For the following exact sequence, i am trying to proof that it is not split $0\to 2\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$
I want to show that for injective function $i : 2\mathbb{Z} \to \mathbb{Z}$
if there exists left inverse $q :\mathbb{Z} \to 2\mathbb{Z} $ , then it would contradict and therefore sequence cannot be split.
but i am having trouble proving it, since $q$ can be $n\mapsto 2n$
pls help :')
If the sequence would split you would obtain an isomorphism $$\mathbb Z \cong 2 \mathbb Z \oplus \mathbb Z / 2 \mathbb Z.$$ But the right hand side contains an element of order two $(0, 1)$, whereas $\mathbb Z$ does not.
The answer that red_trumpet gave is essentially what you are looking for. But I want to add a slightly different approach, which directly contradicts the existence of a left-inverse $q$ to the map $\iota : 2\mathbb{Z} : \mathbb{Z}$.
Suppose there was such a $q$. As $q$ has a right-inverse (namely $\iota$), it is surjective. Now ask yourself where $1$ goes under $q$. Clearly $1$ has to go to a generator of $2\mathbb Z$. But that means that $q(1) = \pm 2$. As any choice uniquely determines the map $q : \mathbb Z \to 2\mathbb Z$, you can now check that for both of these choices $q \circ \iota(2) = q(2) = \pm 4$ is not the identity, hence $q$ is no left-inverse to $\iota$.
