Sentences about Sylow subgroups

[mathmari]

Show that it is a Sylow subgroup

Hey!

I am looking at the following exercise:

If $G$ is finite and $f:G\rightarrow H$ is a group epimorphism, show that if $P\in \text{Syl}_p(G)$ then $f(P)\in \text{Syl}_p(H)$. I have done the following:

Suppose that $|G|=p^km$, where $p\not\mid m$.
Since $P\in \text{Syl}_p(G)$ we have that $|P|=p^k$.
$f: G\rightarrow H$ is a group endomorphism, so it is a bijective map. Therefore, we have that $|G|=|f(G)|=|H|\Rightarrow |H|=p^km$.
So, every subgroup of $H$, say $S$, with $|S|=p^k$ is a $p$-Sylow subgroup of $H$, right? (Wondering)
Therefore, we have to show that $|f(P)|=p^k$, or not? (Wondering)
Does it stand that $|f(P)|=p^k$ because $f$ is bijective? (Wondering)

 

[Deveno]

Re: Show that it is a Sylow subgroup

Hey!

I am looking at the following exercise:

If $G$ is finite and $f:G\rightarrow H$ is a group epimorphism, show that if $P\in \text{Syl}_p(G)$ then $f(P)\in \text{Syl}_p(H)$. I have done the following:

Suppose that $|G|=p^km$, where $p\not\mid m$.
Since $P\in \text{Syl}_p(G)$ we have that $|P|=p^k$.
$f: G\rightarrow H$ is a group endomorphism

Epimorphism, which means surjective (one has to be careful, here. In the category of rings, epimorphisms are not necessarily surjective, and it is a non-trivial theorem of group theory that epimorhisms are surjective). An endomorphism is a homomorpism (not necessarily injective) $G \to G$.

...so it is a bijective map.

No, not even close to true.

Therefore, we have that $|G|=|f(G)|=|H|\Rightarrow |H|=p^km$.

No, epimorphisms are not, in general, bijective. Bijective homomorphisms are isomorphisms.

So, every subgroup of $H$, say $S$, with $|S|=p^k$ is a $p$-Sylow subgroup of $H$, right? (Wondering)
Therefore, we have to show that $|f(P)|=p^k$, or not? (Wondering)
Does it stand that $|f(P)|=p^k$ because $f$ is bijective? (Wondering)

No, and no.

If $g \in P$, for some $P \in \text{Syl}_p(G)$, then $g$ has order $p^k$ for some $k$. Thus $g^{p^k} = e_G$.

It follows since $f$ is a homomorphism that $f(g)^{p^k} = f(g^{p^k}) = f(e_G) = e_H$. Thus the order of $f(g)$ divides $p^k$, so it is $p^m$ for some $0 \leq m \leq k$.

Now $\langle f(g)\rangle$ is a $p$-subgroup of $H = f(G)$ since $f$ is surjective. Thus $f(g) \in f(P)$ and $f(P) \subseteq f(H)$.

Your goal is now to show that $f(P)$ is some Sylow $p$-subgroup of $H$.It is obvious it is a subgroup of a Sylow $p$-subgroup, so assume it is not a maximal $p$-subgroup, and derive a contradiction.

 

[mathmari]

Re: Show that it is a Sylow subgroup

Epimorphism, which means surjective (one has to be careful, here. In the category of rings, epimorphisms are not necessarily surjective, and it is a non-trivial theorem of group theory that epimorhisms are surjective). An endomorphism is a homomorpism (not necessarily injective) $G \to G$.



No, not even close to true.
No, epimorphisms are not, in general, bijective. Bijective homomorphisms are isomorphisms.

No, and no.

Ah ok... (Thinking)

Now $\langle f(g)\rangle$ is a $p$-subgroup of $H = f(G)$ since $f$ is surjective.

We have that the order of $f(g)$ is a power of the prime $p$. Does this mean that it generates a cyclic subgroup? I got stuck right now... (Wondering)
What exactly do we get from the fact that $f$ is surjective? (Wondering)

It is obvious it is a subgroup of a Sylow $p$-subgroup

Why do we have that? (Wondering)

 

[mathmari]

I thought about it again... (Thinking)

Could we show it also as follows? Since $P\in \text{Syl}_p(G)$, we have that $P$ is a subgroup of $G$. Do we have from the correspondence theorem that $f(P)$ is a subgroup of $H$ ? (Wondering)

Suppose that this is true.

From the correspondence theorem we also have that $[G:P]=[H:f(P)]$, right? (Wondering)

Since $P\in \text{Syl}_p(G)$, we have that $[G:P]$ is coprime with $p$.

That means that $[H:f(P)]$ is also coprime with $p$. Do we conclude in that way that $f(P)$ is a $p$-Sylow subgroup of $H$ ? (Wondering)