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To put this in context, this is my first week abstract algebra.

Let $(G,\cdot,e)$ a finite abelian group. And $a\in G$. Show that:

$$\prod_{g\in G}g=\prod_{g\in G}(a\cdot g)$$

Here are my thoughts: Let $G=\{g_1,...,g_n\}$. Then we obtain:

$$\prod_{g\in G}(a\cdot g)=ag_1\cdot...\cdot ag_n=a^n g_1\cdot...\cdot g_n=a^n\prod_{g\in G}g$$

I don't know how to show that $a^n$ must be $e$.

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@JavaMan We haven't covered subgroups or Langrange's Theorem yet. –  Kasper Feb 18 '13 at 19:17

3 Answers 3

up vote 3 down vote accepted

As JavaMan has 'hinted' in the comments, in any finite group $G$ with $|G| = n$, it is the case that $g^n = e$ for every $g\in G$. But since you're only in your first week of abstract algebra, I'm not sure whether you have the tools (e.g. Lagrange's theorem) to prove that.

However, what you do know are the group axioms. From these, you can prove that for any $a\in G$, the function $\phi_a: G\rightarrow G$ given by $\phi_a(g) = a\cdot g$ for all $g\in G$ is a bijection, and this allows you to prove the result.

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Is this correct: $\prod_{g\in G}(a\cdot g)=\prod_{g\in G}\phi_a(g)=\prod_{\phi_a(g)\in G}\phi_a(g)$. This last equality is justified by the fact that G is abelian, so it doensn't matter in which order the group members are multiplied. –  Kasper Feb 18 '13 at 19:30
    
Yes, it's correct (though you should give the justifications of course and define $\phi_a$), but then you'd want a final $= \prod_{g\in G} g$ to finish showing what you wanted to show. –  Tara B Feb 18 '13 at 19:32
    
Okay, I understand, thanks for your answer –  Kasper Feb 18 '13 at 19:34

Multiplication by $a\in G$ is an action on $G$, i.e. it is just a permutation of the elements in $G$. Since the group is abelian, permutation of the elements do not change the product hence the product remains the same.

Edit: As Javaman pointed out, using Lagrange's theorem seems simpler to follow from your work!

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It would be, except that they won't have studied cosets etc yet in the first week of an abstract algebra course! Anyway, this approach is simpler because it deduces the result very straightforwardly from just the group axioms. –  Tara B Feb 18 '13 at 19:26

Suppose that $G$ has $n$ elements. Consider the subset of $G$ of elements of the form $a^{-1}g$. If $a^{-1}g = a^{-1}g'$ then $g = g'$, so this subset has as many elements as all of $G$. Since $G$ is finite, this subset must be all of $G$. Thus you can write

$$\prod_{g \in G} a\cdot g = \prod_{a^{-1}g \in G} (a \cdot a^{-1} g)$$

Do you know how to finish the proof from here?

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Thanks for this answer, helped me to understand it better. –  Kasper Feb 18 '13 at 19:33

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