is H since H contains elements from G. So x*x*y*y = x*y*x*y = (x*y)*(x*y) = e, x*y is in H. So H is closed under multiplication.

Thus H is a subgroup of G.

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I like Mariano's answer. Question for anyone: consider the map f: C[x_1,...,x_n] --> C[x_1,...,x_n] where f(x_i)=x_i^2. Then ker f = <x_1^2,....,x_n^2>. Is there a geometric notion behind this kernel? – math-visitor May 19 '12 at 1:19
Your map is actually injective. – Mariano Suárez-Alvarez May 19 '12 at 1:42