Prove: If $G = \{a_1, a_2, ..., a_n\}$ is a finite abelian group, then $(a_1 a_2...a_n)^2 = e$, where $e$ is neutral in $G$.
I don't even now how to begin this.
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1Hint: every element has an inverse distinct from itself unless it has order $1$ or $2$. – lulu Nov 19 '17 at 14:41
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In the product $(a_1a_2...a_n)^2$, try to pair each element $a_i$ to its inverse. – edm Nov 19 '17 at 14:42
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@lulu Even simpler: Each element has a unique inverse, and no other element has the same inverse. – Arthur Nov 19 '17 at 14:44
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@Arthur Ha! good point. – lulu Nov 19 '17 at 14:47
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As $G$ is a group, every element $a_k$ has its inverse $a_k^{-1}$, which is one of the members of $G$, say $a_{k(n)}$ where $k:\{1,2,\cdots,n\}\rightarrow\{1,2,\cdots,n\}$ is a bijection (because every element has a unique inverse and the inverse of the inverse is the element itself), hence a permutation. Hence the set $\{a_1^{-1},a_2^{-1},\cdots,a_n^{-1}\}=\{a_1,a_2,\cdots,a_n\}$. Therefore $$(a_1a_2\cdots a_n)^2=(a_1a_2\cdots a_n)(a_1^{-1}a_2^{-1}\cdots a_n^{-1})=(a_1a_2\cdots a_n)(a_1a_2\cdots a_n)^{-1}=e$$
QED
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