I think the proof goes as follows (though I'm used to asking questions here, not helping ).

G is finite and with a prime order. Therefore it's cyclic (have you studied that? It's pretty basic).

A cyclic group is of course commutative (abelian).

Therefore, for every sub-group H in R, the normalizer in G N(H) is actually G itself...

So obviously H is a subgroup of N(H)=G...