Answer to Question #177264 in Abstract Algebra for Abhijeet

Question

Which of the following statements are true? Give reasons for your answers. (10)

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 .

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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

• 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 isormorphic to one of its proper sub group under the map

f: Q^* -> Q^*

f(x) = X³

Hence G is not equals to Z

Option (i) is wrong.

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

For eample 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 )

Hence option (ii is wrong).

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

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

-> These dooes not exist any non abelian group of prime order

hence option (iii) is wrong

•  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 )

Hence option (iv) is wrong.

•  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

Hence option (v) is correct