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 $N \unlhd HN$?

Question

Note: groups $N, H, G$ are all non-abelian.

This is what I've known:

• If $NH\cong N\rtimes_{\tt A} H$ (which means $NH$ is also a group and ${\tt A}$ is a group action), then:

  • $HN=NH$, criteria for the product of subgroup to become a subgroup

  • the two statements below are equivalent:

    1. elements of $NH$ can be represented uniquely as $nh$, where $(n,h)\in N\times H$;

    2. $N\cap H=1$

Now: if $N\cap H = 1$ is given, then

• the two statements below are equivalent:

  1. $N\trianglelefteq HN$

  2. $nh=hn$

And I also know the 1st/2nd/3rd group isomorphism theorem, so please be free to refer them.

BTW, I have an idea but not sure it is correct: I draw a picture to represent the 2nd group isomorphism theorem ($\tfrac{H}{H\cap N}\cong \tfrac{HN}{N}$ when $H\leq G$ and $N\trianglelefteq G$), which is shown below:

2nd group isomorphism theorem

As you can see: the box in the left represent the group $HN$, where the intersection of its bottom surface (which is $N$) and its right surface (which is $H$) is $H\cap N$. Now: if we apply the canonical map $\pi: hn \mapsto hn(H\cap N)$ on the box on the left, we will obtain the plane on the right hand side, just like the box on the left is smashed to a $2$-dimensional plane.

Answer 1

One way to define an (internal) semidirect product $K\cong X\rtimes Y$ (which is isomorphic to an external semidirect product) is that all of the following hold:

• $X\cap Y=1$,
• $X\unlhd K$, $Y\le K$, and
• $K=XY$.

Therefore, the answer must be "yes".