Show that $\mathbb{Z}_9$ is not the homomorphic image of $\mathbb{Z}_3 \times \mathbb{Z}_3$.

Question

We solved a similar question in class: $\mathbb{Z}_8$ is not the homomorphic image of $\mathbb{Z}_{15}$ because if $f \colon \mathbb{Z}_{15} \to \mathbb{Z_8}$ is a homomorphism, then $|Im(f)|$ divides $15$ and since $\gcd (15, 8)=1$, the homomorphism is trivial.

I can't use this method here because $\gcd(9, 9)= 9.$

Answer 1

Suppose we have a homomorphism $G \to H$ between two groups of equal size. Then $H$ is isomorphic to a quotient of $G$, so we must have $G \cong H$ because $G$ is the only quotient of $G$ having the same number of elements of $H$.

So it remains to show that $\mathbb{Z}/9\mathbb{Z}$ and $(\mathbb{Z}/3\mathbb{Z})^2$ are not isomorphic.

Note that $\mathbb{Z}/9\mathbb{Z}$ has an element of order $9$, whereas $(\mathbb{Z}/3\mathbb{Z})^2$ does not. Alternatively, you can count the number of subgroups of order $3$.

Answer 2

Suppose we have a surjective homomorphism $\phi:\mathbb{Z}_3 \times \mathbb{Z}_3 \rightarrow \mathbb{Z}_9$.

From the isomorphism theorems, we will have that $( \mathbb{Z}_3 \times \mathbb{Z}_3) / \ker(\phi) \cong \mathbb{Z}_9$. For this to hold, the order of the quotient group on the left must match the order of $\mathbb{Z}_9$ on the right, which forces $\ker(\phi) = \{id\}$.

This is a contradiction because...