Question in Abstract Algebra?

Question

Let a and b be elements of a group G. Assume a is not equal to e, |b| =2, and bab^-1 = a^2. How can I find |a|??

Answer 1

Let's do some regular algebra on this until we get something that helps.

bab^-1 = a^2

Squaring both sides we get:

(bab^-1)(bab^-1) = (a^2)(a^2)

ba^2b^-1 = a^4

Now plug in the original equation for a^2 on the left to get:

bba(b^-1)(b^-1) = a^4

Now since |b| = 2, we get

a = a^4

So a^3 = e

Edit: we know that if a^n = e, then the order of a must divide n. So |a| must be 1 or 3. However, if |a| = 1, then that means a = e, but we are told that is not the case, so |a| = 3.

By the way, it does not make sense to say |a| = 0, since the order of a is the smallest positive n for which a^n = e.

Answer 2

Hmm... this is probably not the best method, but this is what I got after playing around with it a bit.

First, since |b| = 2, you know that b^2 = e, or b = b^-1. You know |a| is not 0, since a is not e. Now if a^2 = e, you have that bab^-1 = e, or bab = e, and so ba = b, or a = b^2 = e, a contradiction. So |a| is not 2.

Now we know bab = a^2, so
(bab)^2 = a^4
babbab = a^4
ba^2b = a^4
a^2 = ba^4b
bab = ba^4b
a = a^4
e = a^3

So |a| = 3.