Product of all elements in finite abelian group equals its own inverse

Question

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 $e$.

Also, $e = a_i^{-1}a_i \to (a_i^{-1})^{-1} = a_i$, not $a_i^{-1} = a_i$, correct? — AbstraktAlgebra, Feb 11 '15 at 11:58
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". — Ofir Schnabel, Feb 11 '15 at 12:00
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 $a_1a_2...a_ma_1a_2...a_ma_{m+1}...a_na_{m+1}...a_n$. — Srinivas K, Feb 11 '15 at 12:00
I was able to piece together the right proof. Thank you guys. — AbstraktAlgebra, Feb 11 '15 at 15:59
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

lhf · Accepted Answer · 2015-02-11T15:20:40.933

12

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.

edited Feb 11 '15 at 15:20

answered Feb 11 '15 at 15:15

lhf

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Product of all elements in finite abelian group equals its own inverse

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