to prove that HN is a sub group, you need to show that HN is closed under the group operation, it contains the identity and has inverse for each element (associativity is given from G). this is all straight forward with . for example - closure where

2. again, by definition for every nk in NK and in NH you have

.

so you get that

3. I guess you forgot to write the N1,2 are normal in H1,2. anyway the proofs would probably be like the one above