about groups?

Question

prove that no group can have exactly two elements of order 2

Answer 1

Suppose that a and b are the only distinct elements of order 2, and let e be the identity.

Now (bab^-1) (bab^-1) = ba(b^-1b)ab^-1 = ba^2b^-1 = bb^-1 = e, so bab^-1 is of order 1 or 2. So we have three cases:

(i) bab^-1 = e: then ba = b, so a = e, contradicting that a has order 2.

(ii) bab^-1 = a: then ba = ab, so (ab)^2 = abab = baab = bb = e, so ab is of order 2 (it can't be of order 1 since aa = e and ab = e implies a = b). Then ab = a or ab = b, so we must have b = e or a = e, contradicting that they have order 2.

(iii) bab^-1 = b: then ba = bb, so a = b, contradicting that they are distinct.

Hence no group can have exactly two elements of order 2.

[Taranto: There's no particular reason to suppose that ab (ba)^-1 = 1. I think you might be getting confused with ba (ba)^-1 = 1 or ab (ab)^-1 = 1.
Also, there is no "multiplicative group of the integers minus zero"; this is almost a group but there are no inverses for any elements except 1 and -1.]

Answer 2

Suppose that there are two elements a and b of order two. Then a^2 = 1 and b^2 = 1

Consider c = a*b.

Now consider c^2 = abab

But note that a = a^(-1) and b = b^(-1)

So c^2 = a*b*a^(-1)*b^(-1)

But a^(-1)*b^(-1) = (ba)^(-1) -- so we get:

c^2 = a*b*(ba)^(-1) = 1

This is a contradiction -- unless c = a or c = b. But that leads to another contradiction.

Note that there are groups with two elements of order 2. In the multiplicative group of the integers minus zero, 1 and -1 are both of order 2.

Answer 3

An element a is said to be of order 2 if a*a = e, the identity element. Now, that means this element is the inverse of itself. Only the identity element can be the inverse of itself.