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Let $G = \{e, a_1, a_2, \dots, a_n\}$ be a finite abelian group and let $S$ be the set of all the elements of $G$ which are not equal to their own inverse. The set $S$ can be divided up into pairs so that each element is paired off with its own inverse. Prove that $(a_1 a_2 \dots a_n)^2 = e$.

Every $x$ in $G$, outside $S$ abides by $x^2 = e.$

Let $a_i$ be in $S$. Since $G$ is abelian, we have

$a_ia_i^{-1} = a_i^{-1}a_i$

$e = a_i^{-1}a_i$

Since $a_i^{-1} = a_i$, $a_i^2 = e.$

Thus, $\dots a_i^2a_j^2\ldots a_n^2 = e.$

Does that make sense?

edit:

We need to prove $a_1a_2...a_na_1a_2...a_n = e$, right? Let $a_i$ be in $G \setminus S$. Then $(a_i)^2$ is $e$, is it? If we rearrange the elements on the left side of the equation above according to what belongs to $S$ and what belongs to $G \setminus S$ and take the product of all the $(a_i)^2$, then every element belonging to $G \setminus S$ will be cancelled in the right side of $a_1a_2...a_na_1a_2...a_n = e$. So, we only need to consider the elements in $S$. Suppose $a_j \in S$. Let $a_k$ be the inverse of $a_j$. We need to prove $(a_j)^2 \cdot (a_k)^2 = a_j \cdot a_j \cdot a_k \cdot a_k = e$. Since $S$ is abelian , $a_j \cdot a_j \cdot a_k \cdot a_k$ can be rearranged as $a_j \cdot a_k \cdot a_j \cdot a_k$ which is $e$.

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  • $\begingroup$ But outside of $S$, $a_i^2\neq e$. $\endgroup$ – Ofir Schnabel Feb 11 '15 at 11:56
  • $\begingroup$ Also, $e = a_i^{-1}a_i \to (a_i^{-1})^{-1} = a_i$, not $a_i^{-1} = a_i$, correct? $\endgroup$ – AbstraktAlgebra Feb 11 '15 at 11:58
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    $\begingroup$ Since this group is commutative you can change the order in $(a_1a_2...a_n)^2$ such that any element will next to its inverse. For elements which are not in $S$ you "do it twice". $\endgroup$ – Ofir Schnabel Feb 11 '15 at 12:00
  • $\begingroup$ Rewrite the element like this $a_1a_2...a_na_1a_2...a_n$. Then Let's say the first $m$ elements belong to $S$.So,this is rewritten as $a_1a_2...a_ma_1a_2...a_ma_{m+1}...a_na_{m+1}...a_n$. $\endgroup$ – Srinivas K Feb 11 '15 at 12:00
  • $\begingroup$ I was able to piece together the right proof. Thank you guys. $\endgroup$ – AbstraktAlgebra Feb 11 '15 at 15:59
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Try this simpler argument:

  • The map $\phi(x)=x^{-1}$ is a homomorphism and so $\phi(a_1 a_2 \cdots a_n) = \phi(a_1) \phi(a_2) \cdots \phi(a_n)$.

  • The map $\phi$ is a bijection and so $\phi(a_1) \phi(a_2) \cdots \phi(a_n)= a_1 a_2 \cdots a_n$.

  • Hence, $a_1 a_2 \cdots a_n$ is its own inverse, which implies that $(a_1 a_2 \cdots a_n)^2=e$.

Note that the first two parts rely on the group being abelian.

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