Using Sylow's Counting to Classify Groups of Order 44

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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.
 
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How many subgroups of order 11 are there? (There are a bunch of other threads about Sylow counting arguments just like this one.)
 
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?
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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