Identity involving 2- and 3-cycles in group $S_3$

Question

I would like to prove that a product of any 2-cycle $\sigma_2$ in symmetric group $S_3$ with any 3-cycle $\sigma_3$ in $S_3$ satisfies the identity $\sigma_2 \sigma_3 = \sigma_3^{-1} \sigma_2$. I don't yet have deep enough an insight into symmetric groups, so I would appreciate a hint.

Answer 1

Okay, let's see... This is my solution- it is general, but is a rather long answer. Let the two-cycle be x = (x1 x2), then inv(x) = (x1 x2), since (x1 x2)^2 = (1). Since x, y are in S3, |x| = 2, |y| = 3, y^2 = inv(y), and x^2 = (1). Thus, any two-cycle in S3 is its own inverse. Consider |xy|. |xy|= 1 or 2 or 3, but if it is 1, then x = inv(y), which is a contradiction, since they are of different orders. We now consider the order being three, then (x1 x2)(y1 y2 y3) = (z1 z2 z3) so (x1 x2) = (z1 z2 z3)(y1 y3 y2). Now, either (z1 z2 z3) = (y1 y2 y3), or (z1 z2 z3) = inv(y1 y2 y3), since there are only 2 elements of order 3 in S3. Either ways, xy= (z1 z3 z2) or (1), that is it is either of order 3 or 1, a contradiction, since (x1 x2)=x = zy is of order 2. Thus, |xy| = 2, so xy = inv(xy), that is x*y = inv(y) * inv(x), but since inv(x) = x, x*y = inv(y)*x, which is the required result.