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? 
 A: 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.
A: Let $K$ denote the kernel of $f$. By the correspondence theorem, $f$ induces a bijection between subgroups of $H$ on one hand, and subgroups of $G$ which contain $K$ on the other hand. This correspondence preserves indices.
Let $Q \in Syl_p(H)$. Let $L$ denote the unique corresponding subgroup of $G$ which contains $K$, specifically $L= f^{-1}(Q)$. Note that $f(L) = Q$ since $f$ is surjective.
As $Q \in Syl_p(H)$, the index $|H:Q|$ is not divisible by $p$. Since indices are preserved under the correspondence theorem, we have $|H:Q| = |G:L|$, so the latter is also not divisible by $p$. Therefore $L$ is divisible by the full power of $p$ dividing $G$. This means that there is some $P \in Syl_p(G)$ which is contained in $L$. 
Now, $f(P) = f(PK) \simeq PK/K$, so $|H:f(P)| = |G:PK|$, which is not divisible by $p$ since $|G:P| = |G:PK||PK:P|$. On the other hand, $|f(P)| = |PK|/|K| = |P|/|P \cap K|$, which is a power of $p$.
Therefore, $f(P) \in Syl_p(H)$, and since $f(P) \leq f(L) = Q$ and $Q \in Syl_p(H)$, this forces $f(P) = Q$.
We have thus shown that every $Q \in Syl_p(H)$ is equal to $f(P)$ for some $P \in Syl_p(G)$, so $|Syl_p(G)|$ must be at least as large as $|Syl_p(H)|$.

Answering questions raised in the comments:


Why does it follow that $f(L) = Q$ from the fact that $f$ is surjective?

In general, if $f:A \to B$ is a surjection, and $C$ is a subset of $B$, then $f(f^{-1}(C)) = C$. This is true even if $A,B,C$ are ordinary sets, not necessarily groups.

So that the inverse of $f$ exists, shouldn't $f$ be 1-1 and onto? In this case we have only that $f$ is onto, right?

$f^{-1}(Q)$ is the inverse image of $Q$, in other words, the subset (subgroup) of $G$ consisting of all elements which are mapped into $Q$ by $f$. The inverse image exists whether or not $f$ is invertible.

Ah ok... Is it known that when $Q \in Syl_p(H)$, then the index $|H:Q|$ is not divisible by $p$? Or do we have to prove it?

Sure, if $Q \in Syl_p(H)$, then by definition $|Q|$ is the highest power of $p$ dividing $|H|$. Then, since $|H|=|Q||H:Q|$, the index $|H:Q|$ cannot also be divisible by $p$.

At the part "Therefore $L$ is divisible by the full power of $p$ dividing $G$", we have that if $|G|=p^ma$ with $p \not\mid a$, then $|L| = p^m b$ with $p \not\mid b$, right? Do we conclude that there is $p$-Sylow subgroup of $G$, say $P$, that is contained in $L$, because $|L|\geq|P|=p^m$?

Yes, apply the Sylow existence theorem to $L$ to conclude that $L$ contains a subgroup $P$ of order $p^m$, i.e. $P \in Syl_p(L)$. Since $p^m$ is the highest power of $p$ dividing $|G|$, in fact $P \in Syl_p(G)$ as well.

Also, could you explain to me why the following holds? $f(P)=f(PK)\simeq PK/K$

$P \leq PK$, so $f(P) \leq f(PK)$. On the other hand, an arbitrary element of $PK$ is of the form $pk$ with $p \in P$ and $k \in K$. Then $f(pk) = f(p)f(k) = f(p)$ since $k$ is in the kernel of $f$. Therefore, $f(PK) \leq f(P)$. We conclude that $f(P) = f(PK)$. Now observe that $K$ is contained in $PK$, so the kernel of $f$ restricted to $PK$ is still $K$. Therefore by the first isomorphism theorem, applied to $f|_{PK}$, we conclude that $f(PK) = f|_{PK}(PK) \simeq PK/K$.
Alternatively, if we restrict $f$ to $P$, then the kernel of the restriction $f|_P$ is $P \cap K$. Then by the first isomorphism theorem applied to $f|_P$ we get $f(P) = f|_P(P) \simeq P / (P \cap K)$. But by the second isomorphism theorem, $P / (P \cap K) \simeq PK / K$.

Ah ok... I think I got it... I have also an other question... We have that $P \leq L$, does this imply that $f(P)\leq f(L)=Q$? Also how do we conclude that $f(P)=Q$?

$P$ is contained in $L$, so $f(P)$ is contained in $f(L)$. Then, since $P$ and $L$ are subgroups, so are their images, so in fact $f(P)$ is a subgroup of $f(L)$. We observed that $f(L) = Q$ in the second paragraph. See also my answer to your first question in the comments.
Therefore, $f(P) \leq f(L) = Q$. Since $f(P)$ and $Q$ are both $p$-Sylow subgroups of $H$, they both have the same number of elements, so the containment must in fact be an equality: $f(P) = Q$.

We have that $f(P)$ and $Q$ are both $p$-Sylow subgroups of $H$. Aren't they then conjugate? I haven't really understood why they must be equal...

Yes, $p$-Sylow subgroups are always conjugate. However, in this case we have $f(P)$ is contained in $Q$, and they have the same order (size), so they must be equal (so they are trivially conjugate, by the identity element). This is true even for sets - forget about groups for a minute. If $A$ and $B$ are sets of the same finite size and $A$ is contained in $B$ then $A$ must equal $B$.

I have also an other question... Suppose that $f:G \to H$ is a group epimorphism and $A \leq G$. Does it hold that  $|A| = |f(A)|$?

No, this does not hold. For a counterexample, let $H$ be the trivial group and let $f$ map everything in $G$ to the identity element of $H$. Then $f$ is a group epimorphism but $|A| = |f(A)|$ does not hold except when $A$ is the trivial subgroup.
What you can say is that if $K$ is the kernel of $f$, then $|A| = |f(A)||K \cap A|$. This follows from the first isomorphism theorem, because the kernel of $f|_A$ (the restriction of $f$ to $A$) is $K \cap A$, so $f(A) = f|_A(A) \simeq A / (K \cap A)$, and therefore $|f(A)| = |A| / |K \cap A|$.
