How to prove a statement that contains two biconditionals: Exercise in Pinter's A Book of Abstract Algebra

Question

In Pinter's A Book of Abstract Algebra, the reader is asked in Chapter 14 exercise D4 to prove the following statement:

I have never been asked to prove a statement that contains two iffs. I just wanted to confirm that the below interpretations are all correctly capturing the idea:

1. "$H$ is normal" is equivalent to "$ab \in H \iff ba \in H$"
2. $[((ab \in H \implies ba \in H) \land (ba \in H \implies ab \in H)) \implies H$ is normal] $\land \ [H$ is normal $\implies ((ab \in H \implies ba \in H) \land (ba \in H \implies ab \in H))]$

So it seems to me that I essentially need to prove 3 different "sub-statements" in order to prove the original statement. Those 3 sub-statements are:

A. $H$ is normal $\implies (ab \in H \implies ba \in H)$

B. $H$ is normal $\implies (ba \in H \implies ab \in H)$

C. $(ab \in H \iff ba \in H) \implies H$ is normal

A and B will be paired together. Is this correct?