# Thread: Sylow Subgroups

1. ## Sylow Subgroups

I'm not sure with the following:

(1) What is the order of every Sylow 5-subgroup of $A_5 \times A_5$ ?

(2) Let G and H be arbitrary finite groups. Prove that every sylow p-subgroup of $G \times H$ has the form $P_1 \times Q_1$ where $P_1$ is a sylow p-subgroup of G, and $Q_1$is a sylow p-subgroup of H.

(3) Find the number of sylow 5-subgroups of $A_5 \times A_5$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For (1) i think it's 25, as its's 5.5 ?

For (2) I think I use the Conjugacy sylow theorem somehow?

For (3) I think it might be 26 because $A_5 \times A_5$ is simple?

Thanks!

2. Originally Posted by Jason Bourne
I'm not sure with the following:

(1) What is the order of every Sylow 5-subgroup of $A_5 \times A_5$ ?

(2) Let G and H be arbitrary finite groups. Prove that every sylow p-subgroup of $G \times H$ has the form $P_1 \times Q_1$ where $P_1$ is a sylow p-subgroup of G, and $Q_1$is a sylow p-subgroup of H.

(3) Find the number of sylow 5-subgroups of $A_5 \times A_5$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

For (1) i think it's 25, as its's 5.5 ?

For (2) I think I use the Conjugacy sylow theorem somehow?

For (3) I think it might be 26 because $A_5 \times A_5$ is simple?

Thanks!
Hi Jason Bourne.

(1) Your answer is correct.

(2) Determine the maximum power of $p$ dividing the order of $G\times H.$

(3) No, it can’t be 26. The order of $A_5\times A_5$ is $3600 = 2^4\times3^2\times5^2.$ The number of Sylow 5-subgroups must divide $2^4\times3^2=144.$ 26 does not divide 144.

3. for part (3) use part (2) and the fact that $A_5$ has 6 Sylow 5-subgroups. to prove that $P \times Q$ is a Sylow p-subgroup of $G \times H$ if $P$ is a Sylow p-subgroup of $G$ and $Q$ is a Sylow p-subgroup $H$

do as TheAbstractionist suggested. but the converse is less trivial and it's the only fairly interesting part of your problem!