prove that (a_1a_2...a_n)^2=e for every finite abelian group.

**VonNemo19** · October 13th 2011, 12:19 PM

OK. I'm trying, but I keep running into problems. I've recently proven that

$ab=e\Rightarrow{ba}=e$ and $a=a^{-1}\Leftrightarrow{a}^2=e$ . What I'm asked to prove seems to be a generalization of of these two statements, but I can't quite see it. Is there any way someone could point me in the right direction here?

**Drexel28** · October 13th 2011, 12:48 PM

Originally Posted by VonNemo19

OK. I'm trying, but I keep running into problems. I've recently proven that

$ab=e\Rightarrow{ba}=e$ and $a=a^{-1}\Leftrightarrow{a}^2=e$ . What I'm asked to prove seems to be a generalization of of these two statements, but I can't quite see it. Is there any way someone could point me in the right direction here?

This isn't true $1+0=1$ in $\mathbb{Z}_2$ . The correct statement is that the product over all the elements in an abelian group of odd order is the identity. Hint: pair things with their inverse (how do you know the inverse isn't equal to the original element)?

EDIT: Misread problem, my apologies.

**Deveno** · October 13th 2011, 12:50 PM

what you are actually being asked to prove is that, if you multiply every group element (in an abelian group, the abelian-ness is important, here) together, and square the result, you get the identity.

there are, actually, LOTS of ways to prove this.

here is one approach:

prove that $(a_1a_2\dots a_n)^{-1} = (a_1)^{-1}(a_2)^{-1}\dots(a_n)^{-1}$ (this uses the fact that G is abelian).

then show that $(a_1)^{-1}(a_2)^{-1}\dots(a_n)^{-1} = a_1a_2\dots a_n$ (why? show that map $a_i \to (a_i)^{-1}$ is bijective)

and now....you're done (why?).

EDIT: alex, you missed the sqaured.

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