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We consider finite algebras for a given signature, in the sense of universal algebra (for example, they might be groups, rings, or lattices). An algebra $A$ is irreducible when $A \cong B \times C$ implies that $B$ or $C$ is the one-point algebra.

Is it the case that a $\Sigma$-algebra can be expressed as a cartesian product of irreducible algebras in an essentially unique way, i.e., unique up to permutation of factors? I suspect that this is either a theorem with somebody's name attached to it, or there is a counterexample in groups.

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I'm not sure if it helps, but in finite group theory, the decomposition of a $\mathbb C$-representation into irreducible ones is not unique, unless all their multiplicities are $0$ or $1$. For instance, the decomposition of the regular representation of a finite group $G$ is not unique, unless $G$ is abelian. – Denis Serre Nov 11 '10 at 9:39
In terms of groups the question is this: can a finite group be a direct product of irreducible groups in two essentially different ways? (A group is irreducible if it cannot be written as a non-trivial direct product.) I suspect Jordan–Hölder theorem is relevant here, but I am not an algebraist. Is a simple group the same thing as an irreducible group? – Andrej Bauer Nov 11 '10 at 13:40
It seems that the Krull-Schmidt theorem, see, answers my question positively for groups. – Andrej Bauer Nov 11 '10 at 14:44
A side note is that the term "indecomposable" might be preferable to "irreducible". Usually "irreducible" means having no nontrivial quotients, e.g., for group theory, irreducibles are simple groups. The group $S_3$ is an indecomposable which is not irreducible. – Todd Trimble Nov 11 '10 at 16:11
If you can get it, Chapter 5 of "Algebras, Lattices, Varieties" has a lot to say about unique factorization, and contains theorems and examples similar to what Gerald Edgar posted. My advisor Ralph McKenzie was one of the authors. Lovasz's proof of the cancellation theorem is one of the most amazing results I have seen, and a version of it is in the chapter. Gerhard "Ask Me About System Design" Paseman, 2010.11.16 – Gerhard Paseman Nov 17 '10 at 1:56
up vote 3 down vote accepted

Let $A, B$ be the algebras with two elements $0,1$ under addition mod $2$ and unary operation $x'=x$ in $A$ and $x'=1-x$ in $B$. Then $A \times B \cong B \times B$, though $A$ and $B$ are not isomorphic.
B. Jonsson

Construct a 12-element commutative semigroup which does not have the unique factorization property.
R. McKinsey

(exercises on p. 170 of Birkhoff, Lattice Theory (3rd edition))

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So, I wonder if "R. McKinsey" is "Ralph McKenzie"... – Gerald Edgar Nov 17 '10 at 4:18
I suspect the R is not an R. Even so, there is at least one McKinsey who did some structure theory. I will wager a small amount of money that it is not Ralph McKenzie. Gerhard "Ask Me About System Design" Paseman, 2010.11.18 – Gerhard Paseman Nov 18 '10 at 21:36
Looks like I might lose the wager. ALV in chapter 5.1 has a generalized version of the commutative semigroup exercise above, credited to R. McKenzie. (Not to be confused with Robert MacKenzie or John McKinsey.) Gerhard "Betting Windows Are Now Closed" Paseman, 2010.11.18 – Gerhard Paseman Nov 19 '10 at 2:01
Bjarni Jónsson's example is very nice. But it seems to me that the group operation is not necessary; the unary function suffices. – Goldstern May 4 '14 at 18:14

This is not true as stated. If you take the empty signature, or any signature with no constants, then the empty set is an algebra, and this will mess things up.

This issue was raised in the paper

M. Barr, The point of the empty set, Cahiers 13:1-12, 1972.

I think that this is all that can go wrong. If you restrict to non-empty signatures, or to non-empty algebras then things are ok; but if you are also considering multisorted theories, then you need to make sure that each sort is non-empty. This issues was also discussed in the Barr paper. See also

G.M. Kelly and A. Pultr, On algebraic recognition of direct-product decompositions, Journal of Pure and Applied Algebra 12:207--224, 1978.

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Excellent, thank you. I did not think of the empty set, but yes, it obviously messes things up. – Andrej Bauer Nov 17 '10 at 9:21

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