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