A question about the definition of semidirect product of groups

Question

Question

Given $N,H$ two subgroups of $G$ : if there exists an isomorphism between $NH$ and $N\times H$ (where the operation over the set $N\times H$ can be anything, not just the operation of the direct product and the semidirect product, i.e., $(n_1,h_1)(n_2,h_2)\mapsto \cdots$ and $N\trianglelefteq NH$, is it possible to obtain $N\cap H=1$?

(PS: groups $N, H, G$ are all non-abelian.)

My main purpose is: since the definition of semi-direct product $K=X\rtimes Y$ of group $X$ and $Y$ is

• $K=XY$,
• $X\cap Y = 1$, and
• $X\trianglelefteq K$ and $Y\leq K$,

then for the reverse: given $XY\cong X\times Y$ where the operation over $X\times Y$ is uncertain, is it possible to obtain $X\cap Y=1$ by $X\trianglelefteq XY$?

What I know so far:

You can find them in here:

For two subgroups $H, N$ of a non-abelian group $G$, if $HN\cong H\rtimes_{\tt A} N$, can we deduct $H\cap N=1$ if we know $N \unlhd HN$?

I'm the same person who asked that question.

Answer 1

No, you cannot conclude that $N\cap H=\{e\}$ in $G$, since you are allowing any operation on the underlying set $N\times H$.

Here is an example: take $G=F\times(\mathbb{Z}/2\mathbb{Z})$, where $F$ is the free group on two letters, $x$ and $y$. Let $N=\langle (x,0+2\mathbb{Z}),(y,0+2\mathbb{Z})\rangle$, and let $H=\langle (x,1+2\mathbb{Z}),(y,1+2\mathbb{Z})\rangle$. Their intersection is nontrivial, since it contains all elements of the form $(w,0+2\mathbb{Z})$, where $w$ is an element of $F$ that has an even number of letters. In addition, though it was not specified, $N$ is normal in $G$.

Now, $NH$ is countably infinite (it equals $G$), as is $N\times H$; thus, there exists a bijection $\psi\colon NH\to N\times H$. Fix that bijection, and define an operation $\circ$ on $N\times H$ by transport of structure: $$(n_1,h_1)\circ (n_2,h_2) = \psi\big(\psi^{-1}(n_1,h_1)\cdot\psi^{-1}(n_2,h_2)\bigr).$$ Then $(N\times H,\circ)$ is a group, $\psi$ is an isomorphism between $NH$ and $N\times H$, but $N\cap H\neq\{e\}$.

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If both $H$ and $N$ are finite, though, then since $|NH||N\cap H|=|N|\,|H|=|N\times H|$, the existence of any bijection between $NH$ and $N\times H$ implies that $|N\cap H|=1$.

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I will note that the nonabelianness of $G$, $N$, $H$ is completely irrelevant. Not sure why you believe it matters. The real issue here is whether $NH$ can be bijected with $N\times H$ when $N\cap H\neq\{e\}$.