I am having a little difficulty with what is seemingly a simple exercise, I must show that in a permutation group $S_{n}$ where $\sigma = (p_1 p_2 ... p_n)$ is a k-cycle and $i<j$ that $(p_i p_j) \sigma $ is a product of two disjoint cycles.
I am not sure how to approach this, can I simply rewrite as $(1 \space 2 \space 3 ... p_i)(p_j, p_k)$? Or have I misunderstood the question.
On
Yes, it is true that composing a cyclic permutation $(p_1,p_2,\ldots,p_k)$ with a transposition $(p_i,p_j)$ will give a product of two disjoint cycles if $p_i$ and $p_j$ occur in the cyclic permutation. For example, it can be verified that $(p_1,\ldots,p_i,p_{i+1},\ldots,p_j,\ldots,p_k)(p_i,p_j)$ $= (p_1,\ldots,p_{i-1}, p_j, \ldots, p_k)(p_i,p_{i+1},\ldots, p_{j-1})$.
More generally, composing a permutation with a transposition either increases or decreases the number of disjoint cycles by $1$. If the two points in the transposition occur in the same cycle of the permutation, composing will split up this cycle into two disjoint cycles. If the two points in the transposition occur in different cycles of the permutation, composing will combine the two disjoint cycles into a single cycle.
Hint: For all $\tau\in S_n$, $\tau(p_1\cdots p_n)\tau^{-1}=(\tau(p_1)\cdots\tau(p_n))$.
What do you get with $\tau=(p_i p_j)$?