Problem: Let a, b be two elements of a group G such that a^{5} = e and

ab a^{-1} = b^{2}, where e is the identity of G. Show that 0(b) = 1 or 0(b) = 31.

**Solution:** Given that ab a^{-1 }= b^{2} ————————–(i)

(ab a^{-1})^{2} = (b^{2})^{2}

⟹ (ab a^{-1}) (ab a^{-1}) = b^{4}

⟹ ab (a^{-1}a) ba^{-1} = b^{4}

⟹ ab e ba^{-1} = b^{4}

⟹ ab^{2} a^{-1} = b^{4} ————————————(ii)

Similarly,

ab^{4} a^{-1} = b^{8} —————————————(iii)

ab^{8} a^{-1} = b^{16} ————————————–(iv)

ab^{16} a^{-1} = b^{32} ————————————-(v)

Now, b^{32} =ab^{16} a^{-1}

^{ } = a(ab^{8}a^{-1}) a^{-1} [by (iv)]

^{ } = a^{2} b^{8} a^{-2}

^{ } = a^{2} (ab^{4}a^{-1}) a^{-2} [by (iii)]

^{ } = a^{3} b^{4} a^{-3}

^{ } = a^{3} (ab^{2}a^{-1}) a^{-3} [by (ii)]

^{ } = a^{4} b^{2} a^{-4}

^{ } = a^{4}(ab a^{-1})a^{-4} [by (i)]

^{ } = a^{5} b a^{-5}

^{ } = e b e^{-5} [Since, a^{5} = e]

^{ } = b

Thus b^{32} = b

⟹ b^{31} = e

⟹ 0(b)| 31.

*Since 31 is a prime number.*

We have 0(b) = 1 or 0(b) = 31. (Proved)