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Let $\sigma=(1\ 3),\ \tau=(2\ 4\ 5),\ \pi=(2\ 3\ 4) \in S_{5}$. Find $\pi\circ\tau\circ\sigma$.

I know the solution is $(1\ 4\ 5\ 3)$.

What i'm doing now is writing the permutations in the "expanded" form,

for example $\tau=\begin{pmatrix} 1\ 2\ 3\ 4\ 5\\ 1\ 5\ 3\ 2\ 4 \end{pmatrix}$,

and then "follow the numbers" over the composition. This way, i find $(1\ 3\ 5\ 2)$, which is incorrect. What am i doing wrong?

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since permutations are generally not commutative, you should be careful about what conventions you follow, in this case its first do \sgima, then \tau and then \pi – Dactyl Dec 27 '10 at 20:02
I had to do about a thousand of these in abstract algebra, and the way that worked best for me was to do each digit and think about where it went. So I'd start with 1, and then (for your example) I'd say, "okay, where does sigma take 1...okay, where does tau take that..." and so on. So, if I had something like (12)(13) and I'd want to see what that was, I'd say, "Well, 1 goes to 3 in (13), then 3 goes to itself under (12). So far I have (...13...) as my solution. Then 2 goes to itself in (13) and then to 1 in (12). So I have (...213...) as my solution. But this is it, so it's (213)." – james Dec 27 '10 at 22:15
up vote 12 down vote accepted

When you read a composition of functions written in the usual notation for permutations, you must remember to read them from right to left. Thus, when you try to compute the composition you must start by looking successively at what does each permutation in the composition do to each integer from $1$ to $5$ (in this case), but from right to left.

So you can start by writing $$\pi \circ \tau \circ \sigma = \begin{pmatrix} 2 &3 &4 \end{pmatrix} \circ \begin{pmatrix} 2 &4 &5 \end{pmatrix} \circ \begin{pmatrix} 1 &3 \end{pmatrix}$$

So you start with $1$. The rightmost permutation sends $1 \mapsto 3 $ and then the middle permutation sends $3 \mapsto 3$ and finally the leftmost permutation sends $3 \mapsto 4$. So in the end the total result is that the composition of the three of them sends $1 \mapsto 4$. In the same way you'll proceed for the other integers and in the end you'll get the answer

$$ \pi \circ \tau \circ \sigma = \begin{pmatrix} 1 &2 &3 &4 &5 \\ 4 &2 &1 &5 &3 \end{pmatrix}$$ or in cycle notation $\pi \circ \tau \circ \sigma = \begin{pmatrix} 1 &4 &5 &3\end{pmatrix}$.

I checked doing the computation from left to right and the "answer" I got from doing that was $\pi \circ \tau \circ \sigma = \begin{pmatrix} 1 &3 &5 &2\end{pmatrix}$ which is what you originally had as your answer.

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thanks! this is what i was looking for:) – user5148 Dec 28 '10 at 11:07

You're probably composing from the left to the right, instead of from the right to the left.

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This is easy to check, and in fact turns out to be the case :-) – TonyK Dec 27 '10 at 20:06
no, i'm composing from right to left, sigma, then tau, then pi. EDIT -- nope!, wait, you were right. ahah! thank you:) – user5148 Dec 27 '10 at 20:07

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