Group Theory Proof -- Simple?
Suppose that a and b are elements of a group G with a having odd order. Assuming that aba^-1 -= b^-1 prove that b^2 = e.
should be aba^-1 = b^-1 not aba^-1 -= b^-1
- DukeLv 71 decade agoFavorite Answer
You know of course that (xyz)⁻¹ = z⁻¹ y⁻¹ x⁻¹ for any x, y, z of G, so inverting
aba⁻¹ = b⁻¹ we get (aba⁻¹)⁻¹ = (b⁻¹)⁻¹ or
(1) ab⁻¹a⁻¹ = b.
Let a's order is 2n+1 /odd number for n = 0, 1, 2, , /, that means
a²ⁿ⁺¹ = e, but then, of course
(a⁻¹)²ⁿ⁺¹ = (a²ⁿ⁺¹)⁻¹ = a⁻²ⁿ⁻¹ = e also.
Multiply aba⁻¹ = b⁻¹ by a from left, by a⁻¹ from right:
aaba⁻¹a⁻¹ = ab⁻¹a⁻¹, but the right side here according (1) is equal to b, i.e:
a² b a⁻² = b, repeat once again and apply the identity:
a³ b a⁻³ = aba⁻¹ = b⁻¹, go on the same way:
a⁴ b a⁻⁴ = b;
a⁵ b a⁻⁵ = b⁻¹;
. . . . . . . . . and, after ODD NUMBER OF TIMES
a²ⁿ⁺¹ b a⁻²ⁿ⁻¹ = b⁻¹
But a²ⁿ⁺¹ = a⁻²ⁿ⁻¹ = e, so on the above text line we have b = b⁻¹.and then
b² = bb⁻¹ = e as required.
P.S. dCromley, it is correct above: a^ord(a) = a^|a| = e
- dCromleyLv 51 decade ago
I sent an email to Duke asking:
isn't a²ⁿ⁺¹ = a not =e?