Constructing a Surjective Homomorphism to Satisfy The First Isomorphism Theorem

Question

Let $H=\langle (1,2)\rangle$ be a subgroup of $\mathbb{Z}_4\times\mathbb{Z}_8$ and $\left|(\mathbb{Z}_4\times\mathbb{Z}_8)/H\right|=32/4=8$. By the First Isomorphism Theorem, $(\mathbb{Z}_4\times\mathbb{Z}_8)/H$ is isomorphic to $\mathbb{Z}_8$ if there exists some surjective homomorphism $\phi:\mathbb{Z}_4\times\mathbb{Z}_8\rightarrow\mathbb{Z}_8$ with kernel $H$. How would one find/construct such a homomorphism?

Answer 1

Consider the multiplication-by-$2$ map $t \colon \Bbb Z_4 \to \Bbb Z_8$, that is, $t([a]_4) = [2a]_8$ for all $a \in \Bbb Z$ (where $[b]_n$ is the class of $b \in \Bbb Z $ in $\Bbb Z_n$).

Now try the map $\phi \colon \Bbb Z_4 \times \Bbb Z_8 \to \Bbb Z_8$ defined by $\phi(x,y) = y-t(x)$. It is surjective? What is its kernel?

Answer 2

Actually, there exists only one group of order $2$ up to isomorphism, so you know that the group you construct this way is isomorph to $\mathbb{Z}_2$ (without knowing an explicit one, though).

The First Isomorphism Theorem rather states that if you find a surjective homomorphism $\mathbb{Z}_4 \times \mathbb{Z}_8 \to \mathbb{Z}_2$ with kernel $G$, then $(\mathbb{Z}_4 \times \mathbb{Z}_8)/G$ and $\mathbb{Z}_2$ are isomorph.

I assume there is a typo in your post and that $G = H$, but it seems to me that $|G| = 4$ in this case (the order of $(1, 2)$ in $\mathbb{Z}_4 \times \mathbb{Z}_8$ is $4$). I'll assume that it is rather $G = \langle (1, 0), (0, 2) \rangle$ from now (but you can correct your post if this is not what you thought about).

Then, you can see that $G = \mathbb{Z}_4 \times J$ where $J = \langle 2 \rangle$ as a subgroup of $\mathbb{Z}_8$. Therefore, to construct $\phi$, you will do :

$$ \phi : (a, b) \in \mathbb{Z}_4 \times \mathbb{Z}_8 \mapsto b \in \mathbb{Z}_2 $$

where, by abuse of notation, you associate to $b \in \mathbb{Z}_8$ its unique representant in $\mathbb{Z}_2$.