# Binary operations defined on sets of groups?

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. :-) –  Brian M. Scott Oct 26 '11 at 3:30
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? –  Tobi Lehman Oct 26 '11 at 3:35
@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 $n$. –  Srivatsan Oct 26 '11 at 3:38
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 $f:S\to G$. –  anon Oct 26 '11 at 3:41
@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. –  Zev Chonoles Oct 26 '11 at 3:45

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.