Let $G$ be group and $N$ is a normal subgroup of $G$. Does $G$ have a copy of $G/N$?
I really know that it is not true. However, I have a solution but I think that there exists an error.
Since $1\rightarrow N\rightarrow G \rightarrow G/N\rightarrow 1$ is an exact sequance , it implies that $G\cong N \rtimes G/N$. Hence, $G$ has a subgroup isomorphic to $G/N$.
This is not true. Think for example to $G=\mathbb{Z}$, $N = 2\mathbb{Z}$ and $G/N=\mathbb{Z}/2\mathbb{Z}$. Then all the subgroups of $G$ have the form $n\mathbb{Z}$ for some $n\in \mathbb{Z}$, and $G/N$ doesn't have this form.
The reason your argument with the short exact sequence doesn't work is that the sequence is not (or does not) split.