Let G be a group and H a subgroup of G of index 2. Show that H contains every element of G of odd order

Let G be a group and H a subgroup of G of index 2. Show that H contains every element of G of odd order

Since H is a subgroup of index 2 in G, it is normal in G. The factor group G/H is cyclic of order 2. Let e be the identity in G. Since H is a subgroup of G, e belongs to H. So it suffices to consider the elements of G of odd order that are not the identity. Let g be of order 2k - 1 in G, for some k > 1. Then g^(2k - 1) = e and

(gH)^(2k - 1) = g^(2k - 1)H = eH = H.

Since (gH)^(2k - 1) = H, by multiplying both sides by gH, we get (gH)^(2k) = gH. Since G/H is of order 2, (gH)^2 = H. Since

(gH)^(2k) = [(gH)^2]^k

and H^k = H, we deduce that

H = gH.

This means that g belongs to H.