Thread: Abelian Subgroup

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

Question: 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.

Attempts: 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!

suppose g is an element of G not in H (any such g is of order 2). since gH ≠ H; for any h in H, gh is not in H, hence has order 2. thus:

ghgh = e, whence
ghg = h^-1, so that
ghg^-1 = ghg = h^-1, for all h in H.

now suppose h,k are any two elements of H.

hk = (gh^-1g)(gk^-1g) = g(h^-1)(gg)(k^-1)g = g(h^-1k^-1)g = g(kh)^-1g = kh,

so H is abelian.