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I am told that finite groups have unique factorization under direct product. That is, call a nontrivial group "indivisible" if it is not isomorphic to a direct product of nontrivial groups. Then every finite group can be "factored" (by direct product) into a unique collection of indivisible groups.

In particular, if $G$ and $H$ are finite groups so that $G\times G\cong H\times H$, then $G\cong H$.

Can anyone provide a reference to a proof of these results? What is known in the infinite case? Thanks.

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    $\begingroup$ Recent related question: mathoverflow.net/questions/45668/… . Uniqueness of roots among finite structures with one-element substructures was proven by Lovasz, along with a cancellation theorem. For infinite structures, check some of the references. Gerhard "Ask Me About System Design" Paseman, 2010.11.19 $\endgroup$
    – Gerhard Paseman
    Commented Nov 19, 2010 at 22:37
  • $\begingroup$ Btw, a term often used in a case like this is "directly irreducible": Start with a category $C$ with finite products and call an object of $C$ directly irreducible if it is neither a terminal object nor isomorphic to the product (in $C$) of two non-terminal objects. In fact, this is a special case of a much more general definition of "irreducible": The isomorphism (iso) classes of $C$ form a commutative monoid $\mathcal V(C)$ under the binary operation that maps a pair $(\mathfrak a, \mathfrak b)$ of iso classes to the iso class of the product $A \prod B$ for some $A \in \mathfrak a$ [...] $\endgroup$
    – Salvo Tringali
    Commented Sep 15, 2022 at 6:44
  • $\begingroup$ [...] and $B \in \mathfrak b$ (the universal property of products implies that this is a well-defined operation). But in a monoid $H$ (written multiplicatively) we do have a natural notion of irreducible: It is an element $a \in H$ such that (i) $a$ is a $\mid_H$-non-unit, i.e., $a \nmid_H 1_H$ (where $\mid_H$ is the divisibility preorder on $H$, i.e., $x\mid_H y$ means that $x\in H$ and $y\in HxH$) and (ii) if $a\ne bc$ for all $\mid_H$-non-units $b, c$ such that $a\nmid_H b$ and $a \nmid_H c$. In particular, the $\mid_{\mathcal V(C)}$-irreducibles are the directly irreducible objects of $C$. $\endgroup$
    – Salvo Tringali
    Commented Sep 15, 2022 at 6:47
  • $\begingroup$ I think "directly indecomposable" is a common terminology, possibly more self-explanatory. This paper (in French) contains a number of references on these issues. $\endgroup$
    – YCor
    Commented Sep 15, 2022 at 7:49
  • $\begingroup$ @YCor Maybe so, but the term "irreducible" seems more in line with a certain general philosophy (see, e.g., en.wikipedia.org/wiki/Irreducible_ring). $\endgroup$
    – Salvo Tringali
    Commented Sep 15, 2022 at 10:03

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About the first fact see this page (the Krull–Remak–Schmidt theorem). For infinite (even finitely generated) groups the situation is different because there exists an infinite f.g. group isomorphic to its direct square.

Update. Hirshon, found two non-isomorphic finitely generated nilpotent (infinite) groups $G,H$ such that $G\times G\cong H\times H$.

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    $\begingroup$ @Tim: I did not know that. What are these $G$ and $H$? I think that if $G$ is finite, then it is exactly the set of all torsion elements in ${\mathbb Z}\times G$. So $G$ must be isomorphic to $H$ if ${\mathbb Z}\times G\cong {\mathbb Z}\times H$. But perhaps I am wrong and do not see something obvious. $\endgroup$
    – user6976
    Commented Nov 19, 2010 at 23:50
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    $\begingroup$ @Tim: It is proved here: springerlink.com/content/k7p548016pvq0163 . There, $G$ and $H$ are finitely generated and nilpotent. $\endgroup$
    – user6976
    Commented Nov 20, 2010 at 1:12
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    $\begingroup$ @Mark: One simple example is $G=C_{11}\rtimes{\mathbb Z}$ with the generator of ${\mathbb Z}$ acting as $x\mapsto x^2$ and $H=C_{11}\rtimes{\mathbb Z}$ with the generator of ${\mathbb Z}$ acting as $x\mapsto x^7$. Then ${\mathbb Z}\times G\cong{\mathbb Z}\times H$ but $G\not\cong H$, if I am not mistaken. $\endgroup$
    – Tim Dokchitser
    Commented Nov 21, 2010 at 1:26
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    $\begingroup$ @Mark: The $C_{11}$ is unique in $G$ (torsion subgroup), and the quotient $G/C_{11}\cong Z$ has two canonical elements (generators), $\pm1\in Z$. They act by $2,2^{-1}=6\in(Z/11Z)^\times$ on $C_{11}$. For $H$, the corresponding pair of numbers is different, $7,7^{-1}=8\in(Z/11Z)^\times$, so $G\not\cong H$. The important thing is that both $2$ and $7$ generate the whole of $(Z/11Z)^\times$, so the image of the action is the same subgroup. Otherwise $G\times Z\cong H\times Z$ has no chance to work... $\endgroup$
    – Tim Dokchitser
    Commented Nov 21, 2010 at 11:24
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    $\begingroup$ ...Now $G\times Z=C_{11}\rtimes (Z\times Z)$ with the two generators $e_1, e_2$ of $Z\times Z$ acting by conjugation as $x\mapsto x^2$ and $x\mapsto x$. Choose another basis of $Z\times Z$, say $f_1=7e_1+2e_2, f_2=10e_1+3e_2$. The determinant of this transformation is 7*3-2*10=1, so it is a basis. Then $f_1$ and $f_2$ act by $x\mapsto x^{2^7}=x^7$ and $x\mapsto x^{2^{10}}=x$, so $G\times Z\cong (C_{11}\rtimes \langle f_1\rangle)\times \langle f_2\rangle\cong H\times Z$, as promised. (To make a long story short, an extra $Z$ keeps the action image but may mess up the generators.) $\endgroup$
    – Tim Dokchitser
    Commented Nov 21, 2010 at 11:42

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