Show that $\alpha$ is a homomorphism. Also, find $\text{ker}(\alpha)$, and decide if $\alpha$ is onto.

Question

Let $\alpha: \mathbb{Z}_9 \times \mathbb{Z}_{27} \rightarrow \mathbb{Z}_{27}$ be given by $\alpha((a, b)) = 3b$, for all $a\in \mathbb{Z}_9,b\in \mathbb{Z}_{27}$. Show that $\alpha$ is a homomorphism. Also, find $\text{ker}(\alpha)$, and decide if $\alpha$ is onto.

My attempt: Let $(a_1,b_1), (a_2,b_2)\in \mathbb{Z}_9 \times \mathbb{Z}_{27}$.

Then $$\alpha((a_1,b_1)+(a_2,b_2))=\alpha(a_1+a_2,b_1+b_2)=3(b_1+b_2)=3b_1+3b_2 =\alpha((a_1, b_1))+\alpha((a_2, b_2))$$

thus, $\alpha$ is a homomorphism.

$$\text{ker}(\alpha)=\{(a,b)\in \mathbb{Z}_9 \times \mathbb{Z}_{27}:\alpha((a,b))=0\}=\mathbb{Z}_9 \times\{0\}$$ (In particular $\alpha$ is not one-to-one).

My question is in determining if $\alpha$ is onto.

If $b\in \mathbb{Z}_{27}$, then $\alpha\left ( \left (a,\frac{1}{3}b \right ) \right )=b$

Does this argument help me to say that $\alpha$ is onto?

Answer 1

Your answer for the kernel is wrong -- $(a,9)$ and $(a,18)$ are also in the kernel, so $$\ker\alpha=\mathbb Z_9\times\{0,9,18\}\cong\mathbb Z_9\times\mathbb Z_3.$$ So by the isomorphism theorem, $$\operatorname{im}\alpha\cong(\mathbb Z_9\times\mathbb Z_{27})/(\mathbb Z_9\times\mathbb Z_3),$$ which (looking at orders) cannot possibly be $\mathbb Z_{27}$.

Your last claim doesn't really make sense -- how to interpret $\frac{1}{3}$ in $\mathbb Z_{27}$? You can't really, because $3$ and $27$ aren't coprime.

Answer 2

You have fixed the homomorphism proof.

Continuing from there:

α((a,b))=e

Whenever 3b = e

This is true for b=0,9,18.

So $ker(\alpha) = Z_{9}\times (0,9,18)$