I am studying for a qualifying exam and am working on the following problem.
If G is a group of order 2010=2*3*5*67, then G has a normal subgroup of order 5.
I know that the number of 67-Sylow subgroups is 1, lets's call it H. So H is normal in G. We also know that the order of H is 67.
Since sylow-p subgroups exist, there is a subgroup K with order 5.
Since H is normal and K is a subgroup, HK is a subgroup of G.
Since the orders of H and K are relatively prime, their intersection is trivial so the order of HK is 5*67.
Also since the orders of H and K are prime, they are cyclic, and again because they are relatively prime, HK is cyclic and hence abelian.
Since abelian groups have a subgroup the size of each divisors of its order, then HK has a subgroup of order 5, call it M.
This is where I am stuck. My study buddy says HK is normal, but can't give me a reason.
He states exactly that "M is a subgroup of HK, HK is normal in G, and since HK is cyclic, we can conclude that M is normal in G."
I don't see why HK is normal and how you can conclude that M is normal in G because HK is cyclic.