Abstract algebra question from IIT JAM 2019

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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