If a and b are elements of a group G, and |a| = 5 and abaˉ¹ = b², what is |b|?

Question

|a| stands for the order of the element a in G, which means "a" raised to its order gives the identity element, e

Answer 1

Nice one! I hadn't seen this before.
Now b^2 = aba^-1
So ab^2a^-1 = (aba^-1)(aba^-1) = b^2.b^2 = b^4
and hence:
ab^4a^-1 = (ab^2a^-1)(ab^2a^-1) = b^4.b^4 = b^8
ab^8a^-1 = (ab^4a^-1)(ab^4a^-1) = b^8.b^8 = b^16
and so on.

Now, b = a^5ba^-1 = a^4(aba^-1)a^-4
= a^4(b^2)a^-4
= a^3(ab^2a^-1)a^-3
= a^3(b^4)a^-3
= a^2(b^8)a^-2
= a(b^16)a^-1
= b^32

So b^31 = e. Since 31 is prime, either b = e or |b| = 31.

Answer 2

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Answer 3

ohh ok, nvm, ill give it a crack...

we know that
aba^-1 = bb and a^5 = e and a, b, and e are elements of set H let me think...
we want to find n where b^n = e correct?

Answer 4

that was on my State test today

Answer 5

Do your own homework.