Show that there isn't such element $\tau$ so $\tau^2=(123456)$ I'd like to know if my solution is correct:
Assume there exists a permutation $\tau$ so that $\tau^2 = (123456)$
$\tau$ is a permutation, so it can be written as a multiplication of transpositions. The permutation $\tau^2$ will contain $2$ times the transpositions in $\tau$. That leads us to conclude that $\tau^2$ is an even permutation, hence it can't be equal to $(123456)$, which is an odd permutation.
 A: Yes, the reasoning is correct.
If you have a homomorphism $\operatorname{sgn}:S^6\to\{\pm 1\}$ (the sign homomorphism), you could say that $\operatorname{sgn}(\tau^2)=\operatorname{sgn}(\tau)^2=(\pm 1)^2=1$, so $\tau^2$ is even.
I am not sure your particular reason for why $(123456)$ is odd, maybe you should mention it (I suspect it is because it is a cycle and you are appealing to a theorem).  But it is not too hard to write it as a product of transpositions: $(12)(23)(34)(45)(56)$.

A sign-determination trick I particularly like is to draw the permutation graphically, adjust the lines until the intersections only involve two lines at a time, and then count off the number of crossings mod 2.  As a bonus, it gives a decomposition as a product of transpositions.  The following diagrams are read bottom-to-top:

A decomposition of $(123456)$ as $\tau^2$ would involve moving the lines around until there are equal numbers of crossings above and below some horizontal line, but the number of crossings is odd.
