Find all homomorphisms $\phi: \mathbb{Z}_{21} \oplus \mathbb{Z}_{4} \to S_3$

Question

I have attempted to solve this question using the first isomorphism theorem of groups, that is, if $\phi: \mathbb{Z}_{21} \oplus \mathbb{Z}_{4} \to S_3$ is a group homomorphism, then we must have that $$\frac{\mathbb{Z}_{21} \oplus \mathbb{Z}_4}{\text{ker} \, \phi} \cong \text{img} \, \phi.$$ Since $\text{img} \, \phi \leq S_3$, Lagrange's Theorem implies $|\text{img} \, \phi| \in \left\{1, 2, 3, 6\right\}$.

1. If $|\text{img} \, \phi| = 1$, then $\text{img} \, \phi = \left\{\text{id}\right\}$, so $\phi(x, y) = \text{id}$.

2. If $|\text{img} \, \phi| = 2$, then $\text{img} \, \phi = \left\{ \text{id}, \, \sigma \right\}$, where $\sigma \in S_3$ is a $2$-cycle. This is where I get stuck. I know that $|\text{ker} \, \phi| = 42$. It would make intuitive sense that $\text{ker} \, \phi = \left\{ (a, b) : \, a \in \mathbb{Z}_{21}, \, b = 0, 2 \right\}$. But why is this true and what should the map $\phi$ be in this case?

3. If $|\text{img} \, \phi| = 3$, then $\text{img} \, \phi = \left\{ \text{id}, \, (1 \; 2 \; 3), \, (1 \; 3 \; 2) \right\}$. Again, I know that $|\text{ker} \, \phi| = 28$.
4. If $|\text{img} \, \phi| = 6$, then $\text{img} \, \phi = S_3$. Again, I know that $|\text{ker} \, \phi| = 14$.

Answer 1

Of course $\Bbb Z_{21}\otimes \Bbb Z_4\cong\Bbb Z_{84}$ is cyclic.

So a homomorphism is completely determined by where a generator, say $1$, goes.

The choices are anything with order dividing $84$. Thus a $2$-cycle or a $3$-cycle.

There are thus $6$ (including the trivial one). That's the generator can go anywhere.