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Product in finite abelian groups

Let $a\in A$ where $A$ is a finite abelian group.

How do I prove that $\Pi_{x\in A}ax=\Pi_{x\in A}x$?

Thoughts:
If every element is not its own inverse, we see that $\Pi_{x\in A}x=e$. If there are elements that are their own inverse, then it is the product of these elements.
How do I get the identity above?

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    Show that the sets $\{x\mid x\in A\}$ and $\{ax\mid x\in A\}$ are equal. – 2017-02-22
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    @TobiasKildetoft How does the equality follow then? – 2017-02-22
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    Jose, since those sets are equal, up to commutativity, the equality becomes $x_1x_2\cdots x_n = x_1x_2\cdots x_n$, where $A = \{x_1,\ldots, x_n\}$. – 2017-02-22

2 Answers 2

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The set $M:=\{ax\mid x\in A\}$ is equal to $A=\{x\mid x\in A\}$:

We have $M\subset A$. Also, for $y\in A$, we have $y=a(a^{-1}y)\in M$.

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Here's another way you could do it. You can group together all the factors of $a$ in the product $\prod_{x\in A}ax$ to get $a^{|A|}\prod_{x\in A}x$. Now what can you say about $a^{|A|}$?

The answer is hidden below.

We always have $a^{|A|}=1$, since the order of $a$ divides $|A|$.