Is \( f(P) \) a \( p \)-Sylow Subgroup of \( H \)?

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In summary, we have shown that if $G$ is finite and $f:G\rightarrow H$ is a group epimorphism, then $f(P)\in \text{Syl}_p(H)$ for all $P\in \text{Syl}_p(G)$. This is because $f(P)$ is a $p$-subgroup of $H$ and its order divides $p^k$, and $f$ being surjective implies that $f(P)$ is a subgroup of $H$. Additionally, the correspondence theorem shows that $f(P)$ has the same index as $P$, making it a maximal $p$-subgroup and thus a $p$-Sylow subgroup of
  • #1
mathmari
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Show that it is a Sylow subgroup

Hey! :eek:

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)
 
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  • #2
Re: Show that it is a Sylow subgroup

mathmari said:
Hey! :eek:

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.
 
  • #3
Re: Show that it is a Sylow subgroup

Deveno said:
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)
Deveno said:
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)
Deveno said:
It is obvious it is a subgroup of a Sylow $p$-subgroup

Why do we have that? (Wondering)
 
  • #4
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)
 
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FAQ: Is \( f(P) \) a \( p \)-Sylow Subgroup of \( H \)?

1. What is a Sylow subgroup?

A Sylow subgroup is a type of subgroup that plays a significant role in group theory. It is a subgroup of a finite group that has the highest possible order for its size, meaning it is the largest subgroup of that size that can be found within the group.

2. How are Sylow subgroups useful in group theory?

Sylow subgroups are useful in group theory because they provide important information about the structure of a finite group. They can help determine the number of subgroups of a certain size within a group, and they can also be used to prove the existence of normal subgroups.

3. What is the Sylow theorems?

The Sylow theorems are a set of theorems that provide valuable insights into the properties of Sylow subgroups. These theorems state that for any finite group, there will exist a Sylow subgroup of each possible size, and the number of these subgroups will be related to the order of the group.

4. How are Sylow subgroups related to the concept of conjugacy?

Sylow subgroups are closely related to the concept of conjugacy in group theory. This is because any two Sylow subgroups of the same size are conjugate to each other, meaning they are equivalent in terms of their properties and structure.

5. Can Sylow subgroups be used in other areas of mathematics?

Yes, Sylow subgroups have applications in other areas of mathematics such as number theory and geometry. They can also be used in the study of finite fields and Galois theory, as well as in the construction of error-correcting codes in computer science.

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