Injective and surjective homomorphisms between non cyclic group of order $4 $ to $Z_8$

Question

Let $G$ be a non cyclic group of order $4$. Consider the following statements:

$I:$ There is no one-one map homomorphism from $G$ to $Z_8$

$II:$ There is no onto homomorphism from $Z_8$ to $G$

Then which of these statements are true?

Okay, So by classification, $G\simeq Z_2\times Z_2$

Consider statement $I$

If $\phi $ is $1-1$ homomorphism $\phi:G\to^{1-1}\to Z_8\rightarrow \frac{G}{ker\phi}\simeq H \leq Z_8$, where $H$ is a subgroup of $Z_8$ and hence cyclic. But here $ker\phi = \{e\}$ so, $G\simeq H \leq Z_8$ which is not possible since $G$ is not cyclic.

So $I$ is true.

Consider statement $II$

if $\phi$ is onto homomorphism from $\phi: Z_8\to G$, then $\frac{Z_8}{ker\phi}\simeq G$

But $\frac{Z_8}{ker\phi}$ is cyclic since $Z_8$ is cyclic, so $\frac{Z_8}{ker\phi}\simeq G$ is not possible since $G$ is non cyclic. So $II$ is true.

So both statements are true. Is this correct?

Answer 1

I think your arguments are correct, but can be shortened.

Part I: Suppose that there is such a homomorphism $\phi$. Then $\phi(G)\cong G$ is a subgroup of $C_8$, hence cyclic, a contradiction.

Part II: Suppose that $\phi$ is such a homomorphism. Then $\phi(C_8)\cong G$, where $\phi(C_8)$ is cyclic, because $C_8$ is cyclic, a contradiction.