If H is a normal subgroup of a finite group G and |H|=p^K for some prime p, show H is contained in every p-Sylow subgroup of G. I am totally lost on this one, if anyone could help me it would be great
so here is what I have so far:
Let G be a finite group with H a normal subgroup of G such that |G|=p^k for some prime p. By Sylow's second theroem, H is contained in some Sylow p-subgroup of G. Sylow's third thereom says that any two Sylow p-subgroups of G are conjugate.
I am not sure after this, is that any two Sylow 0-subgroups being conjugate enough to show that H is contanied in ever p_Sylow subgroup of G since we know it is contained in one of the p-Sylow subgroups of G?
Sorry i am kind of slow on some of this stuff, i really appreciate the help.
Suppose $\displaystyle H\subseteq P$, where $\displaystyle P$ is some $\displaystyle p$-Sylow subgroup. Then, if $\displaystyle Q$ is another $\displaystyle p$-Sylow subgroup, we can find $\displaystyle g\in G$ so that $\displaystyle Q=gPg^{-1}$. Then, we know that $\displaystyle gHg^{-1}\subseteq Q$, but what is this?
If H is a normal subgroup of G than aH=Ha for all a in G. so aH(a^-1)=H, which means that if H conjugate is contained in Q, than H is contained in Q, right? So than we can apply this to every p-Sylow subgroup of G to show that a normal subgroup H of G is contained in every p-Sylow subgroup of G?