Let $Q$ be normal $p$-subgroup in $G$ where $|G|=n<\infty$. Then the natural surjection $G \to G/Q$. we want to show that there is a bijection between
$$G'=\{\text{Conjugacy classes of G whose elments have p' order}\} \to G'/Q=\{\text{Classes of G/Q whose elements have p' order}\}$$
(p' order means order not divisible by p).
Now, the map that seems most natural is $f:G' \to G'/Q$ where
$$(g) \mapsto (gQ)$$
Problem
I have an issue showing surjectivity that makes me think that we might actually need $Q$ to be in $Syl_p(G)$. Let $(hQ) \in G'/Q$. Clearly, $h \mapsto hQ$. Now, if $h$ has $p'$ order then we're done. If not, then $h=x \cdot q$. However, $x$ doesn't necessarily have $p'$ order either and I don't see a way to force it too. Any ideas?
Let $(xQ)$ be an a conjugacy class of $G/Q$ with elements of $p'$-order. Clearly, $x \mapsto xQ$ where $x \in G$. Let $|x|=p^{a}m$ where $p \not| m$. Notice that $x^m \in Q$ since $xQ$ has $p'$-order and $|xQ| \bigg| |x|$. Now, since $p^a$ and $m$ are relatively prime we have $1=r\cdot m + s\cdot p^a$. For any $q \in Q$ we have
$$xQ=x^{s\cdot p^a}x^{r\cdot m}Q=x^{s \cdot p^a}Q$$
Since $x^{s \cdot p^a}$ is of $p'$-order we conclude that our function is indeed surjective.