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I feel very silly for asking this question:

Let $G$ be an abelian group. Show that $H=\{a\in G:a*a=a\}$ is a subgroup of $G$.

I did not get this question right, but this is what I managed to observe:

Notice that$$a*a=a,$$$$a^{-1}*a*a=a^{-1}*a,$$$$e*a=e,$$$$a=e,$$where $a^{-1}$ is the inverse of $a$, and $e$ is the identity element. Therefore, it must be the case that $H=\{e\}$, which is a subgroup of $G$.

The professor marked that conclusion wrong, however, and corrected it by commenting that this only showed that $e\in H$.

I am a little confused: look at how $H$ is defined. I believe this is just not an intuitive exercise; I never used the fact that $H$ is abelian, so I clearly must be missing something.

wjmolina
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  • To me, it seems $H={e_G}$. The cancellation law tells me that. –  Feb 21 '12 at 23:55
  • Hence, $H$ is a subgroup of $G$ and not a subset as you'd write at the end of your proof. –  Feb 21 '12 at 23:56
  • @JosuéMolina There are no non-trivial idempotents in a group. –  Feb 22 '12 at 00:45

2 Answers2

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Your professor is wrong. You have showed that if $a$ is in $H$, then $a=e$. What you haven't (explicitly) showed is that $e$ is in $H$, but that is easy.

One way to make the question more interesting, and to make the assumption that $G$ is abelian relevant, would be to instead define $H$ as $\{ a \in G:a*a=e\}$.

Chris Eagle
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I assume $*$ is the group operation. You might want to say "let $a$ be in $H$" to begin, but requiring that is pretty pedantic. Also you're conclusion should say "subgroup" rather than "subset", but it is clear what you meant. You're argument is correct, $H=\{e\}$. I have no idea why the professor objected. Here and here are proofs as citations.

Alex Becker
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    Thank you for your input. In the test, I wrote that $H$ is a subgroup of $G$; I typo'd the last sentence here. Also, the professor wrote "$e\in H$ but $H\neq{e}$!" I believe that he must have thought that he asked us something else, as Gerry mentioned above. – wjmolina Feb 22 '12 at 00:18