Let G be a p group K is a normal subgroup of G of order p to the power a. Then prove that the no of normal subgroups of a fixed order containing K is congruent to 1 mod p.
I've proved that the no of subgroups of a fixed order containing K is congruent to 1 mod p.
I have solved it. Check the answer. I've proved that the no of subgroups of a fixed order containing K is congruent to 1 mod p. Now K is normal in G & G is a p group. Then we can act on the set S, containing all the subgroups of fixed order containing K(see, the set is non empty!!), by conjugation of elements of G. From the class equation derived from this action we have p|(|S|-|Sg|) [as G is a p-group], where Sg={H : g.H=H}. So Sg here is nothing but the set containing all normal subgroups of fixed order containing K. So, |Sg| is congruent to 1 (mod p). Proved.