Product of cycles in symmetric group S7

Question

I have this question and although I don't want the answer directly, how do I go about finding other ways of expressing the product of two cycles in S7? Thanks

Answer 1

You express each of them as a product of disjoint cycles. For instance$$\begin{pmatrix}1&3&2&5&7\end{pmatrix}\begin{pmatrix}4&2&6&7\end{pmatrix}=\begin{pmatrix}1&3&2&6\end{pmatrix}\begin{pmatrix}4&5&7\end{pmatrix},\tag1$$whereas$$\begin{pmatrix}1&3&6&7\end{pmatrix}\begin{pmatrix}2&5&4\end{pmatrix}\tag2$$ is already written as a product of disjoint cycles (as are all the other possible answer). But you have distinct cycles in $(1)$ and $(2)$, right? So$$\begin{pmatrix}1&3&2&5&7\end{pmatrix}\begin{pmatrix}4&2&6&7\end{pmatrix}\neq\begin{pmatrix}1&3&6&7\end{pmatrix}\begin{pmatrix}2&5&4\end{pmatrix}.$$

Answer 2

The idea behind cycles is that each cycle is a bijection from the set to itself. For instance, let the set be S = {a,b,c,d,e}, then the cycle (a b c) is the function f from S to S, such that f(a) = b, f(b) = c, f(c) = a, f(d) = d, f(e) = e. Now, when you have two cycles, say, (a b) (b c d), the operation is just function-composition. In this case, let us call the two functions f and g, where f represents (a b) and g represents (b c d). Now, g takes d to b, and f takes b to a, so, their composition takes d to a. That idea applied to each of the elements should tell you what the composition function looks like.

In your question, you need to check for how your composition function looks and compare them with the options you have.

Interestingly, you can prove that any permutation can be written as the product of disjoint cycles, so, when you write the permutation in your question in that form, the problem immediately becomes much simpler.

Hint: You can eliminate Option (a) since, in the question, the function takes 7 to 4, while in the first option, the function takes 7 to 1.