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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.2012-01-23
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    @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.2012-01-23
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    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.2012-01-23
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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.2012-01-23

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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 $31\times32$?2012-01-23
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    @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 $32$.2012-01-23
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    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.2012-01-23
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    Cam you give motivation for $32(31-1)+1$ comment.2012-01-23
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    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" :)]2012-01-23
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    @Dylan: But it must be congruent to $1$ modulo $32$ as well; so it is either $1$ or $32$.2012-01-23
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    @simplicity: Let's count how many elements are in the thirty two $31$-Sylow subgroups. Any two of them intersect trivially; each one of them has $31-1 = 30$ nonidentity elements (each of order $31$); and none of them is in more than one $31$-Sylow subgroup. So for each $31$-Sylow subgroup, there are 30 elements that are not in any other $31$-Sylow subgroup, plus the identity. So if you take the union of the thirty two $31$-Sylow subgroups, that *set* has $32(31-1)+1$ elements: 30 elements for each of the 32 groups, plus the identity.2012-01-23
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    @Dylan: Actually, I learned from Herstein, in Spanish; but my graduate course *was* from Lang. And, yes, in general, if $n_p$ is the number of Sylow subgroups corresponding to $p$, and $|G|=p^mr$, with $\gcd(p,r)=1$, then $n_p|r$, not just $|G|$, simply because $\gcd(n_p,p)=1$.2012-01-23
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So this line of proof shows that there are 960 elements of order 31 and 1 of order 1, leaving only 31 elements left over for the 2-Sylow subgroup, so these elements together with 1 are THE only 2-Sylow subgroup which then has to be normal, so G can't be simple. It is interesting that there IS a non-super-solvable group with this description. GL (5,2) contains elements of order 31. Select one, call it $\phi$. Then $\phi$ can be viewed as a map from $Z_2^5$ to $Z_2^5$, and so one can construct the semi-direct product of $Z_2^5$ and $Z_31$, using $\phi$ and its powers to construct the map of the semi-direct product. This group has 960 elements of order 31 and a normal subgroup of order 32. I call this MG-31, where MG stands for "Mersenne Group", since 31 is a Mersenne number.