Amazing isomorphisms

Question

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?

Answer 1

I don't know if this is really amazing, but I was quite surprised when I did discover these isomorphisms:

• $\mathbb R^n$ and $\mathbb R^m$ are isomorphic as abelian groups (where $n,m \geq 1$). In fact, pick any Hamel basis of $\mathbb R^n$ and $\mathbb R^n$ as $\mathbb Q$ vector space. Both basis are in bijection with $\mathbb R$, so any bijection between them give an isomorphism of $\mathbb Q$ vector spaces, in particular of abelian groups.
• A really surprising example (for me at least): the free group with countably many generator is isomorphic to a subgroup of $F_2$, the free group with two generators! This fact comes from the covering $\sin : X \to X'$ where $X = \mathbb C \setminus \{\frac{\pi}{2} + k\pi, k \in \mathbb Z\}, X' = \mathbb C \setminus \{-1,1\}$, and the more general fact that for any covering between nice space the induced maps on fundamental groups is injective.
• $L^2(\mathbb S^1) \cong \ell^2(\mathbb Z)$ via Parseval equality.
• The space $\mathcal M_k$ of modular forms of weight $2k$ is finite dimensional (which is already non-trivial) and moreover, if $\mathcal M = \bigoplus_{k \in \mathbb N} \mathcal M_k$, we have the a graded algebra isomorphism $\mathcal M \cong \mathbb C[x,y] $ where $x$ has degree $2$ and $y$ has degree $3$ (they represents Eisenstein series $G_4(z)$ and $G_6(z)$ respectively). This is quite surprising that the algebra of all modular forms is simply isomorphic to an algebra of polynomials !
• Finally, a cute example: the group of direct isometries which preserve a cube are isomorphic to the symmetric group $\mathfrak S_4$. In fact, this is exactly the permutation group of the big diagonals $d_1,d_2,d_3,d_4$ of the cube.

Answer 2

My favorite, and a classical example, is the Outer automorphism of $S_6$, which loosely arises from the 'coincidence' that ${6\choose 2}=15=\frac{1}{2^3}\frac{6!}{3!}$ — that is, the number of (unordered) pairs of elements of $\{1,2,3,4,5,6\}$ is exactly the same as the number of partitions into three pairs. John Baez has a nice essay that offers more details on exactly how the automorphism can be defined.

Answer 3

One of my favourites (and certainly one of the things that introduced me to the concept of isomorphism, and perhaps an obvious one) is the isomorphism of $\Bbb C$ and $(M, \times)$, where $$M = \left\lbrace \begin{pmatrix}a& -b \\ b &a \end{pmatrix} : a, b \in \Bbb R, a^2 + b^2 \neq 0\right\rbrace$$

given by $\varphi:\Bbb C\setminus\{ 0 \} \to (M, \times)$, with $(a + bi) \mapsto \begin{pmatrix}a &-b\\ b & a \end{pmatrix}.$

You learn early on that the complex numbers are geometric in nature, and you also learn that matrices encode geometric features. This was a nice connection for me and was in fact one of the things that got me interested in mathematics in the first place!

Answer 4

Consider groups $\mathrm{PSL}_2(\mathbb{F}_p)=\mathrm{SL}_2(\mathbb{F}_p)/Z$, $Z$ being the center of the group. Then $$\mathrm{PSL}_2(\mathbb{F}_4)\cong \mathrm{PSL}_2(\mathbb{F}_5).$$ Another example, consider commutator subgroup of $\mathrm{SL}_2(\mathbb{Z})$, it is isomorphic to free group on $2$ letters.

Answer 5

If $M=\prod_{n=1}^{\infty}\mathbb{Z}$ and $R=\mathrm{End}_{\mathbb{Z}}(M)$, then $R\simeq R^2$ as $R$-modules.

Answer 6

"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 $\ell^2(\mathbb Z)$.

Answer 7

There is an order isomorphism between the linearly ordered set of all rational numbers and the linearly ordered set of all real algebraic numbers. Georg Cantor discovered that.

Answer 8

According to this thread here: Splitting in Short exact sequence, we have that $\mathbb R \cong \mathbb R \oplus\mathbb R/\mathbb Q.$

It is also a well known fact that $\mathbb R \cong \mathbb R / \mathbb Q$.

We can put these together to create a small handful of odd isomorphisms. For example, $\mathbb R / \mathbb Q \cong \mathbb R \oplus \mathbb R / \mathbb Q. $. Personally, it seems strange to me as it looks like the right side has "more stuff" than the left - it has a copy of itself for every real number.

We can take this a step further, and find an example of a group which is isomorphic to a sum of two copies of itself, $ \mathbb R / \mathbb Q \cong \mathbb R / \mathbb Q \oplus \mathbb R / \mathbb Q$, which is perhaps also strange, but it's reasonable if you already know that $\mathbb R^n \cong \mathbb R^m$ for all integers $n,m$.

Answer 9

Put $R = C^\infty(S^n)$ for $n > 0$ even. The space $M = \Gamma(TS^n)$ of sections of the tangent bundle (i.e., vector fields) of $S^n$ is a finitely generated $R$-module, but it is not free. On the other hand, the normal bundle $\nu \to S^n$ is trivial, and the isomorphism $TS^n \oplus \nu = \mathbb{R}^{n+1}$ gives an isomorphism $M\oplus R = R^{n+1}$. (And far from being a novelty, this is the start of algebraic and topological $K$-theory.)

Answer 10

Via the existence of Hamel basis over $\mathbb{Q}$ one can show that the real numbers $\mathbb{R}$ and the complex numbers $\mathbb{C}$ are isomorphic as abelian groups. In a similar vein, if $k$ is an uncountable field, then for every $n \in \mathbb{N}$, $k \cong k[X_1, \ldots, X_n]$ as abelian groups.

Answer 11

My favorite isomorphism is the one used in De Rham's theorem, given by integration.

The surprise is that objects that arise in analysis turn out to be purely topological invariants. (This is one of the great examples of two objects that look different turning out to be equivalent.) To me this is comparable to the surprise of Gauss's Theorema Egregium.