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Let $G$ be a finite abelian group. Prove that $(a_1 a_2 \cdots a_n)^2=e$.

My proof: $$\forall a \in \{a_1,a_2,\cdots,a_n\} \exists ! a^{-1} \in \{a_1,a_2,\cdots,a_n\}:a^{-1}a=aa^{-1}=e$$ hence, $$(a_1a_2\cdots a_n)^2=(a_1a_2\cdots a_n)(a_1a_2\cdots a_n)=(a_1a_2\cdots a_n)(a^{-1}_1a^{-1}_2\cdots a^{-1}_n)=(a_1a^{-1}_1)(a_2a^{-1}_2)\cdots (a_na^{-1}_n)=ee\cdots e=e$$

I tried showing that all $n-1$ elements in $G\setminus \{e\}$ contain a unique ($\leftarrow$ is even this true?) inverse in $G$. Is this proof valid? Should I show/explain more between the steps $(a_1a_2\cdots a_n)(a_1a_2\cdots a_n)=(a_1a_2\cdots a_n)(a^{-1}_1a^{-1}_2\cdots a^{-1}_n)$?

Thank you.

Patrick Shambayati
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  • If ${a_1 \cdots a_n}$ is a group, then every element has a unique inverse. – lisyarus Aug 20 '14 at 19:54
  • Yes, inverses are unique in any group. You don't need to restrict that statement to $G \setminus {e}$. Also, you should mention in the problem statement that $|G| = n$, i.e. this only works if you take the product of all of the elements. –  Aug 20 '14 at 19:54
  • In fact $(a_1a_2\cdots a_n) = \prod_{a_k \in S} a_k$ where $S = \lbrace a\in G: a^2 = e\rbrace$ – Darth Geek Aug 20 '14 at 19:56
  • In fact if $G$ has a unique element $x$ of order $2$, the product $\prod_{g\in G} g = x$, otherwise $\prod_{g\in G} g = 1$. – Myself Aug 20 '14 at 20:08
  • I'm not going to nominate for closing, but this is a duplicate of http://math.stackexchange.com/questions/768246/if-g-is-a-finite-abelian-group-and-a-1-a-n-are-all-its-elements-show-that – Chris Culter Jul 25 '15 at 08:35
  • Or better yet, http://math.stackexchange.com/questions/53026/finite-abelian-group – Chris Culter Jul 25 '15 at 08:47

2 Answers2

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Firstly, the inverse of any element is unique. Try and prove this by contradiction.

Then, if you want to be more rigorous, since every element has an inverse, there is a bijective map from the group to itself mapping every element to its inverse. You could identify elements and their inverses through this map in your notation, but your proof is correct.

forallepsilon
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Inverses being merely unique is not enough. If every element had only one inverse and it was $a_1$ in every case, your proof would not work. You need to come up with a good reason why that can't happen. Not difficult, but your proof is not complete without it.

About uniqueness of inverses: first of all, your restriction to $G\setminus\{e\}$ is unnecessary, since $e$ too has a unique inverse. Secondly, you do not need contradiction: just suppose that $y$ and $z$ are both inverses of $a$ and consider the product $yaz$.

Ben Millwood
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