I am a little stuck here and would like some minor help.
The quesiton I am dealing with is: Assume in an abelian group G that $<b>{\large\cap} <a>=e$, then the order of $(ab)$ is the lcm of the orders of $a$ and $b$.
So far I have the assumption, then Suppose $(ab)^k=e$, where k is the smallest integer. So $|ab|=k$. Since G is abelian, $(ab)^k=a^k b^k=e$. Thus, $(a^k)^{-1}=b^k$ and likewise the same goes for, $(b^k)^{-1}=a^k$. Therefore $(b^k)^{-1}\in<a>$ and $(a^k)^{-1}\in<b>$. So each of $a^k$ and $b^k$ are the identity. Therefore $k$ is a multiple of n and m, so we can say lcm($m,n$)|$k$.
I know I am close, but where to next?
Lastly, if I get rid of the assumption that G is abelian, how do I show that this is always true: $|ab|=|ba|$.
So far I have let |$ab$|=$n $ and consider the product $b(ab)^n$. So $b(ab)^n=b(ab)(ab)…=(ba)(ba)…(ba)(ba)b=(ba)^nb$. From this could I say that $(ab)^n=(ba)^n$ by cancellation? Thus since $(ab)^n=e, (ba)^n=e$, so (ba) has order n. So $|ab|=|ba|$. Q.E.D.
Best Answer
Hints:
$1.$ For the first question, let the order of $a$ be $m$, and the order of $b$ be $n$. Suppose that $(ab)^k=e$. Then as you observed, we have $(a^{-1})^k=b^k$. So each of $a^k$ and $b^k$ is the identity. It follows that $k$ is a multiple of $m$ and $k$ is a multiple of $n$, so the lcm of $m$ and $n$ divides $k$.
$2.$ For the non-Abelian question, suppose that $(ab)^n=e$. By grouping suitably, show that $(ba)^{n+1}=ba$.