Let $G = \{g_1, g_2, \dots, g_n\}$ be a finite abelian group, prove that for any $x ∈ G$, the product $$xg_1 \cdot xg_2 \cdot \cdot \cdot xg_n = g_1 \cdot g_2 \cdot \cdot \cdot g_n.$$

I can easily see why $g_1 \cdot g_2 \cdot \cdot \cdot g_n = e$, this follows from the group being abelian and having an inverse for every element. But I can't see why $x^n = e$. We haven't looked at many propositions yet. What I have so far is that if $\operatorname{ord}(x) = k$, that I need to prove $n \bmod k = 0$, from which will nicely follow that $\operatorname{ord}(x)$ divides the group order.

Edit: changed some notation mistakes in the product.

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    $\begingroup$ In the title the group contains $n+1$ elements! And $g_1. g_2...\ . g_n=e$ does not always hold because elements may be self-inverse (consider for example $C_4$). $\endgroup$ – Peter Feb 15 '16 at 22:42

The elements $xg_1,\ldots ,xg_n$ are again $n$ different elements of $G$, so that they must coincide with elements $g_1,\ldots,g_n$; hence their product also coincides. Indeed, if $xg_i=xg_j$, then multiplying by $x^{-1}$ gives $g_i=g_j$, so that $$ G=\{g_1,\ldots ,g_n\}=\{xg_1,xg_2,\ldots , xg_n\}. $$ On the other hand, this is equivalent to $x^n=e$, which we have proved now for all $x\in G$. Note that the product of all elements need not be $e$, as the example of $C_2=\{e,g\}$ with $g^2=e$ shows. It equals $e$ if $G$ is abelian and has odd order, see here.

More generally, we have $x^{|G|}=1$ for all $x$ in a finite group $G$.

  • $\begingroup$ $\prod_{g \in G} xg= x^n \prod_{g \in G} g$. Because $\prod_{g \in G} g= e$ (this is not given) I must prove $x^n = e$. But I dont see how exactly from your explanation. $\endgroup$ – user260710 Feb 14 '16 at 17:10
  • $\begingroup$ You wrote: "prove that for any $x ∈ G$, the product $xg_1 \cdot xg_2 \cdot \cdot \cdot xg_n = g_1 \cdot g_2 \cdot \cdot \cdot g_n$." This follows because both sides have the same elements. Also $\prod_gg \neq e$ in general, consider the group $C_2=\{e,g\}$ with $g^2=e$. $\endgroup$ – Dietrich Burde Feb 14 '16 at 17:10
  • $\begingroup$ But is there a way to prove that $\prod_{g in G} xg = \prod_{g in G} g$ by proving that $n$ mod $k$ = 0, if ord$(x) = k$ and $|G| = n$. Because they want us to also show as a consequence that $k$ divides the group order. $\endgroup$ – user260710 Feb 14 '16 at 17:30
  • $\begingroup$ You should do it the other way. First the easiest step - the two products are equal. Then $x^n=e$. Then $k$ divides the group order. $\endgroup$ – Dietrich Burde Feb 14 '16 at 17:33

Actually, what you said about $g_1g_2\cdots g_n=e$ is only true in general if the group has odd order.

Have you covered Lagrange's Theorem in class? Because it's really just a consequence of that.

The other way to see this is that if you multiply every element of $G$ by $x\in G$, you're really just permuting the group. So the set $$\{xg_1,xg_2,\dots,xg_n\}$$ is really the same set as $$\{g_1,g_2,\dots,g_n\}$$

And hence their products are the same. Of course, you have to prove that these two sets are the same.


$xg_1, xg_2, \cdots xg_n$ are $n$ elements in $G$.

If $xg_i=xg_j$ then $g_i=g_j$, so the elements $xg_i$ are all distinct, i.e. they are all the elements in $G$ since they are $n$ elements.

In other words you are just multiplying the same set (namely the whole $G$) in a different order, but a group is abelian!

(I'm considering $e=g_1$, we don't need to distinguish it! And $|G|=n$).


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