Showing that $\beta \alpha \beta^{-1}$ and $\alpha$ both odd or even 
Question: Let $\alpha$ and $\beta$ belong to $S_{n}$. Prove that $\beta \alpha \beta^{-1}$ and $\alpha$ are both odd or even.

$\alpha \alpha ^{-1} =\varepsilon =\beta^{-1} \beta $
where $\varepsilon$ is an even permutation.
Let $\alpha = \sigma _{1}\cdot \cdot \cdot \sigma_{s}$, $\beta = \theta _{1}\cdot \cdot \cdot \theta_{t}$
Observe: 
$\varepsilon =\sigma _{1}\cdot \cdot \cdot \sigma_{s} \theta _{1}\cdot \cdot \cdot \theta_{t}$
Thus, s + t is even implies that both s and t are either odd or even.
$\beta \alpha \beta^{-1}$ has $\left ( 2t + s \right )$ 2-cycles.
A crucial piece of detail could be preventing me from completing this proof. 
Any hint is appreciated.
 A: Note that if:
$\alpha = \sigma_1\sigma_2\cdots\sigma_k$
that:
$\beta\alpha\beta^{-1} = (\beta\sigma_1\beta^{-1})(\beta\sigma_2\beta^{-1})\cdots(\beta\sigma_k\beta^{-1})$.
Now if each $\sigma_i$ for $i = 1,\dots,k$ is a transposition, we will show each $\beta\sigma_i\beta^{-1}$ is likewise a transposition.
Suppose $\sigma_i = (a\ b)$. Let us denote:
$c = \beta(a)$ and $d = \beta(b)$, so that $\beta^{-1}(c) = a$, and $\beta^{-1}(d) = b$.
If $m \not\in \{c,d\}$, then $\beta^{-1}(m) \not\in \{a,b\}$ (because $\beta,\beta^{-1}$ are bijective), and thus:
$\sigma_i\beta^{-1}(m) = \sigma_i(\beta^{-1}(m)) = \beta^{-1}(m)$, since $\sigma_i$ only affects $a$ and $b$.
Therefore, $\beta\sigma_i\beta^{-1}(m) = \beta(\sigma_i(\beta^{-1}(m))) = \beta(\beta^{-1}(m)) = m$.
If $m = c$, we have: $\beta\sigma_i\beta^{-1}(c) = \beta(\sigma_i(\beta^{-1}(c))) = \beta(\sigma_i(a)) = \beta(b) = d$,
and if $m = d$, we have: $\beta\sigma_i\beta^{-1}(d) = \beta(\sigma_i(\beta^{-1}(d))) = \beta(\sigma_i(b)) = \beta(a) = c$.
So, $\beta\sigma_i\beta^{-1} = (c\ d)$.
Thus if $\alpha$ is a product of $k$ transpositions, $\beta\alpha\beta^{-1}$ is likewise, whether $k$ be even or odd.
A: Consider the following homomorphism: $f:S_n \to \{-1,1\}$ given by
$$f(A_n)=1 \, \,, f(S_n\backslash A_n)=-1.$$
One should check that this is indeed a homomorphism, and that it is typically denoted $\mathrm{Sgn}$. If you check that it is a homomorphism, you can see that its kernel is exactly $A_n$, so $A_n$ is a normal subgroup which means that $\sigma A_n \sigma^{-1}=A_n$.
edit: Without explicit use of homomorphisms, you can use the fact that the decomposition into transpositions is not unique, but its parity is. Hence, we can see that if $\sigma:=\sigma_1\dots \sigma_n$, where $\sigma_i$ are transpositions and $\tau:=\tau_1\dots \tau_k$ where $\tau_j$ are transpositions, then $\sigma \tau =(\sigma_1\cdots\sigma_n)(\tau_1 \cdots \tau_k)$, which has some parity independent of its decomposition. You can use this to finish the problem ,and look here for some more in depth discussion.
A: Here's the easiest proof. Take the sign function
$$\mbox{sgn}:S_n\to\{-1,1\}$$
$$\mbox{sgn}(a_1\cdots a_n)=(-1)^n\mbox{ for } a_i=\mbox{ transpositions}$$
It can be proved that this is a well defined function and it is a group homomorphism (with integer multiplication on the right side), i.e.
$$\mbox{sgn}(ab)=\mbox{sgn}(a)\mbox{sgn}(b)$$
In particular $\mbox{sgn}(a^{-1})=\mbox{sgn}(a)^{-1}$, the simple property of any group homomorphism.
Therefore
$$\mbox{sgn}(\beta\alpha\beta^{-1})=\mbox{sgn}(\beta)\mbox{sgn}(\alpha)\mbox{sgn}(\beta)^{-1}=\mbox{sgn}(\beta)\mbox{sgn}(\beta)^{-1}\mbox{sgn}(\alpha)=\mbox{sgn}(\alpha)$$
The second equality makes sense, because multiplication in $\{-1,1\}$ is commutative.
