Confusion regarding proving (by definition) that a group $G$ is a direct sum of $H, K$

Question

For some reason, the distinction between an inner direct sum and an outer direct sum left me a bit confused regarding how one would go about proving that a group $G$ is a direct sum of two other groups, $H,K$.

Here is what I think would be required, and I would like to verify I'm not mistaken:

a. If $H,K$ are subgroups of $G$, then it suffices to prove that they are normal in $G$, that their intersection is trivial and that they span $G$.

b. If $H,K$ are NOT subgroups of $G$ (here is where I'm not really sure...) - I have two options when working by definition:

1. Show that $H,K$ are isomorphic to subgroups of $G$ (e.g. $G=\mathbb{Z}_6$ and $H=\mathbb{Z}_3$, $K=\mathbb{Z}_2$ ) and then use (a) above on the isomorphic copies.

2. Show that each element of $G$ can be uniquely written as a sum of an element in $H$ and an element in $K$ (e.g. when $G=\mathbb{Z}_6$ and $H,K$ are already in the form of those isomorphic copies mentioned in (1)).

Since (a) and (b.2) seem likely to be rare scenarios (?) since they don't seem very interesting - I conclude that it will most likely be accomplished using (b.1).

Is this correct (the methods and the last observation)?

Answer 1

Your conclusion is not correct, for the following reason: You haven't provided enough options. There are still other statements involving $G$, $H$ and $K$ which one can prove are equivalent to the statement that $G$ is an external direct sum of $H$ and $K$.

Here is one such statement that comes up fairly often:

There exists an exact sequence of homomorphisms $$1 \to H \xrightarrow{i} G \xrightarrow{p} K \to 1 $$ and a homomorphism $j : G \to H$ such that the composition $j \circ i : H \to H$ is the identity.

In case you are unfamiliar with exactness, it means that $i$ is injective and $p$ is surjective and $\text{image}(i) = \text{kernel}(p)$.

Assuming the statement above is true, one directly constructs an isomorphism from $G$ to $H \oplus K$ in the following manner.

First, combine the homomorphisms $j$ and $p$ into a single homomorphism $f : G \mapsto H \oplus K$ defined by $f(g) \mapsto (j(g),p(g))$.

Next, apply exactness to prove that $f$ is surjective. To do this, suppose you are given $(h,k) \in H \oplus K$. Use surjectivity of $p$ to choose $g' \in G$ such that $p(g')=k$. Then let $g = i\bigl(h \cdot j(g')^{-1}\bigr) \cdot g'$, and prove that $f(g)=(h,k)$.

Finally, use exactness to prove that $f$ is injective. To do this, assume $f(g)=(\text{id}_H,\text{id}_K)$, and so $j(g)=\text{id}_H$ and $p(g)=\text{id}_K$. It follows that $g \in \text{kernel}(p) = \text{image}(i)$ and so $g=i(h')$ for some $h' \in H'$. But then $\text{id}_H = j(g) = j(i(h')) = h'$. Since $i$ is injective, $g = i(h') = i(\text{id}_H) = \text{id}_G$.