# Problem – 1 : Show that if every element of the group G except the identity element is of order 2, then G is abelian.

Solution: Let a, b ∈ G such that a ≠ e, b ≠ e

Then a^{2} = e, b^{2} = e.

Also ab ∈ G and so (ab)^{2} = e

Now (ab)^{2} = e

⟹ ab ab = e

⟹a (ab ab) = a e b

⟹ a^{2} ba b^{2} = ab

⟹ e ba e = ab

⟹ ba = ab

Hence G is abelian

### Problem – 2: If a, b be any two elements of a group G, then ab and ba have the same order.

Solution: ab = e(ab), where e is the identity of G

= (b^{-1}b) (ab), since b^{-1}b = e.

Thus ab = b^{-1}(ba) b

Now o(ba) = o[b^{-1}(ba)b] = o(ab).

Hence, ab and ba have the same order.

## Problem – 3: If a, b be any two elements of a group G such that a^{5} = e and aba^{-1} = b^{2}, where e is the identity of G. Show that o(b) = 1 or o(b) = 31.

Solution: We have, ab a^{-1} = b^{2} ——————— (1)

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

⟹ (ab a^{-1}) (ab a^{-1}) = b^{4}

⟹ ab(a^{-1}a)ba^{-1} = b^{4}

⟹abe ba^{-1} = b^{4}

⟹ ab^{2}a^{-1} = b^{4} ——————————————(3)

Similarly, ab^{4}a^{-1} = b^{8} —————————————– (4)

ab^{8}a^{-1} = b^{16} ——————————————(5)

ab^{16}a^{-1} = b^{32} ——————————————(6)

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

= a (ab^{8}a^{-1}) a^{-1}

= a^{2}b^{8}a^{-2} , by —– (4)

= a^{2} (ab^{4}a^{-1})a^{-2}

= a^{3} b^{4} a^{-3}, by —– (3)

= a^{3} (ab^{2}a^{-1}) a^{-3}

= a^{4} b^{2} a^{-4}, by —– (2)

= a^{4 }(ab a^{-1}) a^{-4}

= a^{5} b a^{-5}, by —– (1)

= e b e^{-5}, since a^{5} = e

= b

Thus b^{32} = b ⟹ b^{31} = b

⟹ o (b)|31. Since 31 is a prime number.

We have o (b) = 1 or, o(b) = 31.