Proof of no simple group of order 992 
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.  
 A: 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.
A: 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. 
