Suppose that a and b are elements of a group G such that a^-1ba = b^-1 and b^-1ab = a^-1. Prove that a^2= b^-2

Answer 1

From the equation b⁻¹ab = a⁻¹, we deduce:

b⁻¹a = a⁻¹b⁻¹
ba = (a⁻¹b⁻¹)⁻¹ = (b⁻¹a)⁻¹ = a⁻¹b

So from the first equation, we have that:

b⁻¹ = a⁻¹ba = a⁻¹a⁻¹b

So b⁻²=a⁻². Now, a = (a⁻¹)⁻¹ = (b⁻¹ab)⁻¹ = b⁻¹a⁻¹b, so:

a² = b⁻¹a⁻¹bb⁻¹a⁻¹b = b⁻¹a⁻²b = b⁻¹b⁻²b = b⁻²

Which was to be demonstrated. Note that since we also showed b⁻²=a⁻², we have as a corollary of this that a⁴ = e and b⁴=e, so the order of a and b is at most 4.

Answer 2

i've got have been given faith you will in all threat have a typo indoors the 1st question. via fact the order of the id is one million, that's unusual, it particularly is achievable that a = e. if so, we'd have b^(-one million) = ab^(-one million)a = eb^(-one million)e = b(-one million) This holds actual for any b^(-one million) in the kind of group, regardless of if or no longer b^2 = e. i've got have been given faith what you meant to sort grew to grow to be aba^(-one million) = b^(-one million)

Answer 3

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