Let x and y be elements of a group G....

Let x and y be elements of a group G. Assume that each of the elements x,y and xy has order 2. Prove that the set H = {1,x,y,xy} is a subgroup of G, and that it has order 4.

I know it has order 4 because it has 4 elements, but the subgroup is throwing me off.

we need to show that H is closed under multiplication (since H is finite, this will assure that it also contains inverses). we have 16 products to verify.

clearly x^2, y^2 and (xy)^2 are all in H, since each has order 2, and 1 is in H. obviously, any product involving the identity is also in H, which takes care of 1*x, x*1,1*y,y*1,1*xy,xy*1, and 1*1.

that leaves 6 products to verify:

x*y = xy, which is in H.
x*xy = x^2y = 1y = y, which is in H
xy*y = xy^2 = x1 = x, which is in H.

the first one we run into trouble with is yx. we somehow need to show that yx is in H.

well, what could yx be? it can't be the identity since that would mean x = y^-1, but y (being of order 2) is its own inverse. yx cannot be y nor x, since neither is the identity. so the only thing it might possibly be is xy.

so what we really need to show, is that yx = xy. let's start here:

(xy)(xy) = 1 (we know this)
x(yx)y = 1 (by the associative law)
x(x(yx)y) = x (multiply both sides on the left by x)
(xx)(yx)y = x (more associativity)
(yx)y = x (because x is of order 2, so xx = 1)
(yx)(yy) = xy (multiplying both sides on the right by y)
yx = xy (ta da!)

so yx is in H, since it is the same as xy. now we can finish:

xy*x = (xy)x = x(yx) = x(xy) = (xx)y = y, which is in H
y*xy = y(xy) = (yx)y = (xy)y = x(yy) = x, which is in H.

so H is closed, and thus a group. note that H is even abelian.