Abelian Group elements and inverses

Question

Let G be a finite abelian group, say, $G={e,a_1,a_2...a_n}$ Prove the following:

a)$(a_1a_2...a_n)^2=e$

b)If there is no element $x \neq e$, $x=x^{-1}$, then $a_1a_2...a_n=e$

c)If there is exactly one $x\neq e$, $x=x^{-1}$, then $a_1a_2...a_n=x$

So these are pretty simple, but I'm not sure how to write out the logic mathematically. For example for part a all I could think to do was this: $a_1a_1a_2a_2...a_6a_7a_6a_7...ee=e$ In that the first group represents the elements that equal their own inverse, and the second part those who don't. What's a more elegant way to do it? Bonus points for telling me how to do $x^{-1}$ in $LaTeX$.

For proving $a)$, I suggest you use the property that any element in a group has an inverse. The result follows in a couple of lines. — user88595, Oct 14 '13 at 22:27
Related (though not a duplicate): http://math.stackexchange.com/questions/474214/product-of-a-finite-abelian-groups-elements/ — Tobias Kildetoft, Oct 15 '13 at 10:57

score 3 · Answer 1 · answered Oct 15 '13 at 10:39

For the sake of not having unanswered questions:

a) $$\forall\,a_i\in G\;\exists !\,a_j\in G\;\;s.t.\;\;a_ia_j=e$$

so pair up the elements with their inverses (you can since $\;G\;$ is abelian):

$$(a_1\cdot\ldots\cdot a_n)^2=(a_1\cdot a_1^{-1})\cdot\ldots\cdot (a_n\cdot a_n^{-1})=e\cdot\ldots\cdot e= e$$

b) If there are no involutions in $\;G\;$ then

$$\forall\;i\in\{1,2,...,n\}\;\exists!\,i'\in\{1,2,...,n\}\;,\;i\neq i'\;,\;\;s.t.\;\;a_ia_{i'}=e$$

and again pair up each element with inverse:

$$(a_1a_{1'})\cdot\ldots\cdot(a_na_{n'})=e\cdot\ldots\cdot e=e$$

c) Do exactly as in the two parts above but this time there is one unique $\;e\neq a_k\in G\;$ which is an inverse to itself (i.e. $\;a_k^2=e\;$, an involution), so in the pairing described above this element remains "unpaired"...

Abelian Group elements and inverses

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