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Suppose σ = (3, 4, 5)(2, 4, 5) ∈ S5.

(a) Express σ as the product of disjoint cycles.

(b) Find the order of σ.

(c) Is σ even or odd?

(d) Express σ using matrix notation.

(e) Find σ-1

Im not sure that I am doing this correctly... for matrix notation I got:

2 3 4 5

4 4 5 3

but I didn't that that it could be a permutation because it is not isomorphic, and also, shouldn't 5 go to 2? Can someone please explain to me how to set up the permutation in matrix form? From there it should be easy to find the disjoint cycles and transpositions and from there we can find whether it is even or odd. Basically I just need someone to explain to me how to set up the matrix....

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  • $\begingroup$ Please use latex when posting on this forum meta.math.stackexchange.com/questions/5020/… $\endgroup$ Dec 17, 2014 at 6:44
  • $\begingroup$ let's simplify your $\sigma$ to begin with. $\endgroup$ Dec 17, 2014 at 6:48
  • $\begingroup$ Before you can do anything else, you need to know in which order you apply the cycles: do you apply first $(3,4,5)$ and then $(2,4,5)$, or do you apply first $(2,4,5)$ and then $(3,4,5)$. Both orders are fairly common, so you’ll have to tell us what your convention is. \\ I can tell you right off, though, that your matrix notation is not correct: it doesn’t specify what happens to $1$, and it sends $2$ and $3$ to the same place, $4$, meaning that it doesn’t represent a permutation. $\endgroup$ Dec 17, 2014 at 6:52
  • $\begingroup$ From what I understand you start from the left and go right... So because it is an element of S5, the top row should read 1 2 3 4 5, but I don't know how to find the bottom row because 1 isnt in the permutation multiplication, and 2 apparently goes no where... $\endgroup$
    – Kaitlyn
    Dec 17, 2014 at 6:56
  • $\begingroup$ @Kaitlyn: Neither of the $3$-cycles affects $1$, so it goes to itself. The first $3$-cycle leaves $2$ alone — i.e., sends it to itself — and the second then sends it to $4$. The first sends $3$ to $4$, and the second then sends $4$ to $5$, so the composition sends $3$ to $5$. Can you finish it from there to complete the matrix form, which so far has $\binom{1\;2\;3}{1\;4\;5}$? $\endgroup$ Dec 17, 2014 at 7:00

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Now that you have the matrix form of $\sigma$, you can easily write it as the product of disjoint cycles. Start with the element $1$, and see what cycle it generates; since $\sigma(1)=1$, it generates the $1$-cycle $(1)$. Now go on to the first element of $\{1,2,3,4,5\}$ that isn’t already in a cycle: in this case that’s $2$, and it generates the $2$-cycle $(24)$, since $\sigma(2)=4$ and $\sigma(4)=2$. Repeat until every element of $\{1,2,3,4,5\}$ is in one of the disjoint cycles that you’ve constructed. In this case you get $(1)(24)(35)$; it’s customary to omit $1$-cycles and write it as $(24)(35)$.

Now that we know that $\sigma=(24)(35)$, it’s pretty easy to find its order: how many times must you apply $\sigma$ to get the identity map? In other words, what’s the smallest positive integer $n$ such that the matrix of $\sigma^n$ is $$\binom{1\;2\;3\;4\;5}{1\;2\;3\;4\;5}\;?$$

It’s also easy to say whether $\sigma$ is even or odd, provided that you know the definitions of those terms in this context. Finally, if you got the order of $\sigma$ right, finding $\sigma^{-1}$ will be very easy.

Added: More generally, you can quite easily find the inverse of a permutation from its matrix form. Suppose that you have a permutation

$$\pi=\binom{1\;2\;3\;4\;5}{2\;5\;3\;4\;1}\;;$$

then $\pi$ is the function from $\{1,2,3,4,5\}$ to $\{1,2,3,4,5\}$ that sends $1$ to $2$, $2$ to $5$, and so on. The inverse of that function just reverses the arrows, sending $2$ to $1$, $5$ to $2$, and so on. To get its matrix, flip the matrix for $\pi$ upside down to get

$$\pi=\binom{2\;5\;3\;4\;1}{1\;2\;3\;4\;5}\;,$$

and then sort the columns to put the top row in order:

$$\pi=\binom{1\;2\;3\;4\;5}{5\;1\;3\;4\;2}\;.$$

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