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Let $G$ be an abelian group. Prove that $H = \{x \in G \mid x=x^{-1}\}$ is a subgroup. ($x^{-1}$ is $x$'s inverse)

to check if the identity is an element of the subgroup: $ex=x$; $(e^{-1})ex=(e^{-1})x$; $ex=(e^{-1})x$; $e=e^{-1}$.

To check closure: this is where I'm not sure I know what I am doing. To check closure, I need to show that... $(ab)x=(ab)^{-1}(x^{-1})$? Let $a,b\in H$, $(ab)(ab)= e$ if $(ab)=(ab)^{-1}$. From that we can get $(aa)(bb)=e$ because we have commutativity and associativity from our group. We know $a=a^{-1}$, $b=b^{-1}$, so $(aa^{-1})(bb^{-1})=e$, $ee=e$; $e=e$. So... $(ab)=ab^{-1}$. If I plug that into $(ab)x=(ab)^{-1}(x^{-1})$, and say that $(ab)x=(ab)(x^{-1})$. If I left multiply with a $(ab)^{-1}$, I get $x=x^{-1}$. Which is what I started with so... I'm done?

To check inverses: the group includes inverses by definition of the group.

H is closed, includes the identity, and inverses, so it is a subgroup of $G$.

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For identity, note that by definition $e*e=e$, so it follows directly that $e=e^{-1}$. –  Dario Sep 27 '12 at 14:13
    
Why the $x$'s? In the definition of $H$ the letter $x$ is just a label for an element of $G$ satisfying some property... –  fretty Sep 30 '12 at 11:26

2 Answers 2

If you know about homomorphisms, then $H=\ker \phi$, where $\phi(x)=x^2$, which is a homomorphism because $G$ is abelian. (Actually, iff $\phi$ is a homomorphism $G$ is abelian.)

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To check closure you need to show that given any $x,y\in H$ that $xy \in H$. For $xy$ to be in $H$, it must be such that

$$(xy)^{-1} = (xy).$$

Now $(xy)^{-1} = y^{-1}x^{-1}=yx = xy$ by definition of $x,y \in H$ and $G$ being abelian. This shows $H$ is closed under the group operation.

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