Here is the setup of a problem in group theory I can trying to solve. $G$ is a group, and there is a short exact sequence $$ 0 \rightarrow \mathbb{Z}/2 \rightarrow G \rightarrow^{\varphi} \mathbb{Z}/k \rightarrow 0. $$ Suppose that $\varphi(g) = 1$ for $g \in G$. Then the order of $g$ is either $k$ or $2k$.
The solution that I've read makes sense to me, but seems to include an implicit assumption that $g$ has finite order. This is what I'm able to understand, with the step I find problematic bolded. (I'm using multiplicative notation in $\mathbb{Z}/k$, but I acknowledge this is non-standard since the group is abelian. This is a bit easier for me to see understand some of the steps.)
Let $n = |g|$ in $G$. Then $g^n = e$, so $\phi(g^n) = \varphi(g)^n = 1^n = 0$. So the order of $1$ divides $n$. That is, $k$ divides $n$ since $1$ has order $k$ in $\mathbb{Z}/k$. The existence of the aforementioned short exact sequence implies that $G$ contains a normal subgroup $K$ with $K \cong \mathbb{Z}/2$ and that $G/K \cong \mathbb{Z}/k$. So $|G/K| = |\mathbb{Z}/k| = k$. As $K \cong \mathbb{Z}/2$, $|K| = 2$. So by Lagrange's theorem, $$ |G/K| = \frac{|G|}{|K|} = \frac{|G|}{2} = k, $$ so $|G| = 2k$. As $g \in G$, $|g|$ divides $2k$. So $|g|$ is a multiple of $k$ and a divisor of $2k$. The only options are $k$ and $2k$.
Several steps of the proof, unless I am misunderstanding them, rely on $g$ having finite order or, more generally, on $G$ being a finite group. If I knew the latter, then $g$ would certainly have finite order. I don't know how I can figure out that this must be the case. The short exact sequence definition, as I understand it, is:
- The map $f: \mathbb{Z}/2 \to G$ is injective ($\text{ker}(f) = \{0\}$)
- The map $\varphi: G \to \mathbb{Z}/k$ is surjective.
- $\text{Im}(f) = \text{ker}(\varphi)$.
This is supposed to imply that there exists a subgroup $K$ of $G$ that is isomorphic to $\mathbb{Z}/2$ and $G/K \cong \mathbb{Z}/k$. The solution I'm reading takes this to be the kernel of $\varphi$. The kernel is certainly a normal subgroup, though I don't fully understand why $K$ must be the kernel. Is it just a convenient choice?
Using your notation we have $$ \varphi(g^k) = \varphi(g)^k = 1^k = 0, $$ so $g^k \in \ker \varphi = \operatorname{im} f$, and then $$ g^k=f(a) $$ for some $a \in \Bbb Z/2$.