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I have an Abelian Group G, with gamma an arbitrary element of G. I am asked to prove the following: $$\prod_{g\in G}g=\prod_{g\in G}\gamma g.$$

I have absolutely no idea how to prove this or why it is true. Could someone give me a hint (not the full answer preferably. just a small hint)?

Community
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user56834
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3 Answers3

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The map $G\rightarrow G: g\mapsto \gamma g$ is a permutation of $G$.

studiosus
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First off, your group has to be finite, say $|G|=n$, in order for the product to make sense. Second, there are two approaches to this.

  1. Write out the product as $\gamma g_1\gamma g_2\cdots \gamma g_n$, gather all the $\gamma$ into one $\gamma^n$ term. What is $\gamma^n$?
  2. The elements $\gamma g_i$ are exactly the same as the elements $g_j$, only in a different order. Therefore you get the same result when you multiply them all together.
Arthur
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I assume that $G$ is finite. In that case

$$\prod_{g \in G} \gamma g = \gamma^{|G|} \prod_{g\in G} g = \prod_{g \in G} g,$$

where the second equality uses the abelian property and the third uses Lagrange's Theorem.

Zestylemonzi
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