Let |G| = (p^n)m where p is prime and gcd(p,m) = 1. Suppose that H is a
normal subgroup of G of order p^n. If K is a subgroup of G of order p^k,
show that K is subgroup of H.
Okay, I wonder if there is more I need to do, or if I need to prove they
are finite. I feel like I am missing something...but here is what I got
p^k has to be less than p^n because if p^k was bigger than p^n then p^k
would not divide the order of G because p and m are relatively prime and K
could not be a subgroup of G. The order of a subgroup must divide the order
of the group.
Both H and K are subgroups of G, they both are closed under the same
operation as G, and because n>k, p^k divides p^n and thus because K is
closed under the operation of H and K's order divides the order of H, K
must be a subgroup of H.
Thanks
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