# Defining more than one operation on the set so that with respect to every one of the operations the set is a group

I guess this could be a hard question but since I am no expert, and I am not even close to an expert I really do not know.

The basic idea is that we start with some group $(G,*)$. Because $(G,*)$ is a group it has over itself defined operation $*$ which satisfies all the group axioms.

Now, suppose that we take the set $G$ and that we seek to find another operation $o$ such that $(G,o)$ is also a group.

It seems to me that for some sets we will be able to find another operation under which the set is a group and for some sets we will not (but I may be mistaken).

So the question would be:

Are there any necessary, or sufficient, or necessary and sufficient conditions on the set $G$, or on the set $G$ and operation $*$, such that under these conditions there exist (or do not exist) at least one more operation (call it $o$) such that $(G,o)$ is also a group? What is known on these matters?

• If $|G|>1$ then there will be more than one such operation. There must be at least $|G|$ of them, because any of the elements of $G$ could be made into the identity element. – Derek Holt Apr 17 '16 at 17:13
• @DerekHolt Hi Derek, could you elaborate a little more on that and write it as an answer, if you wish to? – Farewell Apr 17 '16 at 17:16
• More interesting is asking for the other structure to not be isomorphic to the original. – Tobias Kildetoft Apr 17 '16 at 17:17
• @TobiasKildetoft I am at the first page of some book that has to do with group theory, it could be that I will have more interesting questions if I go more into the theory of groups, right now I only learned the definition of the group. – Farewell Apr 17 '16 at 17:23
• Another question is: how do the two operations interact? The problem is "easier" if the operations have "nothing to do with each other", but then far less "useful". What Derek is suggesting is just a "re-labeling" of the elements (and we can use any set-bijection to accomplish this). – David Wheeler Apr 17 '16 at 17:23

Every group structure $(S,*)$ on a set of cardinality $n$ gives rise to up to $n!$ other isomorphic group structures, given by transfer of structure: pick a bijection $f:S\to S$ and define $s\circ s'=f{-1}(f(s)f(s'))$. There are $n!$ choices of $f$, and $\circ$ is distinct from $*$ if and only if $f$ is not a homomorphism for $*$, so this gives $n!/|\text{Aut}S|$ total structures. Derek's comment points out that this quantity is at least $n$, since automorphisms of $S$ must preserve the identity (so that the automorphism group has order at most $(n-1)!$), but this bound is almost never tight for groups of order larger than 2. For example, cyclic groups of prime order $p$ have exactly $p-1$ automorphisms, so there are $p(p-2)!$ distinct cyclic group structures on a set of order $p$.
• The bound on the automorphism group of a group of order $n$ having order at most $(n-1)$! is tight for the cyclic group of order $3$ and the Klein group. (In both cases, any permutation fixing the identity is an automorphism.) – verret Apr 17 '16 at 20:29