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I need to proof for a finite abelian group $G$ that for all $x\in G$ we have $\prod_{g\in G}xg=\prod_{g\in G}g$.

I figured that using the commutative property $\prod_{g\in G}xg=x^n\prod_{g\in G}g$. Which would leave us to proof $x^n=e$, where $e$ is the identity element.

Hopefully I'm just missing something obvious here, because I feel like smashing my head against the table now.

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    Do you have Lagrange's theorem? – 2017-02-21
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    We have not had it yet - so I assume we're not supposed to use it for this problem. – 2017-02-21
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    well either way, x$^{n}$ should give you 'e' because n is the order of the group. – 2017-02-21
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    The best way to show that, and the only one I know, is by Lagrange's theorem. – 2017-02-21

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The map $g\mapsto xg$ is a permutation of $G$. hence $\prod xg$ is the same as $\prod g$, just in a different order. As $G$ is abelian, different order doesn't matter.