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Tuesday , August 22 2017
Home / Modern Abstract Algebra / Lecture – 4| Abstract Algebra

# 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 a2 = e, b2 = e.

Also ab ∈ G and so (ab)2 = e

Now (ab)2 = e

⟹ ab ab = e

⟹a (ab ab) = a e b

⟹ a2 ba b2 = 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-1b) (ab), since b-1b = 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 a5 = e and aba-1 = b2, where e is the identity of G. Show that o(b) = 1 or o(b) = 31.

Solution: We have, ab a-1 = b2 ——————— (1)

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

⟹ (ab a-1) (ab a-1) = b4

⟹ ab(a-1a)ba-1 = b4

⟹abe ba-1 = b4

⟹ ab2a-1 = b4 ——————————————(3)

Similarly, ab4a-1 = b8 —————————————– (4)

ab8a-1 = b16 ——————————————(5)

ab16a-1 = b32 ——————————————(6)

Now, b32 = ab16a-1

= a (ab8a-1) a-1

= a2b8a-2 ,  by —– (4)

= a2 (ab4a-1)a-2

= a3 b4 a-3, by —– (3)

=  a3 (ab2a-1) a-3

= a4 b2 a-4, by —– (2)

= a4 (ab a-1) a-4

= a5 b a-5, by —– (1)

= e b e-5, since a5 = e

= b

Thus b32 = b ⟹ b31 = b

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

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

# PDF| Lecture – 4: Abstract-lecture-4

## Every group is isomorphic to a group of permutations.

PROOF:  To prove this, let G be any group. We must find a group of ...