Showing that $\beta\alpha\beta^{-1}$ is a $5$-cycle Let $G=S_n$, the symmetric group of degree $n$, where $n\geq 5$.

Let $\alpha=(g~h~i~j~k)$ be a $5$-cycle in $G$. If $\beta$ is any element in $G$, show that $\beta\alpha\beta^{-1}$ is a $5$-cycle. 

Here's what I tried to do: Let $\beta\in S_n$ and $\alpha$ be a $5$-cycle, $(g~h~i~j~k)$, for $n\geq 5$. First write $\beta\alpha\beta^{-1}$ as a product of $2$-cycles. Then, if one of the $2$-cycles of $\beta$ is the inverse of one of the $2$-cycles of $\alpha$, they will cancel out, but the $2$-cycle of $\beta^{-1}$ will not cancel with anything. Likewise, if a $2$-cycle of $\beta$ is not the inverse of anything in $\alpha$, then it will cancel with its inverse in $\beta^{-1}$. Thus, every $2$-cycle in $\beta$ will cancel with a $2$-cycle in $\beta^{-1}$ or in $\alpha$. Then there will be the same number of $2$-cycles that we started with. Simplifying, we will have a $5$-cycle. Thus, $\beta\alpha\beta^{-1}$ is a $5$-cycle. 
Is this right and/or is there an easier way to show that $\beta\alpha\beta^{-1}$ is a $5$-cycle? 
 A: Let $\beta\in S_n$, and consider $\beta \alpha \beta^{-1}$. then $(\beta \alpha \beta^{-1})^5 = \beta \alpha^5 \beta^{-1}=\beta (1) \beta^{-1}= (1)$, so that the order of $\beta \alpha \beta^{-1}$ divides $5$.
Since $\beta \alpha \beta^{-1}$ is not the identity, we see it must have order exactly $5$.
But the only elements of order $5$ in $S_n$ of order 5 are products of disjoint 5-cycles.
Now, think about the set of letters fixed by $\alpha$ and you should be able to force $\beta \alpha \beta^{-1}$ to be a single 5-cycle.
EDIT: We can reorder the letters so that WLOG $\alpha = (12345)$. As noted in the comments, we need only show the result for a generating set of $S_n$, so let's use transpositions. 
then consider $\beta = (ab)$, and $\beta \alpha\beta^{-1} = \beta (12345)\beta^{-1}$.
Now, let $i\in \{1,2,\ldots,n\}$. and consider $\beta \alpha\beta^{-1}(i).$
If $\beta^{-1}(i)>5$, then this is $\beta \alpha\beta^{-1}(i)=\beta\beta^{-1}(i) = i$.
If $\beta^{-1}(i)\leq 5$, then there is some $j\in\{1,2,3,4,5\}$ such that $j = \beta^{-1}(i) \Rightarrow \beta(j)=i$.
Thus, $\beta \alpha\beta^{-1}(i) = \beta \alpha(j) = \beta(j+1\mod 5)$ (as this is what $\alpha$ does).
Thus $\beta \alpha\beta^{-1}$, as a permutation, fixes $i$ such that $\beta^{-1}(i)>5$, 
and sends $i = \beta(1)$ to $\beta(2)$, sends $i=\beta(2)$ to $\beta(3)$ etc.
Thus, $\beta \alpha\beta^{-1} = (\beta(1)\beta(2)\beta(3)\beta(4)\beta(5))$.
The only reason I set $\alpha = (12345)$ was for it to be easier to read... 
A: In general cycle structure is invariant under conjugation. This is because $\beta\alpha\beta^{-1}$ sends $\beta(a)$ to $\beta(\alpha(a))$.
This means that if you have:
$$\alpha=({c_1}_1,{c_1}_2\dots {c_1}_{k_1})({c_2}_1,{c_2}_2\dots {c_2}_{k_2})\dots ({c_n}_1,{c_n}_2\dots {c_n}_{k_n})$$
then 
$\beta\alpha\beta^{-1}=(\beta({c_1}_1),\beta({c_1}_2)\dots \beta({c_1}_{k_1}))(\beta({c_2}_1),\beta({c_2}_2)\dots \beta({c_2}_{k_2}))\dots (\beta({c_n}_1),\beta({c_n}_2)\dots \beta({c_n}_{k_n}))$
A: Hint:
$$(\underbrace{\beta\alpha\beta^{-1}}_{\neq 1})^5 = \beta \alpha \underbrace{\beta^{-1} \beta}_{1} \alpha \underbrace{\beta^{-1} \beta}_{1} \alpha \underbrace{\beta^{-1} \beta}_{1} \alpha \underbrace{\beta^{-1} \beta}_{1} \alpha \beta^{-1} = \beta \underbrace{\alpha^5}_{1} \beta^{-1} = \beta \beta^{-1} = 1$$
A: We have that $\alpha$ maps:
$g \to h\\h \to i\\i \to j\\j \to k\\k \to g$
and that $\alpha(m) = m$, for $m \in \{1,2,3,\dots,n\} - \{g,h,i,j,k\}$.
Suppose that $\beta$ maps:
$g \to a\\h \to b\\i \to c\\j \to d\\k \to f$
(we shall see that we are unconcerned with other values of $\beta(m)$).
Note that above means that $\beta^{-1}$ maps:
$a \to g\\b \to h\\c \to i\\d \to j\\f \to k$
We can distinguish two cases:
Case $1$: $m \in \{1,2,3,\dots,n\} - \{a,b,c,d,f\}$
In this case, whatever $\beta^{-1}(m)$ may be, since $\beta^{-1}$ is bijective, we know that:
$\beta^{-1}(m) \neq g,h,i,j,k$.
Thus $\alpha\beta^{-1}(m) = \alpha(\beta^{-1}(m)) = \beta^{-1}(m)$.
So, in this case, $\beta\alpha\beta^{-1}(m) = \beta(\alpha(\beta^{-1}(m))) = \beta(\beta^{-1}(m)) = m$.
Case $2$: $m \in \{a,b,c,d,f\}$. In this case, we have $\beta\alpha\beta^{-1}$ maps:
$a\to g \to h \to b\\b \to h \to i \to c\\c \to i \to j \to d\\d \to j \to k \to f\\f \to k \to g \to a$
So we see that $\beta\alpha\beta^{-1}$ is the $5$-cycle $(a\ b\ c\ d\ f)$.
