Working in $S_6$ compute $(134) \cdot (12).$ 
Working in $S_6$ compute $(134) \cdot (12).$

I know that cycle multiplication is performed from left to right for example $(1324)\cdot(1243) = (142)(3).$ But in this case I'm confused I don't have $3$ or $4$ on the lhs nor $2$ on the rhs. How should I compute this?

Observe that the first cycle is
$$\color{red}{1\to 3, \quad 3 \to 4, \quad 4\to 1}$$
and the second cycle is
$$\color{blue}{1\to 2, \quad 2\to 1}$$
When we compose the permutations, then we get
$$\color{red}{1\to 3, \quad 3 \to 4, \quad 4} \color{green}{\to}\color{blue}{ 2, \quad 2 \to 1}$$
where the green line remark the point when we change the permutation we are working on. Then, the product permutation is $(1 3 4 2)$.
