If $H \subseteq S_n$ is a normal subgroup, and H contains a cycle $\alpha$ of length $r$, then prove $H$ contains every cycle of length $r$. Textbook: A First Course in Abstract Algebra with Applications 3rd Rotman
If $H \subseteq S_n \text{ (the symmetric group) }$ is a normal subgroup, and H contains a cycle $\alpha$ of length $r$, then prove $H$ contains every cycle of length $r$. 
Every example I have seen of a normal subgroup of $S_n$ have been alternating groups. So for my problem, can I assume that $H=A_n$? 
 A: Hint: show that every two cycles of lenth $r$ are conjugate. Use the fact that $\sigma^{-1}(a_1 a_2 \cdots a_r)\sigma=(\sigma(a_1) \sigma(a_2) \cdots \sigma(a_r))$ for any $\sigma  \in S_n$.
A: Here's a hint to get you started.  Suppose $H \trianglelefteq S_5$ and $\sigma = (1\ 2\ 3\ 4\ 5) \in H$.  Since $H$ is normal, then $\tau \sigma \tau^{-1} \in H$ for all $\tau \in S_5$. Say I want to show that $(2\ 3\ 1\ 4\ 5) \in H$; what can I conjugate $(1\ 2\ 3\ 4\ 5)$ by to get $(3\ 1\ 2\ 4\ 5)$?  You can check that
\begin{align*}
(1\ 3\ 2) (1\ 2\ 3\ 4\ 5) (1\ 3\ 2)^{-1} &= (1\ 3\ 2) (1\ 2\ 3\ 4\ 5) (2\ 3\ 1) = (3\ 1\ 2\ 4\ 5)
\end{align*}
which shows that $(3\ 1\ 2\ 4\ 5) \in H$.  Can you use this example to prove the general result?
A: Here is an example with a 3-cycle to show how conjugating a 3-cycle actually works. It's just the same for an $r$-cycle, but there's more subscripts, and it's messier, but the idea is the same.
We want to compute $\sigma(a\ b\ c)\sigma^{-1}$.
First we suppose $\sigma(a) = k,\ \sigma(b) = m,\ \sigma(c) = n$.
Clearly, then, $\sigma^{-1}(k) = a,\ \sigma^{-1}(m) = b,\ \sigma^{-1}(n) = c$.
Case $1$: $x \not\in \{k,m,n\}$. Whatever $\sigma^{-1}(x)$ may be, it is neither $a,\ b$ nor $c$, because permutations are one-to-one.
Therefore, viewing $(a\ b\ c)$ as a function, we have:
$(a\ b\ c)(\sigma^{-1}(x)) = \sigma^{-1}(x)$, since $a,b,c$ are the only elements permuted by $(a\ b\ c)$.
Thus $\sigma(a\ b\ c)\sigma^{-1}(x) = \sigma((a\ b\ c)(\sigma^{-1}(x)) = \sigma(\sigma^{-1}(x)) = x$ for $x \not\in \{k,m,n\}$.
Case $2$: $x \in \{k,m,n\}$. First we apply $\sigma^{-1}$:
$k \mapsto a$
$m \mapsto b$
$n \mapsto c$ (we are doing "all three at once").
Next we apply $(a\ b\ c)$ to get $(a b c)\sigma^{-1}$:
$k \mapsto a \mapsto b$
$m \mapsto b \mapsto c$
$n \mapsto c \mapsto a$
Finally, we apply $\sigma$, to get $\sigma(a\ b\ c)\sigma^{-1}$:
$k \mapsto a \mapsto b \mapsto m$
$m \mapsto b \mapsto c \mapsto n$
$n \mapsto c \mapsto a \mapsto k$
So for all $x\not\in \{k,m,n\}$, we have $\sigma(a\ b\ c)\sigma^{-1}$ fixes $x$, and on the set $\{k,m,n\}$ we have that $\sigma(a\ b\ c)\sigma^{-1}$ acts as: $(k\ m\ n)$, that is:
$\sigma(a\ b\ c)\sigma^{-1} = (\sigma(a)\ \sigma(b)\ \sigma (c))$
So any conjugate of a $3$-cycle is another $3$-cycle (as I pointed out before, the "same idea" works for any $r$-cycle).
The question then becomes: given a particular $3$-cycle, say $(a\ b\ c)$, can I write any OTHER $3$-cycle, say $(k\ m\ n)$, as a conjugate of my first $3$-cycle? That is, can we find an appropriate $\sigma$? And the answer is yes, ANY $\sigma$ with $\sigma(a) = k, \sigma(b) = m$ and $\sigma(c) = m$ will do (choose the images of $\{1,2,\dots,n\} - \{a,b,c\}$ any way you like).
The upshot of all this is: a normal subgroup $H$ of $S_n$ contains ALL examples of any given cycle-type it contains-it is a union of conjugacy classes (for example, $A_n$ is the union of all cycles types that are even).

N.B.: If by $\sigma\tau$ one means "apply $\sigma$, then $\tau$", one uses $\sigma^{-1}(a\ b\ c)\sigma$ instead (the argument is the same, though).
