Let G be a group and a; b ∈ G. Suppose |a| = |b| = |ab| = 2. Then show that ab = ba. 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.
 A: Write $abab=e$ and multiply both sides by $ba$.
A: 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.
A: 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.
A: 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.
A: 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.
A: This is my favorite oneliner: $ab = a(ab)^2 b = aababb = a^2 ba b^2 = ba$. 
(Edit: it's the short form of the answer given by N.F. Taussig.)
