Let $G$ be a group, $N$ a normal subgroup of $G$ such that $G/N$ is infinite cyclic, $C$ an infinite-cyclic subgroup such that $C\cap N=1$, and $\Lambda$ a subgroup of index $2$ of $G$.
Let $N_{2}:=N\cap \Lambda$. Then, $N_{2}$ is a normal subgroup of $G$, so $N_{2}C$ is a subgroup of $G$. I want to show that $N_{2}C$ has finite index in $G$.
Firstly, $|G\colon N_{2}|=|G\colon N||N \colon N_{2}|$ and $|G:N|=\infty$ because $G/N$ is infinite cyclic by hypothesis. On the other hand, $|G\colon N_{2}|=|G: N_{2}C||N_{2}C\colon N_{2}|$ and $|N_{2}C\colon N_{2}|$ is $\infty$ because $$N_{2}C/ N_{2}\cong C/(N_{2}\cap C) \cong C.$$
Nevertheless, from here I cannot conclude that $|G\colon N_{2}C|$ is finite, right? Does anyone have any clue? Thanks in advance!
$NC/N$ is a nontrivial subgroup of the infinite cyclic group $G/N$, so $|G:NC| = |\frac{G}{N}:\frac{NC}{N}|$ is finite.
Now $|NC:N_2C| \le |N:N_2| \le 2$ is finite, so $|G:N_2C| = |G:NC| \times |NC:N_2C|$ is finite.