Seth Warner's "Modern Algebra" (1965), exercise 13.6.
Context: self-study, from a many-years-ago maths degree which did not focus deeply on abstract algebra.
If a group $G$ is the direct composite of subsemigroups $H$ and $K$, then $H$ and $K$ are normal subgroups of $G$.
The object of this exercise is to demonstrate that even though the factor structures are as general as being semigroups, the nature of the direct composite (also known as "internal direct product") constrains those factors to actually being normal subgroups.
Now, the definition given for direct composite is as follows (paraphrased for convenience):
Let $H$ and $K$ be closed subsets of an algebraic structure $(G, \odot)$. Then $(G, \odot)$ is the direct composite of $H$ and $K$ iff the mapping $\phi: H \times K \to G$ defined as $$\phi: (x, y) \to x \odot y$$ is an isomorphism from the cartesian product of $(H, \odot_H)$ and $(K, \odot_K)$.
Now for the purpose of this exercise, I understand that this is all we are allowed to use, that is, we are not given that $(a, b) \odot (c, d) = (a \odot c, b \odot d)$. This is being glossed over in the text book, and taken for granted.
It is straightforward to show that every element of $G$ is of the form $h \odot k$ for some $h \in H$, $k \in K$. This follows directly from the fact that $\phi$ is surjective.
Similarly, it is also easy to show that $H \cap K = e$ (where $e$ is the identity element of $G$) follows from the fact that $\phi$ is injective.
It still remains to prove that both $H$ and $K$ are subgroups of $G$, which will be done by proving that the identity $e$ is the identity of both $H$ and $K$ (not sure what to use to get that to follow), and then that each element of $H$ and $K$ has an inverse in that set, which I haven't even begun to think about.
Once I've got that $H$ and $K$ are subgroups, it is straightforward to prove normality, because I can just copy that bit from a textbook. :-)
In summary, I think I need to show that:
- $(a, b) \odot (c, d) = (a \odot c, b \odot d)$
- Both $H$ and $K$ are groups (from which it will follow that they are normal subgroups of $G$)
unless there is a way to get where I want to go without achieving the above.
Any suggestions as to where to start -- given the constraints?
For the first part use that $\phi$ is an isomorphism, hence in particular an injective group homomorphism. The crucial observation is the following:
$$ \phi((a,b)\odot(c,d)) =\phi((a,b))\odot\phi((c,d)) =(a\odot b)\odot(c\odot d) =\phi((a\odot b,c\odot d)) $$
As $a,b,c,d$ were arbitrary we just proved that the group structure on the cartesian product is the natural one.
For the second part start with noting that if $h\in H$, then $h=h\odot 1_G$ is the unique representation of $h\in H\subseteq G$. This just follows using that $\phi$ is an isomorphism and elements in the product $H\times K$ are equal iff their coordinates coincide. Now write $h^{-1}\in G$ as $h^{-1}=h'\odot k'$. Lift the equalities
$$ h\odot h^{-1} =(h\odot 1_G)\odot(h'\odot k') =1_G =h^{-1}\odot =(h'\odot k')\odot(h\odot 1_G) $$
to $H\times K$ along $\phi$. By the above we then get that
$$ (h,1_G)\odot(h',k') =(hh',k') =(1_G,1_G) =(h'h,k') =(h',k')\odot(h,1_G) $$
which is only possible if $h'=h^{-1}$ and $k'=1_G$. This shows that $H$ is a subgroup and the argument is identical for $K$.
However, this is a very unusual way of approaching direct products.
EDIT: There is a very important point to remark on: $H\times K$ is a group with respect to some operation; otherwise it is not meaningful to talk about isomorphisms (or even homomorphisms for that matter) unless you are using this word in a non-standard way.
So the statement "$\phi\colon H\times K\to G$ is an isomorphism" includes the assertion that $H\times K$ is a group and that $\phi$ is a homomorphism. Although I admit that writing $\odot$ for the operation on $H\times K$ was not the best choice of notation.
Maybe to make it clearer: Suppose that $\phi\colon(H\times K,\odot_{H\times K})\to(G,\odot_G)$ is an isomorphism of groups, then:
\begin{align*} \phi((a,b)\odot_{H\times K}(c,d)) &=\phi((a,b))\odot_G\phi((c,d)) \\ &=(a\odot_G b)\odot_G(c\odot_G d) \\ &=\phi((a\odot_G b,c\odot_G d)) \end{align*}
We know that both entries are still in their respective substructures as those are closed (under composition I assume) by assumption. So this is a valid equation implying that the composition law $\odot_{H\times K}$ such that $\phi$ is an isomorphism is necessarily just given by componentwise compostion using $\odot_G$, i.e.
$$ (a,b)\odot_{H\times K}(c,d)=(a\odot_G b,c\odot_G d)\,. $$