Commutative finite group, $(A_1A_2\dots A_n)^2 = e$ and $A_1A_2\dots A_n=e$

Question

This is a question I am struggling to answer because my TA sent it to me for practice.

Say $G=\{A_1,A_2,A_3,\dots,A_n\}$ a commutative group so that $|G|=n$.

Prove that:

$(A_1A_2\dots A_n)^2 = e$.
If for every $A_i \in G$ which is different from $e$, $A_i^2 \ne e$, then $A_1A_2\dots A_n=e$.
If there is a single element $A_i \in G$ which is different from $e$ so that $A_i^2 = e$, then $A_i = A_1A_2\dots A_n$.

I really don't know how to approach this, and I still don't want a full answer. I am asking for hints that will help me approach this. Thanks!

score 5 · Answer 1 · answered May 24 '14 at 11:03

5

Hint: Remember that any element in $G$ has an unique inverse.

answered May 24 '14 at 11:03

you mean Ai^(-1)? – HaloKiller May 24 '14 at 11:03
1

Yes; it is also an element of the group, so $A_{i}^{-1}=A_j$ for a $1\leq j\leq n$. – May 24 '14 at 11:05
Okay I have solved 1 and I think I know how to solve 2. Am I supposed to use the If statement? – HaloKiller May 24 '14 at 11:07
You need to. Otherwise $\Bbb{Z}/2\Bbb{Z}$ is a counterexample for $(ii)$. – May 24 '14 at 12:25

1 Answers1