Answer to Question #203244 in Abstract Algebra for Anand

Question

Which of the following statements are true? Give reasons for your answers.

 i) If a group G is isomorphic to one of its proper subgroups, then G = Z.

 ii) If x and y are elements of a non-abelian group (G, ∗) such that x ∗ y = y ∗ x, then x = e or y = e, where e is the identity of G with respect to . ∗

 iii) There exists a unique non-abelian group of prime order. 

 iv) If (a, b)∈A× A, where A is a group, then o((a, b)) = o(a)o(b).

 v) If H and K are normal subgroups of a group G, then hk = kh ∀ h ∈H, k ∈K.

Answer 1

Solution:

i) False.

If a group G is isomorphic to one of its proper subgroups, then G = Z.

For example Q^* - Group of non-zero rational under multiplication.

Q^* is isomorphic to one of its proper subgroup under the map

f: Q^* -> Q^*

f(x) = X³

Hence G does not equal to Z

ii) False.

 If x and y are elements of a non-abelian group (G, ∗) such that x ∗ y = y ∗ x,

For example Q8 ={+/- 1 , +/-i , +/-j , +/-k}

( -1 )i = i ( -1 ) = -i

but -1 is not equals to 1 and i is not equals to -i

hence we have xy = yx but ( x is not equals to e and y is not equals to e )

iii) False.

There exists a unique non-abelian group of prime order.

Every group of prime order is cyclic every cyclic group is abelian

=> There does not exist any non-abelian group of prime order

vi) False.

If (a, b)∈A× A, where A is a group, then o((a, b)) = o(a)o(b).

Let A= Z6 let a=2, b=2 and 0 ( a ) = 3 , 0 ( b )

then 0 [ ( a, b ) ] = 3 but not equals to 0 ( a ) 0 ( b )

v) True.

If H and K are normal subgroups of group G, then hk = kh ∀ h ∈H, k ∈K.

If H and K are normal subgroups of G

Then HK= KH

=> hk = kh