Prove or disprove: special case of second isomorphism theorem applied to infinite order group Prove or disprove: given normal subgroups $H, N \trianglelefteq G$, which are both groups of infinite order, $HN / H \cong N, HN / N \cong H \Leftrightarrow HN \cong H \times N$.
This is a special case of second isomorphism theorem when $N, H$ normalize each other.
I first consider the case when at least one of $N$ and $H$ has finite order (assume $N$ is the one of finite order):

*

*By the second isomorphism theorem, there's $HN / H \cong N \cong N / (N \cap H)$.

*By the first isomorphism theorem on the natural homomorphism of $N \cap H$, and $N$ is finite, it is only possible that $N \cap H = \{e\}$.

*Since $N \trianglelefteq HN$, $\forall n \in N, h \in H, \exists n' \in N, nh = hn' \Rightarrow nhn^{-1} = hn'n^{-1}$.

*Since $H \trianglelefteq HN$, $nhn^{-1} = hn'n^{-1}\in H \Rightarrow n'n^{-1}\in H$, by $N \cap H = \{e\}$, $n'n^{-1} = e \Rightarrow n' = n$.

*Now we have $H \cap N = \{e\}$ and $\forall n \in N, h \in H, nh = hn$, which implies $HN \cong H \times N$.

It seems there's no implication to $N \cap H = \{e\}$ when both $N, H$ have infinite order: consider the group $(\mathbb{C}^{\ast}, \cdot)$, where $\mathbb{C}^{\ast}$ is non-zero complex number and $\cdot$ is complex number multiplication, under $\phi: \mathbb{C}^{\ast} \to \mathbb{C}^{\ast}$ defined by $z \mapsto z^2$, there's $\mathbb{C}^{\ast} / \phi \cong \mathbb{C}^{\ast}$ while $\ker \phi = \{1, -1\}$.
Maybe it is just the way I use to prove is dumb and there's an alternative way to prove, or it is not applicable to infinite group. Please help me to prove or disprove.
 A: The claim is false.
Let $G$ be free of countable rank, freely generated by $x_1,x_2,x_3,\ldots$; let $H$ be the normal subgroup generated by $x_1,x_3,x_5,\ldots$ and $N$ the normal subgroup generated by $x_2,x_4,x_6,\ldots$. Being subgroups of a free group they are themselves free, and they must be of countable rank (any finite set of elements would involve only finitely many letters, and thus cannot generated the whole group). So $G\cong H\cong N$, and $G=HN$.
It is not hard to verify that $G/N$ is the free group on $x_1N,x_3N,x_5N,\ldots$, and so is isomorphic to $H$; and $G/H$ is the free group on $x_2H,x_4H,x_6H,\ldots$, and so is isomorphic to $N$.
However, $G=HN$ is not isomorphic to $H\times N$, because no nontrivial free group can be expressed as a nontrivial direct product. (For example, two elements in a free group commute if and only if they are powers of the same word, but in a nontrivial direct product centralizers are much larger). Note that $H\cap N$ is nontrivial: for example, the intersection contains $[x_1,x_2]=x_1^{-1}x_2^{-1}x_1x_2$.
You could have $HN\cong H\times N$ even if $H\cap N\neq\{e\}$. Take $G$ to be the direct sum of countably many copies of $\mathbb{Z}/2\mathbb{Z}$ and countably many copies of $\mathbb{Z}/4\mathbb{Z}$. Let $N$ be the sum of all copies of $\mathbb{Z}/4\mathbb{Z}$ and $k$ copies of $\mathbb{Z}/2\mathbb{Z}$; let $H$ be the sum of all copies of $\mathbb{Z}/2\mathbb{Z}$, and the subgroups of order $2$ in $k$ of the summands isomorphic to $\mathbb{Z}/4\mathbb{Z}$. Then $G=HN$, $G/N\cong H$ and $G/H\cong N$. We have $H\cap N$ is of order $2^k$; and while $HN$ is not itself the direct product of $H$ and $N$, it is isomorphic to $H\times N$, which is just isomorphic to $G$ itself.
