 ## Question:

If A is a group in which every element other than the identity, I, has order 2, prove that A is Abelian. Hence show that if X and Y are distinct elements of A, neither being equal to the identity, then the set {I, X,Y ,XY } forms a subgroup of A.
Deduce that if B is a group of order 2p, with p a prime greater than 2, then B must contain an element of order p.

## Step-by-step

If every element of A, apart from I, has order 2, then, for any element X, ${X}^{2} = I.$
Consider two elements X and Y and let XY = Z. Then
${X}^{2}Y = XZ ⇒ Y = XZ$,
${XY}^{2} = ZY ⇒ X = ZY$ .
It follows that YX = XZZY . But, since XY = Z, Z must belong to A and therefore ${Z}^{2} = I$. Substituting this gives YX = XY , proving that the group is Abelian.
Consider the set S = {I, X,Y ,XY }, for which
(i) Associativity holds, since it does for A.
(ii) It is closed, the only products needing non-trivial examinations being $XY X = XXY = {X}^{2}Y = Y$ and $Y XY = XY Y = {XY}^{2} = X$ (here we have twice used the fact that A, and
hence S, is Abelian).
(iii) The identity I is contained in the set.
(iv) Since all elements are of order 2 (or 1), each is its own inverse.
Thus the set is a subgroup of A of order 4.
Now consider the group B of order 2p, where p is prime. Since the order of an element must divide the order of the group, all elements in B must have order 1 (I only) or 2 or p.
Suppose all elements, other than I, have order 2. Then, as shown above, B must be Abelian and have a subgroup of order 4. However, the order of any subgroup must divide the order of the group and 4 cannot divide 2p since p is prime. It follows that the supposition that all elements can be of order 2 is false, and consequently at least one must have order p.