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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?

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    $\begingroup$ It's not deep but I've always been very fond of this observation: For every group $(G,\cdot)$ define a new group $(H,\circ)$ as follows: as a set, $G = H$ but the product is given by $a\circ b = b\cdot a$. It is easy to verify that this is actually a group. But as it turns out, $G\cong H$ ! $\endgroup$
    – Myself
    Jul 15, 2016 at 21:57
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    $\begingroup$ The circle group $S^1$ is isomorphic to $C^*$, the non-zero complex numbers. Also $\mathbb R$ and $\mathbb C$ are isomorphic as abelian groups. $\endgroup$
    – Arkady
    Jul 15, 2016 at 22:04
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    $\begingroup$ This is possibly the opposite (sorry) of what you asked for and @Myself described, but we can construct the opposite ring $A^{\text {op}}$ of a noncommutative ring $A$: Multiplication is given by $a.b = ba$, and addition is unchanged. For some quite reasonable $A$, the rings $A$ and $A^{\text{op}}$ are not (abstractly) isomorphic. $\endgroup$
    – anomaly
    Jul 15, 2016 at 22:19
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    $\begingroup$ As fields, we have $\mathbb{C} \simeq \overline{\mathbb{C}(X)}$, where the latter denotes the algebraic closure of $\mathbb{C}(X)$ (of course, this isomorphism does not preserve any additional structures such as the topology or $\mathbb{C}$-vector space structure). $\endgroup$
    – Joel Cohen
    Jul 15, 2016 at 22:43
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    $\begingroup$ @SpamIAm: They're both divisible and have isomorphic torsion subgroups, so the only thing left to determine is the rank of the torsion-free parts. $\endgroup$
    – anomaly
    Jul 16, 2016 at 4:25

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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.
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  • $\begingroup$ On the topic of the second item, any subgroup of a free group is also free (the easiest way to prove this is via covering spaces, though it's not hard to prove directly). Choose a "badly behaved" set of $F_2$ and the group it generates is going to be a free group of large rank. $\endgroup$
    – anomaly
    Jul 15, 2016 at 22:17
  • $\begingroup$ I agree but for me that a infinite free group embed inside a finitely generated free group looked strange ! Of course now it seems normal but the first time it surprised me a bit. $\endgroup$
    – user171326
    Jul 15, 2016 at 22:19
  • $\begingroup$ The third one was startling for me, especially before I saw a picture $\endgroup$ Jul 15, 2016 at 22:54
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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.

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    $\begingroup$ $\ldots\,$and the group of all automorphisms of $S_6$ has three subgroups of index $2$, no two of which are isomorphic to each other. (The group of all inner automorphisms of $S_6$ is one of those.) $\qquad$ $\endgroup$ Jul 16, 2016 at 3:38
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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!

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    $\begingroup$ If $k$ is a field, every finite dimensional $k$-algebra $A$ has a representation as matrices of a similar form: map $a\in A$ to the matrix representing the linear map $x\in A\mapsto ax\in A$ with respect to some fixed basis of $A$. $\endgroup$ Jul 15, 2016 at 22:24
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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.

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

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

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    $\begingroup$ Great answer ! Do you know if there is similar isomorphisms with the Wiener algebra $\mathcal A(S^1)$ ? $\endgroup$
    – user171326
    Jul 16, 2016 at 18:27
  • $\begingroup$ Thank you. I don't know about that last item. $\qquad$ $\endgroup$ Jul 16, 2016 at 21:13
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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.

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  • $\begingroup$ Where can i find more about that? $\endgroup$
    – Hemispherr
    Jul 16, 2016 at 4:21
  • $\begingroup$ @user4773863 : Start by googling "Cantor's back-and-forth method". $\qquad$ $\endgroup$ Jul 16, 2016 at 4:27
  • $\begingroup$ Maybe I should add that as far as I know, all such order-isomorphisms are messy; there's nothing like a closed form. $\qquad$ $\endgroup$ Jul 16, 2016 at 17:53
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    $\begingroup$ @user4773863: in fact all countable dense linear orders without endpoints are isomorphic by the same proof. $\endgroup$ Jul 17, 2016 at 0:09
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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 $n,m$.

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  • $\begingroup$ I think this is meant as an isomorphism of vector spaces over $\mathbb{Q}$ (where you even have $\mathbb{R} \simeq \mathbb{Q} ^{\mathbb{R}} $ by picking a basis) . $\endgroup$
    – Joel Cohen
    Jul 16, 2016 at 13:21
  • $\begingroup$ @6005 As $\mathbb{Q}$-vector spaces, and, in particular, as Abelian groups. The quotient topology on $\mathbb{R}/\mathbb{Q}$ is trivial, however. $\endgroup$ Jul 16, 2016 at 18:45
  • $\begingroup$ Thanks @AlexProvost, you're right. I just made a mistake in my thinking about $\mathbb{R}/\mathbb{Q}$. $\endgroup$ Jul 16, 2016 at 18:49
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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.)

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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.

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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.

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    $\begingroup$ Along similar lines, there's the Gauss-Bonnet theorem: For a closed $2$-manifold $\Sigma$ with curvature form $K$, we have $\int_\Sigma K = 2\pi \chi(\Sigma)$. The interesting bit is that the left side depends on the choice of Riemannian metric on $\Sigma$, while the right side is a purely topological invariant. There's also a higher-dimensional extension due to Chern. $\endgroup$
    – anomaly
    Jul 16, 2016 at 15:09

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