TITLE: Amazing isomorphisms
QUESTION [46 upvotes]: Just as a recreational topic, what group/ring/other algebraic structure isomorphisms you know that seem unusual, or downright unintuitive?
Are there such structures which we don't yet know whether they are isomorphic or not?

REPLY [9 votes]: "N.H." has already posted this one, but only very tersely, and it's worth a few comments.
$$
L^2(\mathbb S^1) \cong \ell^2(\mathbb Z)
$$
Let us take the former to mean the set of all complex-valued functions $f$ on $\mathbb S^1 = \mathbb R/(2\pi)$, i.e. periodic functions of period $2\pi$, that satisfy
$$
\int_{\mathbb S^1} |f(x)|^2 \, dx < \infty.
$$
Note that since $f(x)$ is complex and not necessarily real, we need to say $|f(x)|^2$ rather than $f(x)^2$.  Such functions have a Fourier series
$$
f(x) \sim \sum_{n=-\infty}^\infty c_n e^{inx}.
$$
The meaning  of $\text{“}{\sim}\text{''}$ might take some discussion, but for now let us note that Carleson's theorem says the measure of the set of points $x$ at which the series fails to converge to $f(x)$ is $0$, and a much more easily proved theorem says it converges in $\ell^2(\mathbb Z)$.  The space $\ell^2(\mathbb Z)$ is the space of sequences $\{c_n\}_{n=-\infty}^\infty$ for which $\displaystyle\sum_{n=-\infty}^\infty |c_n|^2 < \infty$ with the inner product $\displaystyle \langle b,c\rangle = \sum_{n=-\infty}^\infty b_n \overline{c}_n$, where $\overline c$ is the complex cojugate of $c$.  One $L^2(\mathbb S^1)$ we have the inner product $\displaystyle \langle f,g\rangle = \int_{\mathbb S^1} f(x) \overline{g(x)}\,dx$.
The Riesz–Fischer theorem says these two inner product spaces are isomorphic, and in particular that the transform from $f$ to $\{c_n\}_{n=-\infty}^\infty$ is an isomophism.
Notice that here we have to construe $f$ to mean the equivalence class of $f$ where two functions are equivalent if they're equal almost everywhere.  If we didn't do that then (1) two functions differing on a non-empty subset of measure $0$ in the domain would be at distance $0$ from each other, so we wouldn't quite have an inner product space, and (2) the cardinalities of the underlying sets of the two spaces would differ.  This last fact enables us to say just what the cardinality of the space $L^2(\mathbb S^1)$ is: It's not actually bigger than that of $\ell^2(\mathbb Z)$.