I have the following question which i do not know how to continue.

Let H be a subgroup of G. G and H has order 2n and n respectively. Suppose exactly half of the elements in G has order 2 and there is no elements of order 2 in H, prove that H is abelian.Question:

I started by considering the homomorphism defined by . I know that since f is an homomorphism, order of h = order of . Also, since no elements in H has order 2, their inverse are not themselves. I attempt to prove by contradicting by assuming that H is not abelian. However, i cant proceed on after that. Is the approach correct or should i consider something else? Thank You!Attempts: