- #1
mathmari
Gold Member
MHB
- 5,049
- 7
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)
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)
Last edited by a moderator: