MHB Inequality related to number of p-Sylow subgroups

[mathmari]

Hey!

I want to show that if $G$ is finite and $f:G\rightarrow H$ is a group epimorphism then $|\text{Syl}_p(G)|\geq |\text{Syl}_p(H)|$. I have done the following:

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? (Wondering)

 

[caffeinemachine]

Hey!

I want to show that if $G$ is finite and $f:G\rightarrow H$ is a group epimorphism then $|\text{Syl}_p(G)|\geq |\text{Syl}_p(H)|$. I have done the following:

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? (Wondering)

Let $G$ be a finite group and $N$ be a normal subgroup of $G$. We will show that the number of Sylow $p$-subgroups of $G$ is at least the number of Sylow $p$-subgroups of $G/N$. From here the problem you have posted can be easily solved.
Let $|G|=p^am$, and $|N|=p^bn$, where $m$ and $n$ are relatively prime to $p$. Let $B$ be a Sylow $p$-subgroup of $G/N$. Then there is a subgroup $Q$ of $G$ such that $QN/N=B$. It is easy to see that $|QN|=p^{a}n$. Therefore $QN$ contains a Sylow $p$-subgroup of $G$.

Claim. Let $Q$ be a subgroup of $G$ such that $QN/N$ is a Sylow $p$-subgorup of $G$. Let $P$ be a Sylow $p$-subgorup of $G$ contained in $QN$. Then $PN=QN$.
Proof. Note that $|PN/N|= |P|/|P\cap N|$. Thus $|PN/N|\geq p^{a}/p^{b}$. This inequality cannot be strict and thus we must have $|PN|=p^a n = |QN|$. Since $PN\leq QN$, we conclude that $PN=QN$.

Now suppose $Q_1N/N, \ldots, Q_kN/N$ be all the Sylow $p$-subgroups of $G/N$, where $Q_1, \ldots, Q_k$'s are some subgroups of $G$. For each $i$, let $P_i$ be a Sylow $p$-subgroup of $G$ contained in $Q_iN$. We claim that $P_1, \ldots, P_k$ are pairwise distinct. To see this, suppose $P_i=P_j$. Then $P_iN=P_jN$, which by the above claim gives $Q_iN=Q_jN$. This means $Q_iN/N=Q_jN/N$, forcing $i=j$.

This completes the proof.

 

[Deveno]

You can also (in a non-rigorous way) think of it like this: an epimorphism $f:G \to H$ might "shrink" some of the Sylow $p$-subgroups of $G$ to a subgroup that is no longer "big enough" to be a Sylow $p$-subgroup of $H$ (this depends on the size of $P \cap \text{ker }f$, for $P \in \text{Syl}_p(G)$), but you're never going to get "more" since every subgroup of $H$ comes from a subgroup of $G$ "over $\text{ker }f$".

In other words, just by preserving the multiplicative property, homomorphisms also preserve certain containment relationships between subgroups of $G$ reflected in $H$, this is often expressed by saying we have a "lattice homomorphism" (most texts on group theory don't discuss lattices as algebraic structures, so they express this other ways).

(maybe someone else can draw a picture).