Is the decomposition of an algebra into irreducible components essentially unique? 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.
 A: 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.

A: 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))  
