Show that if $G$ is finite and $f:G\rightarrow H$ is a group epimorphism, $|\text{Syl}_p(G)|\geq |\text{Syl}_p(H)|$. I want to show that if $G$ is finite and $f:G\rightarrow H$ is a group epimorphism, $|\text{Syl}_p(G)|\geq |\text{Syl}_p(H)|$. 
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I have done the following: 
Since $f$ is a group epimorphism we have that $f$ is surjective, i.e. for each $y\in H$ there is a $x\in G$ such that $f(x)=y$. 
So it might be that there are $x\in G$ for which there is no point in $H$ that is mapped to through $f$. 
That means that $|G|\geq |H|$, right? 
To show that $|\text{Syl}_p(G)|\geq |\text{Syl}_p(H)|$ we have to show that the power of $p$ at the prime factorizations of $|G|$ and $|H|$ is greater at $|G|$ than a $|H|$, or not? 
But how could we do that? 
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EDIT: 
Since $f:G\rightarrow H$ is a group epimorphism, from the first isomorphism theorem we have that $H$ is isomorphism to $G/\ker f$. 
So, $|H|=|G/\ker f|=\frac{|G|}{|\ker f|}$, so $|H| |\ker f|=|G|$. 
That means that $|H|\mid |G|$. 
From the prime factorizations of $|H|$ and $|G|$, all power of primes of $|H|$ must be smaller than or equal to the corresponding of $|G|$. 
Let $|H|=p^ma$ and $|G|=p^nb$, where $p\not\mid a$ and $p\not\mid b$. 
Then it must hold that $m\leq n$. 
How is this related to the number of $p$-Sylow subgroups? 

Given any $p$-sylow $P\in\mathrm{Syl}_p(G)$, we can apply $\phi$ to it and get a $p$-subgroup $\phi(P)\le H$. It would be nice if these were also $p$-sylow subgroups, i.e. $\phi(P)\in\mathrm{Syl}_p(H)$ for all $P\in\mathrm{Syl}_p(G)$, since then we would get a map $\psi:\mathrm{Syl}_p(G)\to\mathrm{Syl}_p(H)$ and we could check its surjectivity. But is this necessarily true?
Let $K=\ker\phi$. Suppose the highest powers of $p$ dividing $K,G,H$ are $p^r,p^n,p^s$ respectively. Observe that we have the relation $p^r\cdot p^s=p^n$. Restricting the homomorphism $\phi$ to $P$ has kernel $K\cap P$, so we compute the image's size $|\phi(P)|=|P|/|K\cap P|$. Bound $|K\cap P|$ from above in order to bound $|\phi(P)|$ from below and conclude that $\phi(P)$ is always a $p$-sylow subgroup of $H$.
To check surectivity, use the fact that sylow subgroups are conjugate.
