A Summer of Maths

What I thought is, if I labelled subset as $\{ a_1, a_2, ..., a_p\}$, and defined

$a_i * a_j = a_{(i \times_{p} j)}$

then I think, I get a cyclic group. I assume the operation is associative, but it follows from the associativity of multiplication modulo.

Then, if I define $f(a_i) = i$, I get that

$f(a_i * a_j) = f(a_{(i \times_{p} j)}) = i \times_{p} j = f(a_i) \times_{p} f(a_j)$

and this is clearly bijective, so it is an isomorphism to $\mathbb{Z}_p$.


In this case I get a cyclic group, for all sets with a prime number of elements.
However, I need this to be true for every number of elements, and if it helps, they will be naturals.

So you can choose between making them act additively, so for instance $a_i * a_j = a_{i +_n j}$, in which case it makes no difference whether or not $n$ is prime; or you can choose another group and group operation and choose your isomorphism a different way.

I see. Well, it was the first one that I picked! So, if I choose them additively, then it seems plausible?
I'll have a look at it now, and check the details to see whether it works. Thanks!

Yes. But you seem to be making it harder than you have to: by labelling the elements of your subset from 1 up to n you have already defined an isomorphism from your subset to $\mathbb{Z}_n$, by the method I described two posts ago, since the group structure is inherited from the identification with the elements of $\mathbb{Z}_n$.

A more critical issue is that not all subsets of $\mathbb{N}$ are finite, but I'm sure you can work out a way round that...

More specifically, if $A \subset \mathbb{N}$ is any subset and $(G,*)$ is a group with $\left| A \right| = \left| G \right|$ then any bijection $f : A \to G$ induces a group structure $(A, \cdot)$ on $A$ by defining $a \cdot b = f^{-1}(f(a) * f(b))$.

A more critical issue is that not all subsets of $\mathbb{N}$ are finite, but I'm sure you can work out a way round that...

Found some links: Does every non-empty set admit a group structure on Google. Interesting!

.. a famous example being "every set has a well-ordering".

Does that work?

Why is this called the Axiom of ``Extensionality", expressed as: $\forall x (x \in A \equiv x \in B) \Rightarrow A = B$? It does not 'extend' anything.

It doesn't mean 'extension' as in 'make something bigger', it means extension as in 'the ability to include more things'. The reasoning goes a bit deeper into relations.

Background:
A relation $R$ on a set $X$ can be thought of in a few ways (all are equivalent). One way is as a subset $R \subseteq X \times X$, which means it's a collection of ordered pairs $(x,y)$ where $x,y \in X$. (An example of a relation is the set $X \times X$ itself, which can be thought of as the set of all 'coordinates'.) Relations generalise functions: a function $f$ is a relation for which for each $x$ there is exactly one $y$ such that $(x,y) \in f$, and we denote this $y$ by $f(x)$. In this sense, we can think of a relation as a 'multi-valued function'. (For instance, the set containing all $(x,y)$ where $y=\sqrt{x}$ or $y=-\sqrt{x}$ is a relation, but not a function.) But the nitty-gritty aside, just think of a relation $R$ as being a set containing ordered pairs of elements of some set.

Now, if $(x,y) \in R$ then we say $x$ is an '$R$-predecessor' of $y$. So for instance if $f$ is a function then you could say (although most people don't) that $x$ is an $f$-predecessor of $f(x)$ for each $x$ in the domain of $f$.

The relevant bit:
A relation $R$ is called 'extensional' if each element $y \in X$ is determined by its $R$-predecessors. That is, if for all $x \in X$ $(x,y) \in R \Rightarrow (x,y') \in R$, then $y=y'$. The axiom of extensionality, therefore, is just the assertion that $\in$ is an extensional relation.

where can I learn what all that notation means?

...

The relevant bit:
A relation $R$ is called 'extensional' if each element $y \in X$ is determined by its $R$-predecessors. That is, if for all $x \in X$ $(x,y) \in R \Rightarrow (x,y') \in R$, then $y=y'$. The axiom of extensionality, therefore, is just the assertion that $\in$ is an extensional relation.

I see. Just for clarity, this notion of a ``relation" is as general as it can be.
e.g. I don't attribute it any properties other than being a subset of a Cartesian product?

Is $x \in X$ $(x,y) \in R \Rightarrow (x,y') \in R$ the same as

If for all $x \in X$, we have $x R y \Rightarrow x R y'$, then $y = y'$?

In this case, I get that the axiom of extensionality would be: $R$ is $\in$ and $y = A, y' = B$.

Following this logic, would $=$ and $\equiv$ be extensional relations?

Yes. But, for instance, if we write $H \le G$ to mean that $H$ is a subgroup of a group $G$ (where we consider the groups as being 'up to isomorphism'); then $\le$ is not extensional (for instance, $D_6$ and $C_6$ have the same subgroups).

Oh, yes. I wasn't very sure about the meaning of $=$ in that case, but took it as being the same object.

Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful -- I notice most of these have no prerequisites.
Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )

Promise me you won't just do maths this summer? I'm a bit worried.

Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful -- I notice most of these have no prerequisites.
Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )