Let $G$ be a finite abelian group and let p be a positive prime number that divides the order of $G$. Prove that if H and K are Sylow p-subgroups, then $H = K$.
I'll first assume that $H$ and $K$ are Sylow $p$-subgroups which tells me $H=p^n$ and $K= p^n$...I donʻt even know if that's correct haha. I am trying not to hate group theory, help please (RA > GT). Just need some motivation to set me in right direction. Clearly I know I will use the fact that it is abelian to show $H=K$ at some point haha. :)
Since $H$ and $K$ are Sylow $p$-subgroups, they are conjugate by Sylow's Theorems. Hence $gHg^{-1}=K$ for $g\in G$; that is, $gH=Kg$. But $G$ is abelian, so $gH=gK$. Cancel $g$ by multiplying by $g^{-1}$ on the left.