G contains a normal p-Sylow subgroup

[mathmari]

Hey!

Let $G$ be a non-abelian finite group with center $Z>1$.
I want to show that if $G/Z$ is a $p$-group, for some prime $p$, then $G$ contains a normal $p$-Sylow subgroup and $p\mid |Z|$.

We have that $$|G/Z|=p^n, n\geq 1\Rightarrow \frac{|G|}{|Z|}=p^n\Rightarrow |G|=p^n|Z|$$ That means that there are $p$-Sylow in $G$, right? (Wondering)

Now we have to show that there is only one $p$-Sylow, or not? (Wondering)

How could we do that? (Wondering)

 

[johng1]

Let $P$ be a Sylow p subgroup of $G$. Then $PZ/Z$ is the Sylow p subgroup of $G/Z$; i.e. $G=PZ$. Then easily $P$ is normal in $G$. Then also $Z(P)\subseteq Z(G)$ and so p divides the order of the center of G.

 

[mathmari]

Let $P$ be a Sylow p subgroup of $G$. Then $PZ/Z$ is the Sylow p subgroup of $G/Z$; i.e. $G=PZ$.

Why is $PZ/Z$ the Sylow p subgroup of $G/Z$ and not $P/Z$ ? And why does it hold that $G=PZ$ ? (Wondering)

 

[johng1]

For $G$ any finite group and $p$ any prime,

1. If $G$ is a $p$ group, then $G$ is the Sylow $p$ subroup.

2. If $N$ is any normal subgroup of $G$ and $P$ is a Sylow $p$ subgroup of $G$, then $PN/N$ is a Sylow $p$ subgroup of $G/N$.

So in your problem, $G/Z$ is the Sylow $p$ subgroup of $G/Z$ and $PZ/Z$ is a Sylow $p$ subgroup. So $G/Z=PZ/Z$ and thus $G=PZ$.

 

[mathmari]

1. If $G$ is a $p$ group, then $G$ is the Sylow $p$ subroup.

Do you mean that in that case there is just one Sylow subgroup? (Wondering)

2. If $N$ is any normal subgroup of $G$ and $P$ is a Sylow $p$ subgroup of $G$, then $PN/N$ is a Sylow $p$ subgroup of $G/N$.

Why does it hold that then $PN/N$ is a Sylow $p$ subgroup of $G/N$ ? (Wondering)

 

[mathmari]

Let $P$ be a Sylow p subgroup of $G$. Then $PZ/Z$ is the Sylow p subgroup of $G/Z$; i.e. $G=PZ$. Then easily $P$ is normal in $G$. Then also $Z(P)\subseteq Z(G)$ and so p divides the order of the center of G.

Isn't it as follows? (Wondering)

Since $G/Z$ is a $p$-group, it contains $p$-Sylow subgroups, say $P$.
From the correspondence theorem we have that there is a bijective mapping between the subgroups of $G$ that contain $Z$ and the subgroups of $G/Z$,
$$\phi (A)\mapsto A/Z, \ A\in G$$
So, the corresponding $p$-Sylow of $G$ exists and it is the $PZ$, right? (Wondering)