Does $G \times H = K$ iff $G$ and $H$ are normal in $K$?

Question

Let $G \times H = K$. By applying The Isomorphism Theorem to the homomorphism $(g, h) \rightarrow g : K \rightarrow G$, I get $K/H \cong G$, so $H$ is normal in $K$. Similarly, $G$ is normal in $K$. Is this line of reasoning correct? Is the converse true, that if $H$ and $G$ are normal in $K$ then $G \times H = K$? If not what can be said about $H$ and $G$?

Answer 1

I assume that when you write $H$ in $K$, you mean $\{1\} \times H$ and $G \times \{1\}$. If that is the case, then you are correct. Another way of seeing that is 'manually', i.e. if $(1,h') \in \{1\} \times H$, then for any $(g,h) \in K$ we have that

$$ (g,h)(1,h')(g^{-1},h^{-1}) = (1,hh'h^{-1}) \in \{1\} \times H. $$

The same works for $G$. As for the other direction, you need more hypotheses: for example, if the group is abelian any pair of subgroups is normal, but their product may not be the whole group. A concrete case: $\mathbb{Z} \oplus 0 \oplus 0$ and $0 \oplus \mathbb{Z} \oplus 0$ are normal in $\mathbb{Z}^3$, but their sum (which is the product in this case) is $\mathbb{Z}^2 \not \simeq \mathbb{Z}^3$.

What does hold is the following: if $H,G \triangleleft K$ and $HG = K$, $G \cap H = \{1\}$, then $K \simeq H \times G$. Hint: take the function that maps $(g,h)$ to $gh \in K$, and prove that it is an isomorphism.