# Prove that no group of order 56 can be simple using Sylow theorems

Prove that no group of order $56$ can be simple using steps

●finding sylows number 2-subgroups and sylow 7-subgroups
●explain why any of sylow 7-subgroups must intersect trivially, but this is not the case for sylow 2-subgroup
● use counting argument to show that if there is more than one sylow 7 subgroup, then there can be only one sylow 2 subgroup
● Show that GROUP of order $56$ is not simple.

my question is why does this argument fails to show that no group of order 112 can be simple

• Why do you expect the argument to fail? Commented Dec 12, 2014 at 12:22
• I think because the intersection.but I'm not sure Commented Dec 12, 2014 at 12:25
• I mean, did you try to work through the same method and something failed? Commented Dec 12, 2014 at 12:25
• There exists a group of order $112$ with $|{\rm Syl}_7(G)|=8$ and $|{\rm Syl}_2(G)|>1$, so that part of the argument must fail! Commented Dec 12, 2014 at 12:27
• Yes ..sylow group number is not same Commented Dec 12, 2014 at 12:27

If in a group of order $56 = 2^3 \cdot 7$ we have $8$ Sylow $7$-subgroups, then we get $8 \cdot (7-1) = 48$ elements of order $7$ (as the intersections of Sylow $7$'s are trivial). Thus we have $56 - 48 = 8$ elements left for the Sylow $2$-subgroups, which have order $8$. Thus there must be just one.
In a group of order $112 = 2^4 \cdot 7$ we cannot obtain the same by counting elements. It's possible that $N(7) = 8$ and $N(2^4) = 7$.