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Prove there are no simple groups of order $992$.

Factorise it. $31 \times 2^5 $ so you have $|G|=31 \times 2^5 \geq n_{31}(31-1)+ n_{2}(2^5-1)+1$

Putting it in Sylow theorem. So how do you get the contradiction? Or is this totally wrong.

I need to use Sylow theorem to prove this. Hmm, can someone describe how you prove this. I know the Sylow the theorems well the proof of them.

It seems from the notes you have to start with $n_{2}>1$ and then $n_2=1(mod2)$. However, I don't understand this at all.

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The $2$-Sylow subgroups could have non-trivial intersection, since a group of order $32$ need not be cyclic, so your inequality is not correct. –  Zev Chonoles Jan 23 '12 at 15:57
    
@ZevChonoles What inequality do you use? I got given two ways, one is a formula like that and the other is a method I don't understand. –  simplicity Jan 23 '12 at 16:03
    
How many possible Sylow 31-groups can there be? Remember that groups of order 31 are cyclic, so you can be assured the different Sylows have trivial intersection. –  user641 Jan 23 '12 at 16:15
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I like this particular number, since you can have 31 cyclic Sylow 2-subgroups, but instead of 31*(32-1)+1 = 962 elements of order a power of 2, you can get only 512 elements of order a power of 2. The Sylows overlap quite a bit. –  Jack Schmidt Jan 23 '12 at 16:29
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What you write is not correct: the number of elements of order a power of $2$ is not necessarily equal to $2^5-1$ times the number of $2$-Sylow subgroups because, as Zev Chonoles points out, you do not have that two distinct $2$-Sylow subgroups must intersect trivially (which is what goes behind that particualr inequality).

The number of $31$-Sylow subgroups must divide $31\times 32$ and be congruent to $1$ modulo $31$; so either there is a single $31$-Sylow subgroup (in which case the group is not simple), or there are thirty two $31$-Sylow subgroups.

If there are thirty two $31$-Sylow subgroups, then since any two distinct ones must intersect trivially (the groups are cyclic of prime order, so the only proper subgroup is trivial), they account for $32(31-1) + 1$ elements of $G$.

That means that there are $32\times 31 - 32\times 30 = 32$ elements whose order is not $31$. Since a $2$-Sylow subgroup must contain $32$ elements, there are only enough elements left over for a single $2$-Sylow subgroup, which must therefore be normal.

So $G$ will have either a single $31$-Sylow subgroup, or a single $2$-Sylow subgroup. Either way, it is not simple.

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How do you know there are 32 subgroups. This confused me more. What you using to conclude number of 31-sylow subgroup divides $31\times32$? –  simplicity Jan 23 '12 at 16:28
    
@simplicity: Sylow's Third Theorem states: "If $G$ is a finite group and $p$ is a prime, then the number of $p$-Sylow subgroups of $G$ divides $|G|$ and is congruent to $1$ modulo $p$." Since $|G|=992 = 2^5\times 31=32\times 31$, and the only divisors of $|G|$ that are congruent to $1$ modulo $31$ are $1$ and $32$, then the number of $31$-Sylow subgroups of $G$ is either $1$ or $32$. –  Arturo Magidin Jan 23 '12 at 16:31
    
Oh I feel so silly now. I thought you was doing (31+1), but yeah I get that now. Oh, yes that makes perfect sense now. The proof makes sense. –  simplicity Jan 23 '12 at 16:34
    
Cam you give motivation for $32(31-1)+1$ comment. –  simplicity Jan 23 '12 at 16:42
    
You could immediately say that the number of $31$-Sylows divides $32$, right? It's the index of the normalizer. This is minor, since the right numbers pop out quickly. [Also, did you learn from Lang's book? It's the only one I've seen that uses "$p$-Sylow" :)] –  Dylan Moreland Jan 23 '12 at 16:48
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