If $f:G\to H$ is an epimorphism between finite groups and $K\subset G$ a Sylow p-group then I want to show that $f(K)\subset H$ is also a Sylow p-group.
So we want to show that $|f(K)|=p^m$ where $m$ is the biggest power such that $|H|$ is still divisible by $p^m$. This is the same as saying that $|f(K)|$ is a power of $p$ and $[H:f(K)]=|H|/|f(K)|$ is not divisible by $p$, right?
OK, I'm fine with the fact that $|f(K)|$ is a power of $p$.
Now for the other part: write $N=\ker f$.
Then we have $$[H:f(K)]=[G/N:KN/N]=[G:KN]$$ using the isomorphism theorems. So since $[G:KN]$ is not divisible by $p$ and $[G:KN]\leq [G:K]$, we get the claim that $[H:f(K)]$ is also not divisible by $p$.
I am unsure about this solution, is it correct?
I realise this question has been asked twice before but I couldn't fully make sense of the answers and would like feedback to my solution to find flaws in my thinking.
For last few statements (before bold-faced sentence), instead of inequality, you have to notice the following: since $K\subseteq KN \subseteq G$ and since the index of $K$ in $G$ is not divisible by $p$, hence index of $KN$ in $G$ is also not divisible by $p$.
Do you know the relations between indices $[G:K]$, $[G:KN]$ and $[KN:K]$? A relation is in terms of equality, not an inequality.