Abstract algebra question from IIT JAM 2019

General Mathematics
Question:

Let G be non-cyclic group of order 4. Consider the statements I and II.

I. There is no injective (one-one) homomorphism from G to \mathbb{Z}_8

II. There is no surjective (onto) homomorphism from \mathbb{Z}_8 to G Then,


(A) I is true          (B) I is false
(C) II is true         (D) II is false

Solution:  let suppose that f:G\rightarrow \mathbb{Z}_8 is an injective homomorphism. Then By Fundamental theorem of homomorphism (First isomorphism theorem ) we have, 

\frac{G}{ker(f)}≈f(G)

But, ker(f)=\{e\} (since f is injective i.e. one-one map ) Hence above implies,

G≈f(G) But f(G) is subgroup of \mathbb{Z}_8 and hence must be cyclic(Since, subgroup of a cyclic group is cyclic) So that G is cyclic. Contradiction to fact that G is non-cyclic group and hence our assumption must be wrong! Hence, there is No injective (one-one) homomorphism from G to \mathbb{Z}_8.

So that, statement I is true

Now, if g:\mathbb{Z}_8 \rightarrow G is onto homomorphism then we have g(\mathbb{Z}_8)=G which is contradiction. (Since by the result "homomorphic image of cyclic group is cyclic" we have g(\mathbb{Z}_8) is cyclic whereas G is not cyclic)

Hence our assumption must be wrong!! So,  There is no surjective (onto) homomorphism from \mathbb{Z}_8 to G
Hence statement II is also true.

So that answer is, option (A), (C).

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