Internal Direct Product without Normality I know that if $G$ is a group with normal subgroups $H,K$ such that $G=HK$ and $H\cap K = \langle e \rangle$, then $G\cong H\times K$.
What happens if the normal restriction on $H$ or $K$ or both is removed? Is there still an isomorphism? If not, what fails about the previous isomorphism $f:H\times K\to G$ given by $f(h,k)=hk$? Can conclusions about the structure of $G$ still be made with respect to $H$ and $K$?
 A: Suppose $H$ is normal but $K$ is not normal. Suppose $G = HK$ and $H \cap K = \{ 0 \}$.
It is still the case that every element $g \in G$ can be expressed in the form $ g = hk$ for some $h \in H$ and $k \in K$.
The thing that changes is the multiplication rule. Suppose you want to multiply $g_1 = h_1k_1$ with $g_2 = h_2 k_2$. You get:
$$ g_1 g_2 = h_1 k_1 h_2 k_2 = h_1 (k_1 h_2 k_1^{-1}) k_1 k_2.$$
But since $H$ is a normal subgroup, $k_1 h_2 k_1^{-1}$ is also an element of $H$! Let's call this element $h_2^{k_1} = k_1 h_2 k_1^{-1} \in H$. Thus
$$ g_1 g_2 = (h_1 h_2^{k_1}) (k_1 k_2), $$
which now is manifestly written as an element of $H$ multiplied by an element of $K$.
So you can think of $G$ as follows: The elements of $G$ are uniquely specified by pairs $(h,k)$, where $h \in H$ and $k \in K$. The multiplication rule is
$$ (h_1, k_1).(h_2, k_2) = (h_1 h_2^{k_1}, k_1 k_2).$$
This construction may seem a little weird at first! So why not look at a few familiar examples to get used to it? For example, suppose $G$ is the symmetry group of the triangle. Suppose $H$ is the normal subgroup consisting of rotational symmetries, $H = \{ 1, r, r^2 \}$ and $K$ is a subgroup generated by a single reflection, $K = \{ 1, \sigma\}$. It is true that $G = HK$ and $H \cap K = \{ 1 \}$. And $\sigma r \sigma^{-1} = r^2$ and $\sigma r^2 \sigma^{-1} = r$. It may be instructive to think about how these general remarks about multiplication in semi-direct products apply to this example.
