$G$ be a group with subgroups $H,K$ such that $G=HK$ , $|H \cap K|=1$ and $H$ is normal in $G$ , then is $K$ also normal in $G$?

Question

Let $G$ be a group with subgroups $H,K$ such that $G=HK$ , $|H \cap K|=1$ and $H$ is normal in $G$ , then does this imply that $K$ is also normal in $G$ ? If the answer is no , then what happens if we also assume $G$ is finite ? I feel it need not be true but I cannot come up with a counterexample . Please help .

Answer 1

As others stated in the comments, semidirect product is the key here.

Note that $H\cap K = \{e\}$ implies there is a bijection between $HK$ and $H\times K$ (but not necessarily isomorphism of groups) given by $hk\leftrightarrow(h,k)$.

For $k\in K$, let us define $\varphi_k\in \operatorname{Aut}(H) $ with $h\mapsto khk^{-1}$. It is easy to verify that $k\mapsto \varphi_k=\varphi\colon K\to \operatorname{Aut}(H)$ is homomorphism.

Now, for and $h,h'\in H,\ k,k'\in K$ we have $(hk)(h'k') = hkh'k^{-1}kk' = (h\varphi_k(h'))(kk')$ and thus the above bijection $hk\leftrightarrow (h,k)$ gives isomorphism $G\cong H\rtimes K$.

It is just as easy to verify that if $G\cong H\rtimes K$, then $H$ is normal in $G$ and $H\cap K =\{e\}$.

Thus, all examples and counterexamples will be found studying groups that are constructed as semidirect product, just as the comments suggest. Dihedral groups are, for example, given as semidirect product $\Bbb Z_n\rtimes \Bbb Z_2$, i.e. have presentation $\langle a,b\,|\, a^n = b^2 = e,\ bab^{-1} = a^{-1}\rangle$. It is easy to check that $\Bbb Z_2$ will never be normal in $\Bbb Z_n\rtimes \Bbb Z_2$ (for $n>2$).

Edit:

Actually, we can do this in general. Let $G\cong H\rtimes_\varphi K$ where $\varphi\colon K\to\operatorname{Aut}(H)$.

Let $h\in H, k\in K$. We have $$(h,e)\cdot(e,k)\cdot(h^{-1},e) = (h\cdot\varphi_k(h^{-1}),k)$$ so, for $K$ to be normal, we must have $h\cdot \varphi_k(h^{-1}) = e$, for all $h$ and $k$, which implies that $\varphi$ is trivial, i.e. $G\cong H\times K$. Conversely, if $G\cong H\times K$, $K$ is obviously normal in $G$.

Thus, we have the result:

If $H$ is normal subgroup of $G$, $K$ a subgroup such that $H\cap K=\{e\}$ and $G = HK$, then $K$ is normal in $G$ if and only if $G\cong H\times K$.