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Suppose that $G=MN$ and $G=MP$ are two exact factorization of a finite group $G$. What is the relation between $M$ and $P$? Clearly if $G=MN$ then $G=M(mNm^{-1})$ is another factorization of $G$. Is this the only possibility to change $N$ into $P$?

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It should be noted that the product in this problem is different from direct (Cartesian) product. A theorem of Lovasz gives cancellation of Caretesian product (AB iso AC implies B iso C) for finite algebras with one-element subalgebras. Gerhard "Ask Me About System Design" Paseman, 2011.02.19 –  Gerhard Paseman Feb 19 '11 at 19:55
    
@mathuser, if you're still here, maybe you could accept one of the posted answers, or explain what more needs to be done to provide a satisfactory answer. –  Gerry Myerson Feb 19 '11 at 22:32
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5 Answers

up vote 2 down vote accepted

The answer of the problem is the following:

A group G has two exact factorizations $G = M N = M P$ if and only if there exists a unitary bijective map (just a map, not a morphism of groups) $v : N \to P$ such that $n v(n)^{-1} \in M$, for all $n \in N$ and $v$ satisfy four natural compatibility conditions (not very transparent to write down here -- they are similar, mutatis mutandis, to the one given in Propositions 2.1 in http://front.math.ucdavis.edu/0903.5060 -- for example one of them is that v is a morphism of $M$-sets).

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The answer is no. $G=MN$ is an exact factorisation is equivalent to $N$ acting regularly on the set of right cosets of $M$ in $G$. It is not necessary for two regular subgroups of a group to be isomorphic. Think for example $G=S_n$ and $M=S_{n-1}$. Then any group of order $n$ acts regularly on $n$ points so provides a group for $N$.

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Well, not every subgroup of $S_n $ of order $n$ is regular (take $S_3$ in $S_6$), but there are in general nonisomorphic regular subgroups. Example: $\langle (1,2,3,4) \rangle $ and $\langle (1,2)(3,4),(1,3)(2,4) \rangle $ in $S_4$. –  Frieder Ladisch Feb 11 '11 at 17:01
    
Yes not every subgroup of $S_n$ of order $n$ is regular, but every group of order $n$ acts regularly on itself by right multiplication and so yields a regular subgroup of $S_n$. –  Michael Giudici Feb 12 '11 at 2:36
    
@Michael Giudici: Of course, you're right. Sorry I misunderstood the last sentence of your answer! –  Frieder Ladisch Feb 12 '11 at 12:30
    
An even simpler counterexample takes place with abelian groups, such as with $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$: the first factor $\mathbb{Z}/p\mathbb{Z}$ has $p$ different complements, which clearly cannot be conjugated (being the group abelian). –  Maurizio Monge Feb 20 '11 at 19:00
    
Ah, by the way not every subgroup with order $n$ can be a group in the above construction: take $G=S_6$ and $M=S_5$ (the permutations of the first $5$ elements $\{1,\dots,5\}$): the group generated by $(1,2)(3,4,5)$ is isomorphic to $\mathbb{Z}/6\mathbb{Z}$, but it is even contained in $M$! In any case adding the requirement that $M \cap N = 1$ is enough. –  Maurizio Monge Feb 20 '11 at 19:05
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Does "exact factorization" mean every element of $G$ has exactly one expression as $mn$ with $m$ in $M$ and $n$ in $N$? If not, ignore the rest of this answer. But if so, then let $G$ be the group of the square, let $M$ be the subgroup of rotations, and let $N$ be any one of the four subgroups generated by a flip. Then $G=MN$ but the four available $N$ are not all conjugate to each other.

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Yes, exact factorization means exactly what you have written. Is any general connection between $P$ and $N$? If $M$ is normal then $N$ has to be ismorphic to $P$ but that is not enough. –  mathuser Feb 11 '11 at 11:23
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I am not an expert in group theory.

The problem is interesting and is a particular case to the following, which is a little bit more general: when two bicrossed products of two groups are isomorphic? (a group G has an exact factorization iff it is a bicrossed product of two (sub)groups -- see a Tackeuchi's old theorem).

A partial answer to the problem (in fact a Schreier type theorem for the factorization/bicrossed product) was given in http://front.math.ucdavis.edu/0903.5060

In this elementary paper the problem is solved in the case that the isomorphism fix (stabilize) one of the factors.

PS (edit): I think we can get a full answer of your question if we apply Corollary 3.11 and Theorem 3.3 from the above paper (writen equivalent in the language of bicrossed products, sorry for the categorical stuff used there). The identity is an isomorphism of the group $G$ that fix (satabilize) M. Thus we can apply for it Corollary 3.11.

Cheers! Gigel Militaru

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What a regular action means?, transitive withous fixed points? And what a regular subgroup is? If $M$ is normal in $G$ then both $N$ and $P$ are isomorphic to $G/M$.

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The action of a group on a set is regular if it is transitive and only the identity element fixes a point. A regular subgroup of a permutation group is a subgroup that acts regularly. –  Michael Giudici Feb 13 '11 at 8:21
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