b^5 = b^2 b^3
b^3 = a b^2 a from this a b^3 = b^2 a
b^5 = b^2 (a b^2 a )
b^5 = b^2 a (b^2 a )
can u continue ?
Let G be a multiplicative group (not Abel's group, so it's non-commutative) and a and b are elements that belong to it. Also, a² = 1 and b³ = a^-1 * b² * a. Knowing this, show that b^5 = 1.
From a² = 1, I can see derive that a = a^-1. I have also derived that b² = ab³a and
b³ = ab²a, but I don't know if I'm on the right track. I've tried other ways too, but I always seem to end up where I started. I suppose this task isn't that hard, but I just need a little extra push to be able to solve it. I've been thinking if there's a way to state b in form b = a^n. This would solve probably everything, since a^(2n)=1.
Any help would be appreciated. Thanks in advance!
Ah! So the key was to replace b^n terms with b^(n-1) terms until there were no b terms left. The final phases I got were
(b²a)(b²a) = (b²ab)ba = baba = (bab)a = aa = a² = 1.
Here I used the following equivalences that I got from further manipulating b³ = ab²a:
b²ab = ba
bab = a
I suppose this is correct. Thanks a lot for the hint! =)
I got b²ab = ba in the following way:I can't see from where the above follows, and from this precisely follows that
Tonio
b³ = ab²a <=> b²b = ab²a <=> (ab³a)b = ab²a (replaced b² = ab³a)
<=> ab³ab = ab²a <=> b²ab = ba (multiplied from left with a, then with b^-1)
In this, I also used the following information:
G is a multiplicative group, so there has to exist inverse b^-1 so that b^-1 * b= e.
We already knew that a^-1 = a and that a² = 1, so it must be that the neutral element e = 1.