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 $LaTeX$.
Best Answer
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"...