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$G$ is abelian, $H = \{x \in G : x = x^{-1}\}$ is a subgroup?

I know to prove that a subset $H$ of group $G$ be a subgroup, one needs to (i) prove $\forall x,y \in H:x \circ y \in H$ and (ii) $\forall x \in H:x^{-1} \in H.$

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Careful: the empty set satisfies the properties you listed above, but is not a subgroup! –  Stahl Mar 4 '13 at 20:08
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marked as duplicate by Cameron Buie, Git Gud, azimut, Norbert, Michael Albanese Oct 19 '13 at 16:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

6 Answers

up vote 4 down vote accepted

Unless you know $H$ is a nonempty subset of $G$, we need also to show that the identity $e \in G$ is also in $H$:

(o) Clearly, the identity $e \in G$ is its own inverse, hence $e = e^{-1} \in H$.

$(ii)\;$ $\forall x \in H$, $x\in H \implies x = x^{-1} \in H$. Hence we have closure under inverses.

$(i)$ $x\circ y \in H$? $$\; x, y \in H, \implies x = x^{-1}, y = y^{-1}$$ $$ x \circ y = x^{-1}\circ y^{-1} = y^{-1}\circ x^{-1} \quad\quad\quad\quad\tag{G is abelian}$$ $$y^{-1}\circ x^{-1} = (x\circ y)^{-1} \implies x\circ y \in H$$

Hence we have closure under the group operation.

Therefore, $H \leq G$. That is, $H$ is a subgroup of $G$.


Added: Note that this problem is equivalent to the task of proving that if $G$ is an abelian group, and $H$ is a subset of $G$ such that $H = \{x \in G\mid x^2 = e\}$, then $H$ is a subgroup of $G$. Why? For $x \in G$ such that $x^2 = e$, note that $$x^2 = e \iff x^{-1}\circ x^2 = x^{-1} \circ e \iff x^{-1}\circ (x\circ x) = x^{-1}$$ $$\iff (x^{-1}\circ x) \circ x = x^{-1} \iff e\circ x = x^{-1} \iff x = x^{-1}$$

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Thanks, that was very well explained! Sorry, I'm trying to learn abstract algebra by myself and I don't have too much experience with proofs. My book has answers, but they are for a very small set of exercises. –  user8210 Mar 4 '13 at 20:54
    
Your welcome, and thank you. Yes, textbook solutions are usually given for more computational types of problems, seldom for proofs. It takes some time and effort to get comfortable with proofs. You might want to find a copy of the text by Velleman: How to Prove It: A Structured Approach. It's really worth having, studying, and even simply for use as a reference. –  amWhy Mar 4 '13 at 21:01
    
I hope I didn't confuse you with the "added" part of my answer: it is not at all needed for the proof at hand. It's just that one frequent question that student's encounter, (and a favorite test question), is to prove that if G is abelian, and H is a subset of G containing all elements g of G such that $g^2 = e$, then H is a subgroup of G. And your experience with this question will help you with a question like that! –  amWhy Mar 4 '13 at 21:22
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Consider the map $\varphi:G\to G$ given by $\varphi(x)=x^2$. Since $G$ is abelian, then $\varphi$ is a homomorphism, and it's readily seen that $H$ is precisely the kernel of $\varphi$. Thus, $H$ is a subgroup of $G$.

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This is the best answer. No calculations are needed. –  Martin Brandenburg Mar 4 '13 at 21:07
    
Actually, @Martin, I doubt this will be useful to the OP, at this stage of the game. At this stage of the game, I doubt that the asker has encountered "homomorphisms", let alone what the "kernel" of a homomorphism is! –  amWhy Mar 4 '13 at 21:26
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You know what you need to do, so proceed and do it. For example, to show that if $x\in H$ then $y=x^{-1}\in H$, all you have to do is note that it is given that $x=x^{-1}$, and thus $y=x$, and thus clearly in $H$.

Now, all that is left for you is to start by saying: suppose $x,y\in H$, then I need to show that $xy\in H$.

Now write down what you know: $x^{-1}=?$ and $y^{-1}=?$. And write what you want to show: $(xy)^{-1}=?$. Fill in the question marks, and complete the proof.

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Straightforward insertion helps:

(i) We always have $(xy)^{-1}=y^{-1}x^{-1}$.If we know $x=x^{-1}$ and $y=y^{-1}$, then this becomes $yx$. As $G$ is abelian, ideed $(xy)^{-1}=xy$, so $H$ is closed under multiplication.

(ii) If $x=x^{-1}$ then $(x^{-1})^{-1}=x^{-1}$, so also $x^{-1}\in H$.

(iii) Something is missing. There are subsets that are closed under multiplication and taking inverses, but are not subgroups!

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At first I thought (iii) might be a nitpick about the set being nonempty, but then that still doesn't explain the plural "subsets". Is it just a typo? –  rschwieb Mar 4 '13 at 20:13
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$x,y \in H \implies (xy)^{-1} = y^{-1} x^{-1} = yx = xy$

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It is clear that $e\in H$ as $e=e^{-1}$. So $H\neq $ the empty set.

Pick $x,y\in H$, then consider $xy$. Since $G$ is ablian and elements of $H$ are their own inverse we can see that $(xy)^{-1}=x^{-1}y^{-1}=xy$ so $xy\in H$. So $H$ is closed.

Then pick $x\in H$ since $x\in H$, $x=x^{-1}$ so $xx=x^2=e$ so inverses are in $H$

Therefore $H$ is nonempty, closed subset of $G$ with inverses, so $H$ is a subgroup of $G$.

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