/math/ general

You can be injective but not surjective with the same infinite cardinal.

OOOOOOOOOH

You motherfucker.

You got me man. You fucking got me. I guess I only have finite IQ.

They are the same infinities, but infinity gives you a lot of room to move around.

So let me show you a counterexample. Let G be the free group with 2 generators. That is, the group generated by the elements {a,b} with no extra relations. Elements of this group are words consisting of a, b and their inverses. It is an infinite group. Similarly let H be the free group with 3 generators. Same thing as before, the words in the alphabet {x,y,z} and their inverses.

There is a rather natural injection from G to H, just send a to x and b to y. What about the other way around?

There is in fact, an injection from H to G. One is given by

g(x) = aa
g(y) = bb
g(z) = ab

(Can you prove this morphism is injective?)

These groups are also not isomorphic, as an isomorphism would have to send two elements of H to a and b and so the proper subgroup of H generated by those two would be all of H, which is impossible.

We have a small chat where I sometimes discuss things like this, and people talk about things they are studying. If you want an invite, you can create a disposable email (you can google that) and I can send you one.

I do want an invite. I have no math friends irl.

yucek@apkmd.com

Can I get an invite too, man? I hardly have anyone to talk about math related things.
bicrotrago@housat.com

The answer is "no" for infinite groups.

I'm trying to come up with a counter example.

Does every periodic function have an smallest period?

No. Consider the infinite symmetric group [math]\mathfrak S(\mathbb N)[/math].
We have an injection [math]\mathfrak S(\mathbb N) \hookrightarrow \mathbb Z_2 \times \mathfrak S(\mathbb N)[/math] mapping [math]\sigma[/math] to [math](0, \sigma)[/math].
Conversely, we have a morphism [math]\mathbb Z_2 \times \mathfrak S(\mathbb N) \hookrightarrow \mathfrak S(\mathbb N)[/math] sending [math](k, \sigma)[/math] to [math](1,2)^k \tilde\sigma[/math] where [math]\tilde \sigma(n) = \sigma(n+2)[/math].
These groups are not isomorphic because [math]\mathbb Z_2 \times \mathfrak S(\mathbb N)[/math] has a normal subgroup of order 2, whereas [math]\mathfrak S(\mathbb N)[/math] doesn't. Indeed, a subgroup of order 2 of [math] \mathfrak S(\mathbb N)[/math] is necessarily generated by an element of the center (think about it), but [math]\mathfrak S(\mathbb N)[/math] has trivial center.

Nope! Easy counter example: a constant function

Let's exclude that then, does every nonconstant periodic function have an smallest period?