intersection of sylow subgroup

Question

I am in need of

• Example of a group $G$ a subgroup $A$ which not normal in $G$ and $p$ sylow subgoup $B$ of $G$ such that $A \cap B$ is a $p$-sylow subgroup of $A$.

A detailed solution will be helpful.

Answer 1

I think, you need a group $G$ of order $|G|=p^nm,(p,m)=1$ in which $p$ is a prime and it has only one $p-$sylow subgroup. Such these groups may be found among the groups of order $pq$ for example. If we accept there is a $p$-sylow subgroup of $G$, say $B$, with $|B|=p^n$ then we can conclude that $A\cap B$ is a $p$-sylow subgroup of $G$. If fact,

• $A\leq G$ then $|A|||G|$ and then $|A|=p^sk,~~k\mid m,~~ s\leq n$,

• $|A\cap B|\leq |G|$ so $|A\cap B|=p^t,~~ t\leq n$,

• $B$ is normal in $G$ and so $AB\leq G$ and so $|A B|||G|$,

Expanding $|AB|||G|$, we will lead to this result that $t=s$

Answer 2

If you want to construct an example which is non-trivial under several points of view, that is $A$ and $B$ and $G$ all distinct, and $A$ is not a $p$-group, you may find it as follows.

Find a finite group $G$ which has a non-normal subgroup $A$ which is not a $p$-group, but whose order is divisible by $p$. Let $S$ be a Sylow $p$-subgroup of $A$. Choose $B$ to be a Sylow $p$-subgroup of $G$ containing $S$. (Such a $B$ exists by Sylow's theorems.)

For instance, take $p = 2$, $G = S_{4}$, $A = \langle (123), (13) \rangle$ of order 6, $S = \langle (13) \rangle$, $B = \langle (1234), (13) \rangle$.

PS A simpler, but somewhat trivial example would be $G = B = \langle (1234), (13) \rangle$, $A = \langle (13) \rangle$.