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This is coming from a graduate-level abstract algebra class, for reference. My professor says that given two groups $G,H$ we say $D$ is the direct sum of $G$ and $H$ and write $C = G \oplus H$ if $G$ and $H$ are disjoint except for zero and $C = G+H = \{g+h | g \in G, h \in H\}$.

My issue is that this isn't really a definition, since for arbitrary groups $G$ and $H$ that aren't necessarily disjoint, this isn't defined. For example, I have an assignment question to show that the direct sum of two modules with a certain property still has that property, but I don't see how the direct sum is defined for arbitrary modules. For example, what would $\mathbb{Z}_3 \oplus \mathbb{Z_2}$ be?

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    $\begingroup$ There are two concepts called "internal direct sum" and "external direct sum", which are often confused deliberately because when they both make sense they are "as good as identical" (canonically isomorphic). Google for them. $\endgroup$ – darij grinberg Nov 19 '17 at 2:05
  • $\begingroup$ If the problem concerns disjointness, well, aren't ${\bf Z}_2$ and ${\bf Z}_3$ disjoint? $\endgroup$ – Gerry Myerson Nov 19 '17 at 2:09
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    $\begingroup$ recall that $Z_{2}$ and $Z_{3}$ are different quotient sets of $Z$, so the equivalence class of $1$ in $Z_{2}$ is not the same as that of $1$ in $Z_{3}$. $\endgroup$ – user363464 Nov 19 '17 at 2:16
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    $\begingroup$ Well, both groups have an identity element, as does every group, but there's no reason why they have to be the same identity element. E.g., ${\bf Z}_2$ could be the multiplicative group $\{\,1,-1\,\}$, and ${\bf Z}_3$ could be the rotations of a triangle. $\endgroup$ – Gerry Myerson Nov 19 '17 at 2:16
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    $\begingroup$ Asking whether unrelated groups are disjoint is a set theoretic nightmare. Internal vs. external sums/products is the relevant issue. $\endgroup$ – Guest Nov 19 '17 at 4:31
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For arbitrary modules $M, N$ over a ring $A$, there's a external direct sum module $M\boxplus N = \{(m, n):\, m\in M, n\in N\}$ with operation $(m, n) + (m', n') = (m + m, n + n')$ and $A$-action $a(n, m) = (an, am)$. The resulting module has $M\boxplus N = M\oplus N$ in your notation, embedding $M$ and $N$ in $M\boxplus N$ in the obvious way. Furthermore, if $P = M\oplus N$, then the map $P \to M\boxplus N$ defined by $m + n \to (m, n)$ is a well-defined isomorphism (injectivity following from the fact $M\cap N = 0$). As such, the two types of direct sum usually aren't distinguished in nomenclature or notation.

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