I'm having trouble understanding this question and help would be appreciated.
If |ab|=2 and |a|=2, |b|=2, wouldn't this imply that |a||b|=|ab|=4?
How would I go about proving that this is Abelian?
I do know that an identity element exists(by assumption, e=1) and that an inverse for A and B exists, which I'm also assuming is 1/|2|.
Thank you for the help.
On
Since $|ab| = 2$, we know that
$$\begin{align} (ab)(ab) &= e \\ a^{-1}(ab)(ab)b^{-1} &= a^{-1}eb^{-1} \\ ebae &= a^{-1}eb^{-1} \\ ba &= a^{-1}b^{-1} \\ ba &= (aa)a^{-1}b^{-1}(bb) \\ ba &= a(aa^{-1})(b^{-1}b)b \\ ba &= aeeb \\ ba &= ab \\ \end{align}$$
At least that's how I think it goes; I haven't done group theory in 2+ years.
Also, I am new here, so please excuse the way I typed it; I still have to learn how to use the coding.
On
There are several questions in one question here and I'll try to answer those seperately. First of all on orders of elements: the easiest possible example to se that $|a||b|=|ab|$ does not necessarily hold is in the (additive) group $\mathbb{Z}/(2)\times\mathbb{Z}/(2)$. If $a=(1,0)$ and $b=(0,1)$, both elements have order $2$ and $a+b=(1,1)$ has order $2$ just as well.
For the next question, one has to consider what it means that $|a|=2$. It means that $a^2=1$, i.e. $a^{-1}=a$. It thus follows that $b^{-1}=b$ and $(ab)^{-1}=ab$, while in general $(ab)^{-1}=b^{-1}a^{-1}$. (Can you see why?) Combining these facts, we find $ab=(ab)^{-1} = b^{-1}a^{-1}=ba$.
The main point to remember is that if an element in a group has order $2$, it is its own inverse.
On
Since $|a| = 2$, $a^2 = e$. Similarly, $|b| = 2 \Rightarrow b^2 = e$ and $|ab| = 2 \Rightarrow (ab)(ab) = e$.
If we left-multiply the equation $$abab = e$$ by $a$ and right-multiply it by $b$, we obtain
\begin{align*} aababb & = aeb\\ (aa)(ba)(bb) & = ab\\ e(ba)e & = ab\\ ba & = ab \end{align*}
where we have used the fact that if an element has order $2$, then it is its own inverse.
On
It's worth noting that it looks like you are confusing two similar notations. When talking about the order of an element $a$ in a group, there are a number of ways of writing it, with $|x|$ being pretty common.
At the same time, in the real numbers, we use $|x|$ to denote the absolute value of an element $x$, which satisfies $|xy| = |x||y|$.
However! Just because we are using the same notation doesn't mean that the same properties hold. In particular, it is not in general true that in a group $|ab| = |a||b|$. This does happen some of the time, but by no means is it generally true. For examples, see the other answers.
In short: one should not be tricked by common notation into believing that properties carry over; make sure you are aware of what context that notation is from, and what properties hold in that context.
Write $abab=e$ and multiply both sides by $ba$.