Is $(\sigma\circ \tau)^n=\sigma^n\circ\tau^n$?

Question

Is $$(\sigma\circ \tau)^n=\sigma^n\circ\tau^n?$$

a. It seems equivalent to ask whereas composition of permutation is an homomorphism

This seems to be incorrect as $\sigma \circ \tau \neq \tau \circ \sigma$ for $S_n$ such that $n\leq 2$

b. In particular if we have that ${\rm ord}(\sigma)=n$ and ${\rm ord}(\tau)=n$ can we say that $(\sigma\circ \tau)^n=\sigma^n\circ\tau^n={\rm id}^n\circ{\rm id}^n={\rm id}?$

Answer 1

No.

Consider $\sigma=(12), \tau=(23)$. Then $(\sigma\circ\tau)^2=(123)^2=(132)$, whereas $\sigma^2\circ\tau^2={\rm id}$.