Help with understanding proof: every group of order 6 is cclic or dihedral. I am considering the proof that every group of order 6 is cyclic or dihedral (following some lecture notes). The initial part of the proof, in broad brushstrokes, considers that the elements in the group can only be of order 1,2,3,6. Clearly if there is an element of order 6 the group 6 is cyclic. Now it must have an element of order 2 but cannot have only elements of order 2 (due to the group order of 6), so it would otherwise have an element of order 2 and an element of order 3. So far so good.
Now what I am stuck on is considering the cases for the scenario with an element of order 3, say $r\in G$, and of order 2, say $s\in G$. Now to be dihedral there is not only the fact of having some element of order n and one of order 2, but also the relationship $srs^{-1}=r^{-1}$ must be obeyed. So the proof proceeds to consider these cases.
We know that the cyclic group generated by r, $<r>$, has index 2 and thus is a normal subgroup. So we can have 


*

*$srs^{-1}=srs=e$ which is not possible since this would require $r=e$ which is not the case


*

*W can also have $srs^{-1}=r^2=r^{-1}$ which gives us the dihedral group.

*Finally $srs^{-1}=r$. Now this is the one I am stuck on. The proof says: $srs^{-1}=r \implies sr=rs$ and $sr$ is of order 6$\implies G \cong C_6$
I don't get this last point at all. I have been mulling around with the algebra and just don't see why $srs^{-1}=r$ implies that $sr$ is of order 6. This is the only bit I don't understand. I would greatly appreciate if someone could point me in the right direction (hint or exact answer!)
 A: If $srs^{-1} = r$, then multiplying both sides on the right by $s$ you get that $sr = rs$, i.e., that $s$ and $r$ commute with each other.
Now, show that in ANY group, if you have two elements $a$ and $b$ that commute with each other, then the order of $ab$ divides the LCM of the orders of $a$ and $b$ and is divisible by the GCD of the orders of $a$ and $b$. and is divisible by the (LCM / GCD).
What does that mean if the orders of $a$ and $b$ are relatively prime?
Edit: It seems I can't get my basic facts straight; fixed the fact above.
A: We can just check that $sr$ is a generator; recalling that $s$ has order 2, $r$ has order 3, and using the fact that they commute, we take successive powers:
\begin{align*}
(sr)^1 &= sr \\
(sr)^2 &= srsr = rssr = r^2\\
(sr)^3 &= r^2sr = r^3s = s\\
(sr)^4 &= ssr = r\\
(sr)^5 &= rsr = sr^2\\
(sr)^6 &= sr^2sr = s^2r^3 = e
\end{align*}
So $sr$ has order 6, and the group is cyclic.
A: If you want an exact answer, here is one. Let $|G|=6$. Its Sylow $3$-subgroup has index $2$, thus is normal. Its Sylow $2$-subgroup intersects with the $3$-subgroup trivially, and and their internal product is $G$. Thus $G\cong\Bbb Z_3\rtimes\Bbb Z_2$. Now construct an homomorphism from $\Bbb Z_2$ to $Aut(\Bbb Z_3)$. Argue that there are only $2$ possibilities for this homomorohism. One of them gives you the cyclic group, another gives you the dihedral group. 
