I'm trying to understand the relationship between internal and external aspects of products. Having read this answer, I formulated the proposition below (with an added part) and attempted a proof.
Proposition. For a group $G$, TFAE.
- $G$ is decomposable, i.e $G\cong K_1\times K_2$.
- There exists $N,H\vartriangleleft G$ which are complements in the sense $N\cap H=\bf 0$ and $NH=G$.
- There exist subgroups $K_1,K_2$ with $K_1\vee K_2=G$ such that "multiplication" $K_1\times K_2\to K_1\vee K_2=G$ given by $(a,b)\mapsto ab$ is an injective group homomorphism (hence isomorphism by the universal property of the join of subobjects).
Proof. $[1\implies 2]$ Suppose $G= K_1\times K_2$ and let $j_1,j_2$ denote the unit injections of the factors to the product. If $\pi_1,\pi_2$ are the projections then $j_1=\ker \pi_2,j_2=\ker \pi_1$ so $j_1(K_1),j_2(K_2)$ are normal. Their intersection is trivial because each has a constant coordinate at the identity. Their product is all of $G$ because $(a,b)=(1,b)(a,1)$.
$[2\implies 3]$ Now suppose the second condition. We shall first prove the kernel of the "multiplication" is trivial. Note $(a,b)\in \ker \phi$ iff $a=b^{-1}$, i.e iff $(a,b)=(b^{-1},b)$. Since subgroups are closed under inversion, $b^{-1}\in H$. If $(a,b)\in \ker \phi$ necessarily also $b^{-1}\in N$ and so $b^{-1}\in H\cap N=\bf 0$ meaning $b=1=b^{-1}$. This proves $\ker \phi$ is trivial. The multiplication is a group homomorphism because normal subgroups commute pointwise.
$[3\implies 1]$ By assumption.
First of all, is my proof correct? I am worried I may have missed something and I don't want to miss any subtleties.
Second of all, I am wondering whether a proof of $[2\implies 3]$ can avoid inverses and thus generalize to monoids and perhaps even unital categories in general. If not, what's a more categorical approach? The same argument I gave holds for vector space etc...