Consequences of $f$ is injective and $g$ is surjective in the following sequence?

Question

I am trying to study if there can be a SES of the following form

$$0 \to \mathbb Z_4 \xrightarrow{f} \mathbb Z_8 \oplus \mathbb Z_2 \xrightarrow{g} \mathbb Z_4 \to 0.$$

So, I know if there is one, $f$ must be injective and $g$ must be surjective. I do not understand why $f$ is injective, this will leads that $\mathbb Z_8 \oplus \mathbb Z_2$ contains an isomorphic copy of $\mathbb Z _4$ could someone explain this to me please?

Any help will be appreciated!

Answer 1

An exact sequence is a sequence of morphisms such that the image of one morphism is equal to the kernel of the next. If this sequence is exact (this is never said, but this is the most important part) then:

• the first map from $0$ to $\mathbb{Z}_4$ is injective since the domain has only one element, and its image is only one element, the $0$ of $\mathbb{Z}_4$.
• The image of $0\to \mathbb{Z}_4$ is equal to the kernel of $f$, so $f$ is injective.

Similarly for the second part of the sequence

• The codomain of the last map is a single element $0$, and so the kernel of the last map is the whole group $\mathbb{Z}_4$.
• Because the image of $g$ is equal to the kernel of the last map, $g$ must be surjective.

All of this follows from the sequence being an exact sequence. This creates strong constraints on the functions $f$ and $g$, and in general, is not true.

The moral of the story is that, in an exact sequence, writing $0\to A \xrightarrow{f} B$ is a way to say $f$ is injective, and $A \xrightarrow{g} B\to 0$ is a way to say $g$ is surjective.

Answer 2

Hint: By the isomorphism theorem applied to $g$, we would have $$ \left(\mathbb Z_8 \oplus \mathbb Z_2\right)\ /\ \mathrm{Im}(f) \cong \mathbb Z_4, $$ since $\mathrm{ker}\ g=\mathrm{Im}\ f$. Now go through the few choices for $f$, and see what you find. Note that $f$ is determined by the image of a generator of $\mathbb Z_4$, and this image must have order $4$ in $\mathbb Z_8 \oplus \mathbb Z_2$, so there are only a few possibilities.