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The direct sum $\oplus$ versus the cartesian product $\times$

Is there a difference in the proof in showing that two cyclic groups $C_m\times C_n\cong C_{mn}$ where $m,n$ are relatively prime and showing that $C_m\bigoplus C_n$ for the same $m,n$? Is there a catch?

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    $\begingroup$ Technically, $\times$ means product (with a universal mapping in property), and $\oplus$ means coproduct (with a universal mapping out property). In an additive category like abelian groups, we always have a canonical isomorphism $$A\oplus B \to A\times B$$, given by $(id_A \times 0) \oplus (0\times id_B)$. $\endgroup$ Commented Feb 15, 2012 at 12:37
  • $\begingroup$ @JustinYoung: Thank you. Would you mind explaining what "universal mapping in property" and "universal mapping out property" are? Also, what is the difference between simply an isomorphism and a "canonical isomorphism"? $\endgroup$ Commented Feb 15, 2012 at 13:02
  • $\begingroup$ I would only ever use $\oplus$ when I was using additive notation, and $\times$ if I was using multiplicative notation. So I would talk about $Z_m\oplus Z_n$ but $C_m\times C_n$. I'm not sure if the other way is incorrect, but certainly it doesn't feel right... $\endgroup$ Commented Feb 15, 2012 at 13:24
  • $\begingroup$ See this previous question. $\endgroup$ Commented Feb 15, 2012 at 16:39

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So far as I know, the two notations mean the same thing. There is a difference if, instead of two groups, you have infinitely many.

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  • $\begingroup$ Thank you, Gerry. Hmm, then I really don't know why we are given this question. Anyways, 1) Out of curiosity, what happens when there are infinitely many? 2) Does it make any difference if say we consider the $C_n,C_m$ as rings of the form $Z_n,Z_m$? $\endgroup$ Commented Feb 15, 2012 at 12:31
  • $\begingroup$ In the direct product of a countable infinity of groups, an element is an infinite sequence of elements of those groups. In the direct sum, same thing, only all but finitely many of the terms in the sequence must be the identity element. E.g., $(7,7,7,\dots)$ is in the direct product of infinitely many copies of the integers, but not in the direct sum. $\endgroup$ Commented Feb 16, 2012 at 0:00
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    $\begingroup$ As to why you were given the question, maybe your instructor is using different definitions. Best to ask your instructor. $\endgroup$ Commented Feb 16, 2012 at 0:02
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In any category, a coproduct $A\oplus B$ is characterized by having two maps $A\to A\oplus B$ and $B\to A\oplus B$ with the property that given any pair of maps $f: A\to C$ and $g: B\to C$ there exists a unique map $f\oplus g: A\oplus B \to C$ such that $A\to A\oplus B \to C$ is the map $f: A\to C$ and $B\to A\oplus B \to C$ is the map $g: B\to C$.

Dually, a product $A\times B$ is characterized by having two maps $A\times B \to A$ and $A\times B \to B$ with the property that given any pair of maps $f: C\to A$ and $g: C\to B$ there exists a unique map $f\times g: C\to A\times B$ such that $C\to A\times B \to A$ is $f$ and $C\to A\times B \to B$ is $g$.

In an additive category (in which zero maps make sense), the universal properties give a canonical map $A\oplus B \to A\times B$ given by $\Gamma = (id_A\times 0) \oplus (0\times id_B)$. We also have the composite maps $\alpha: A\times B \to A \to A\oplus B$ and $\beta: A\times B \to B \to A\oplus B$, and since we can add maps we get a map $\Delta = \alpha + \beta: A\times B \to A\oplus B$. Now, we can use the universal properties to see that $\Gamma$ and $\Delta$ are inverse isomorphisms.

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  • $\begingroup$ Here's a discussion of "canonical", I don't have anything insightful to say about it: mathoverflow.net/questions/19644/… $\endgroup$ Commented Feb 15, 2012 at 13:38

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