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?
 A: 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 $(1 3 4 2)$.
A: Recall that, in $S_6$, $(134) \cdot (12)=(134)(12)(5)(6)$. Thus we have
$$\begin{align}
1&\xrightarrow{(134)}3\xrightarrow{(12)}3\xrightarrow{(5)}3\xrightarrow{(6)}3,\\
3&\xrightarrow{(134)}4\xrightarrow{(12)}4\xrightarrow{(5)}4\xrightarrow{(6)}4,\\
4&\xrightarrow{(134)}1\xrightarrow{(12)}2\xrightarrow{(5)}2\xrightarrow{(6)}2,\\
2&\xrightarrow{(134)}2\xrightarrow{(12)}1\xrightarrow{(5)}1\xrightarrow{(6)}1,\\
5&\xrightarrow{(134)}5\xrightarrow{(12)}5\xrightarrow{(5)}5\xrightarrow{(6)}5,\\
6&\xrightarrow{(134)}6\xrightarrow{(12)}6\xrightarrow{(5)}6\xrightarrow{(6)}6,
\end{align}$$
so the answer is $(1342)(5)(6)=(1342)$.
A: The notation for permutations and cycles is confusing similar, so when you write them out, try to keep the distinction clear:
$$P(123456)C(134)=P(324156)$$
$$P(324156)C(12)=P(314256)$$
To write out in shorthand notation, we get $(3421)(5)(6)=(1342)$.
