# Sylow's Counting Argument

1. Feb 22, 2008

### mathusers

Hi last one here. Any hints on this is appreciated too :)

Let G be a group of order 44. Show using Sylow's counting that G has a normal subgroup of order 11. Use the results to classify all groups of order 44.

2. Feb 22, 2008

### NateTG

How many subgroups of order 11 are there? (There are a bunch of other threads about Sylow counting arguments just like this one.)

3. Feb 23, 2008

### mathusers

ok here is what i have managed to get so far

By Sylow's 1st theorem, G has a subgroup of order 11. let $n_p$ be the number of sylow p-subgroups. then $n_{11} = 1(mod 11)$ and $n_{11}$ divides 2^2 (=4) so $n_{11}=1$. Therefore, it must be a normal subgroup (since it has no distinct conjugates).

How can we use this to classify all groups of order 44 though?

4. Feb 23, 2008

### morphism

Well, what are your thoughts? The obvious thing is to think about the Sylow 2s. There's either 1 or 11 of them (why?). If there's just 1, say H, then G is isomorphic to the direct product of H and the Sylow 11 (why?). Now what?

Scenario 2: there's 11 Sylow 2s. Take one of them and play around with it and with the Sylow 11. Remember that the Sylow 11 is cyclic and normal in G -- this will probably be useful in getting a presentation for G.