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Suppose $S$ is a set of groups of order $n$, is there a binary operation $*:S\times S \to S$ that is definable on $S$?

The obvious operations I started with were

  • Cartesian product, but that produces groups of order $n^2$.
  • Intersection, but that produces groups of order $\leq n$.
  • Matrix multiplication of the Cayley table representations, but that doesn't produce another $n\times n$ matrix over the integers $\{0,1,...\,n-1\}$

I'm curious, are there are any known operations one can define that takes pairs of groups and produces one of the same order?

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    Projection on the first or second argument. :-)2011-10-26
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    Oh yes, I didn't think of that. However, I'd like to have an operation that has the properties of associativity, identity and inverses. Any thoughts?2011-10-26
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    @habitmelon So, in essense, do you want $S$ to be a group under $\ast$? Nice question. :) As a side note, the number of groups of size $n$ is varies wildly with $n$, so such a group might not have a simple description. For e.g., it depends on the the factorization of $n$: for prime $n$, there's just one group of order $n$.2011-10-26
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    There will always be only finitely many groups of order $n$, say there are $m$ of them. Any group operation on $S$ will be of the form $X*Y\to f^{-1}(f(X)f(Y))$ for some group $G$ of order $m$ and some bijection $f:S\to G$.2011-10-26
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    @anon, Srivatsan: The OP says groups, not isomorphism classes of groups, so $S$ could be infinite (or even a proper class); although perhaps isomorphism classes are what is intended.2011-10-26
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    @Zev: Hm. Well, I think what I wrote still applies if $m$ is infinite. The problem is it doesn't provide for any 'natural' constructions.2011-10-26
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    Let's assume $n \in \mathbb{N}$, I am only considering finite groups. The number of isomorphism classes is at most $n!$, so $m$ is defined and finite.2011-10-26
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    @SrivatsanNarayanan Yes! For n=32, the number of isomorphism classes is 51, this is the smallest n with the property that the number of isomorphism classes is at least n. After 32, there is 48, 64, 96, 128... you can see where I'm going with this. I want to see if I can construct a group that is isomorphic to one of its elements. I want it to contain itself (up to isomorphism)2011-10-26
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    Closest thing I can think of is http://en.wikipedia.org/wiki/Ext_functor#The_Baer_sum_of_extensions This chooses a normal subgroup and a quotient group rather than just the size.2011-10-26
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    What do you mean by intersection of arbitrary groups?2011-10-26
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    @ccc Intersection on subgroups of the same order, not arbitrary groups.2011-10-26
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    But what does that mean? What's the intersection of, say, $S_3$ and $\mathbb{Z} / 6 \mathbb{Z}$?2011-10-26

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Assuming isomorphism classes of groups is intended, the answer to the question is yes, because we can just make the set of $|S|$ groups of order $n$ into a cyclic group of order $|S|$ in some arbitrary fashion. But if you mean is there some "natural" way of doing it that involves the group structure of the groups in $S$, then the answer is almost certainly no.

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    I will look for some structural feature that gives a unique integer between 0 and n-1, this sounds like a straightforward approach.2011-10-27