# Problem re: Sylow's Theorems

• Jan 17th 2011, 09:12 AM
r45
Problem re: Sylow's Theorems
Hi there,

I had this question on an exam today and I'm not sure how to do it.

"Let G = $\mathbb{Z}_3$ x $A_4$. List all the Sylow 2-subgroups of G, and all the Sylow 3-subgroups of G, giving reasons that your lists are complete"

I found that G has order 36, and so I determined that $n_2$ = 1, 3 or 9 and that $n_3$ = 1 or 4 (where $n_i$ denotes the number of Sylow i-subgroups) but from there I couldn't do much. By brute force I managed to find one subgroup of order 4 and one of order 9, so in the end I just put that $n_2$ = 1 and $n_3$= 1.

Can anyone verify, or correct this?

Many thanks
• Jan 17th 2011, 09:32 AM
Drexel28
Quote:

Originally Posted by r45
Hi there,

I had this question on an exam today and I'm not sure how to do it.

"Let G = $\mathbb{Z}_3$ x $A_4$. List all the Sylow 2-subgroups of G, and all the Sylow 3-subgroups of G, giving reasons that your lists are complete"

I found that G has order 36, and so I determined that $n_2$ = 1, 3 or 9 and that $n_3$ = 1 or 4 (where $n_i$ denotes the number of Sylow i-subgroups) but from there I couldn't do much. By brute force I managed to find one subgroup of order 4 and one of order 9, so in the end I just put that $n_2$ = 1 and $n_3$= 1.

Can anyone verify, or correct this?

Many thanks

$A_4$ has four subgroups of order $3$. For example, $\left|\langle (1,2,3)\rangle\right|=3$. It follows then that $n_3=4$ (these can actually be all found by using the fact that all Sylow $p$-subgroups are conjugate). Similarly, there is one subgroup of order $4$ of $A_4$ namely $\{e,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}$ from where it pretty easily follows that $n_2=1$.
• Jan 17th 2011, 09:48 AM
r45
I thought that when looking for $n_3$ I was trying to find subgroups of order 9? Because $36 = 3^{2}.4$, so in this case Sylow 3-subgroups have order $3^{2}=9$?
• Jan 17th 2011, 09:53 AM
r45
So to clarify on our Sylow 2-subgroup in the context of G as I defined above, would it be:

$\{(1, (e)), (1, (13)(24)), (1, (12)(34)), (1, (14)(23))\}$?

By the way thankyou very much for the response.
• Jan 17th 2011, 10:19 AM
tonio
Quote:

Originally Posted by r45
I thought that when looking for $n_3$ I was trying to find subgroups of order 9? Because $36 = 3^{2}.4$, so in this case Sylow 3-subgroups have order $3^{2}=9$?

You're right. Drexel apparently misread and thought you were talking of $A_4$ and not of $G:=\mathbb{Z}_3\times A_4$...or perhaps

he thought to mention the 3-Sylow sbgps. of $A_4$ as a hint for you.

Anyway, his description gives you some hints: any 3-Sylow sbgp. of G will be the product of $\mathbb{Z}_3$ with a sbgp. of order 3 of $A_4$ ...

Tonio
• Jan 17th 2011, 12:12 PM
Drexel28
Quote:

Originally Posted by tonio
or perhaps

he thought to mention the 3-Sylow sbgps. of $A_4$ as a hint for you.

Anyway, his description gives you some hints: any 3-Sylow sbgp. of G will be the product of $\mathbb{Z}_3$ with a sbgp. of order 3 of $A_4$ ...

Tonio

This is what I was doing, sorry if there was any confusion!
• Jan 18th 2011, 12:07 AM
Swlabr
You can prove there is one Sylow 2-subgroup by proving that your subgroup of order 4 is normal; in general, once you have a maximal subgroup of order $p^i$, conjugate it with everything you can find to find the other Sylow p-subgroups...