Solution for Section G.2 Question 2

We must show that for all x,y Î H, x × y-1 Î H.

We begin by noting that if x,y Î H, then x2 = e and y2 = e.
Thus x = x-1 and y = y-1.

Now to show that x × y-1 Î H we must show (x × y-1)2 = e.
We know that

(x × y-1)2
=
(x × y-1) × (x × y-1)
=
x × (y-1) × (x × y-1))
=
x × (y-1) × ( y-1 × x))
=
x × ((y-1) × y-1) × x)
=
x × (e × x)
=
x × x
=
e

Hence x × y-1 Î H, so H is a subgroup of G.

Back to Section G.2