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Home / Modern Abstract Algebra / Show that 0(b) = 1 or 0(b) = 31

# Show that 0(b) = 1 or 0(b) = 31

Problem: Let a, b be two elements of a group G such that a5 = e and

ab a-1 = b2, where e is the identity of G. Show that 0(b) = 1 or 0(b) = 31.

Solution: Given that ab a-1 = b2 ————————–(i)

(ab a-1)2 = (b2)2

⟹ (ab a-1) (ab a-1) = b4

⟹ ab (a-1a) ba-1 = b4

⟹ ab e ba-1 = b4

⟹ ab2 a-1 = b4 ————————————(ii)

Similarly,

ab4 a-1 = b8 —————————————(iii)

ab8 a-1 = b16 ————————————–(iv)

ab16 a-1 = b32 ————————————-(v)

Now, b32 =ab16 a-1

= a(ab8a-1) a-1       [by (iv)]

= a2 b8 a-2

= a2 (ab4a-1) a-2 [by (iii)]

= a3 b4 a-3

= a3 (ab2a-1) a-3 [by (ii)]

= a4 b2 a-4

= a4(ab a-1)a-4 [by (i)]

= a5 b a-5

= e b e-5 [Since, a5 = e]

= b

Thus b32 = b

⟹ b31 = e

⟹ 0(b)| 31.

Since 31 is a prime number.

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

## Lecture – 4| Abstract Algebra

Problem – 1 : Show that if every element of the group G except the ...