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Let $(G,+)$ be a Group. I want to show that if $a+a=e$ (e being the neutral element) is true for all $a \in G$ then the Group G is abelian.

My first thought is that $a$ must be zero. Sadly I don't have any ideas how to start on this problem. Any tips are welcome.

I know that abelian is just another word for commutative. So when we have $a+b$ then it must be the same as $b+a$ or $a\cdot b =b \cdot a$.

Any tips?

Lisa
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1 Answers1

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Note that $(a+b) + (b+a) = a+ b+b+a = a+e+a = a+a = e$

(This is a hint, as you have asked for, but only one steps remains to show what you want).

Thomas
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  • I guess the last step is to show that (a+b)=(b+a)? I hope this isn't a very stupid question :( – Lisa Jan 19 '22 at 06:31
  • @Lisa The calculation shows that $b+a$ is the inverse of $a+b$. What do you know about the inverse? – Thomas Jan 19 '22 at 06:33
  • I know that for $+$ the Inverse of an element $a$ is $-a$. For $\cdot$ the inverse of $a$ is $a^{-1}$ – Lisa Jan 19 '22 at 06:34
  • You know a bit more - first, you make an assumption about the inverse in your question. Then you have just, correctly, written "the" inverse. (As opposed to "some" inverse). – Thomas Jan 19 '22 at 06:35
  • Oh, maybe since because we know that for $a$ the inverse is $-a$ and with how you've shown that $b+a$ is the inverse of $a+b$ in this case we can conclude $a+b -(b+a) =0$? – Lisa Jan 19 '22 at 06:40
  • By assumption, the inverse of any $a\in G$ is $a$. So the inverse of $a+b$ is $a+b$. My calculation shows that also $b+a$ is inverse to $a+b$. What do you now need to finish? Look at my previous comment. – Thomas Jan 19 '22 at 06:42
  • We assumed the inverse of $a$ being $a$ and showed that this isn't the case and therefore we conclude that $G$ must be abelian? – Lisa Jan 19 '22 at 06:46
  • Let us continue this discussion in chat. – Thomas Jan 19 '22 at 06:47