# About Algebraic Structure on arbitrary sets.

Given a non-empty set $X$. How do we plug a binary operation $*$ on the set $X$ so that $(X,*)$ forms a group?

There are sets, namely the set of natural numbers, set of prime numbers, set of perfect numbers and many for which finding an * to form them group is an extremely difficult job. Although I know some of the * for which the set of natural number forms a Group with repsect to *.

Please correct me if my understanding goes wrong!

• It is trivial for the sets you namend "difficult" because they are countable, so that there is a group structure isomorphic to $(\mathbb{Z},+)$. Jan 1, 2014 at 14:05
• Hello, Give me explicitly a binary operation * on the set of primes, which makes the set Group. Jan 2, 2014 at 4:40
• Since $\mathbb{N}$ is a group with respect to the nim sum $\oplus$ (en.wikipedia.org/wiki/Nim-sum), we can take $p_n \oplus p_m := p_{n \oplus m}$. Jan 2, 2014 at 10:11
• Can you please elaborate the fact that N is a group with respect to the nim sum ⊕ ? And how that leads to the group structure of the set of primes? Jan 2, 2014 at 10:43
• @Fukuzita: you can even have fields of arbitrary infinite cardinality (see math.stackexchange.com/questions/1296889). Transporting the structure on an infinite set $X$ of the same cardinality turns it into a field! Related: math.stackexchange.com/questions/1677199 Aug 22, 2016 at 12:57

Suppose that $X$ and $Y$ are nonempty sets of equal cardinality. If $(Y, \circ)$ is a group, then we can use a bijection $f: X \rightarrow Y$ to construct a binary operation $*$ on $X$ so that $(X, *)$ is a group and $(X, *) \cong (Y, \circ)$. Proving this is a good exercise.

Thus if $X$ is finite, the fact that $X$ has the same cardinality as the cyclic group of order $|X|$ gives us a binary operation $*$ that makes $(X, *)$ into a cyclic group.

What if $X$ is infinite? In this case we need the Axiom of Choice (see this question from MO). It is a consequence of choice that $X$ has the same cardinality as the set $\mathscr{R}$ of finite subsets of $X$. Now the set $\mathscr{R}$ equipped with symmetric difference makes $\mathscr{R}$ into an abelian group, so we are done.

In fact, $\mathscr{R}$ has a nontrivial structure of a ring without unit, when we define the sum of two sets as their symmetric difference, and the product of two sets as their intersection. Note also that finite cyclic groups have a ring structure (with unit).

It depends on the type of group you want. In the case of a collection of integers/rationals, one could create groups akin to $\mathbb{Z}_n$ or $\mathbb{Z}/\langle n \rangle$ or $(\mathbb{Z},+)$ or maybe even $(\mathbb{Q},\cdot)$. However, with many collections of numbers, it is not obvious how to make a group out of them. There are the 'obvious' ways, as I just mentioned. There are also very suprising ways to form groups: the easy example is the way we form a group out of the rational points on elliptic curves (assuming there is a rational point).

For your example concerning a group over the primes, you may want to read a question I asked previously, Are there Groups of Strictly Primes. In general, however, there is no right/wrong way to form a group out of a collection of numbers. In fact, the Axiom of Choice allows us to form a group out of any nonempty collection of numbers of any cardinality. So we know there is always at least one group given a nonempty collection of elements.

• Well. Thanks! What about more strict structures viz. Rings/Fields? Jan 1, 2014 at 6:45
• @SkSarifHassan You have to remember, a ring is just a collection of elements with $2$ 'group' structures, a group $+$ and a semigroup $\cdot$. The Axiom of Choice again should allow us to impose a ring like structure so that these are compatible (though I have not thought this completely through). Fields are just special rings. However, given that they require more than just the general structure of a ring, we cannot guarantee their existence for a set of any cardinality. For example, a set of size $6$ can have a ring structure but will never be a field. Jan 1, 2014 at 6:51
• To be more specific, there is no (finite) field of order not a prime power. See this, math.stackexchange.com/questions/72856/… Jan 1, 2014 at 6:53
• See the problem is with compatibility (distributive law) of those two binary operations. There are many sets having two different Abelian group structures but they fails to follow distributive property. What do you say about it?. Jan 1, 2014 at 6:58
• @SkSarifHassan I don't really see a problem. It's a little late so I'm not sure I'm up to doing this in full but the Axiom of Choice allows us to force the group $(R,+)$. Then by construction we should be able to 'match' the necessary group products for $(R,\cdot)$ so that the distributive law holds and the rest of our axioms hold. Then I would imagine that you can choose the remaining products for $\cdot$ as you wish because $(R,\cdot)$ need not have inverses or even be commutative. So $\cdot$ can be rather 'wild', within reason. But this of course needs more thought. Jan 1, 2014 at 7:02