Prove or give counter example: $G \cong N \times G/N \Rightarrow N$ has a normal complement.

Question

My original question is the following:

Given a short exact sequence $1 \to N \xrightarrow{\iota} G \xrightarrow{\pi} Q \to 1$, we have: \begin{equation*} G \cong N \times Q \ \Longrightarrow \ \text{the sequence is a left split sequence} \end{equation*}

The converse is very easy to prove, but I couldn't prove this direction. Also, when I researched this problem (e.g. see theorem 3.2 here), I saw that if we put some restrictions to the isomorphism between $G$ and $N \times Q$, then we can deduce that the sequence is left split. So I think that my statement is wrong in this form.

To find a counter-example, first, I tried to minimize the variables in the statement. It is not hard to see that the above problem is equivalent to the following:

Given any group $G$ and normal subgroup $N$, we have: \begin{equation*} G \cong N \times G/N \ \Longrightarrow \ N \ \ \text{has a normal complement} \end{equation*}

I looked at a lot of groups, but I couldn't find any counter-example. It would be great if you could give one, or you could give a hint about how to prove it if it is not incorrect.

Thanks for any help.

Answer 1

As a counterexample, let $G$ be the (restricted) direct product of countably many copies of $C_2 \times C_4$: i.e. $$G= {\large \times}_{i>0} (\langle x_1 \rangle { \times} \langle y_i\rangle ),$$ where $x_i$ and $y_i$ have order $2$ and $4$, respectively. Now let $N = \langle x_{2i}, y_1^2 \mid i>0 \rangle$.

Then $G/N \cong G$ and $G \cong N \times G/N$, but $N$ has no normal complement in $G$.