Another Cycle Decomposition Question

In $\displaystyle S_4$ let $\displaystyle \sigma$ be a cycle of length 4, and $\displaystyle \tau$ be a transposition.

Show $\displaystyle \sigma \tau$ is even, and briefly explain why $\displaystyle \sigma \tau\neq id$

Let $\displaystyle \sigma = (abcd)$, then the decomposition would be $\displaystyle (ab)(ac)(ad)$, and let $\displaystyle \tau = (ef)$. This means that $\displaystyle \sigma \tau$ would be $\displaystyle (ab)(ac)(ad)(ef)$, which is even. *Think I've gone about this in the correct way?*

For any transposition $\displaystyle (ij)$, $\displaystyle (ij)^2 = id$, this means that $\displaystyle (ij) = (ij)$.

For our transpositions above, $\displaystyle (ab)\neq(ac)\neq(ad)$, so even if $\displaystyle (ef) = (ab) or (ac) or (ad)$, it would still not equal the identity transposition. *Would this be enough to briefly explain?*

I managed to attempt the first couple of sections of the question, but this last part has got me very confused:

Deduce that $\displaystyle \sigma \tau$ is either a cycle of length 3, or a product of 2 disjoint transpositions. Conclude that $\displaystyle \sigma = \gamma \lambda \tau$ for some not necessarily disjoint permutations $\displaystyle \gamma, \lambda \in S_4$

Would greatly appreciate some help in how to go about this problem.

Thanks in advance