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

Question

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?

Answer 1

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 $(1 3 4 2)$.

Answer 2

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 $(1342)(5)(6)=(1342)$.

Answer 3

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 $(3421)(5)(6)=(1342)$.