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The Whitney graph isomorphism theorem gives an example of an extraordinary exception: a very general statement holds except for one very specific case.

Another example is the classification theorem for finite simple groups: a very general statement holds except for very few (26) sporadic cases.

I am looking for more of this kind of theorems-with-not-so-many-sporadic-exceptions.

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There are many examples coming from Waring's problem: Every sufficiently large number is a sum of at most 7 cubes, for example. But, removing "sufficiently large" requires changing the 7 to a 9. – Andrés E. Caicedo Aug 24 '12 at 0:32
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Theorem: All natural numbers are larger than $10$, except $0,1,2,3,4,5,6,7,8,9$. – Asaf Karagila Aug 24 '12 at 0:35
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@Hans: I know it's not. I'm trying to prod the discussion into helping you shape the definition in your head, so you could write it up. If I thought this is something you're looking for, I'd post it as an answer. – Asaf Karagila Aug 24 '12 at 0:41
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@Asaf We need to add 10 to the list of exceptions, no? – Ragib Zaman Aug 24 '12 at 3:11
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Actually all answers to this question should qualify. – celtschk Aug 24 '12 at 7:15

34 Answers 34

Every automorphism of $S_n$ is inner if $n \neq 6$.

P.S. That $S_6$ has an `essentially' unique outer automorphism is quite a non-obvious fact.

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I've posted this answer to other questions, but it's worth repeating:

The topological manifold $\mathbb{R}^n$ has a unique smooth structure up to diffeomorphism as long as $n \neq 4$. However, $\mathbb{R}^4$ admits uncountably many exotic smooth structures.

Edit: Other thoughts:

  • The h-cobordism theorem holds for $n= 0, 1, 2$ and $n \geq 5$. It is open for $n = 3$ and false (smoothly) for $n=4$.

  • I've heard that many theorems in number theory only work for field characteristic $\neq 2$ (but my number theory background is sadly lacking).

  • $(\mathbb{Z}/n\mathbb{Z})^\times$ is cyclic iff $n = 1, 2, 4, p^k, 2p^k$, where $p$ is an odd prime, $k \geq 1$. (Granted, there aren't a finite number of exceptions here, but I wanted to mention it anyway.)

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Well in Galois theory there is a simple test using the discriminant to see if the Galois group of an irreducible cubic is $A_3$ or $S_3$, but sadly it fails in characteristic 2 because $1 = -1$. – user38268 Sep 3 '12 at 23:21
    
copied from wikipedia – user135041 Apr 8 '14 at 22:00
    
@Herbert: Guilty as charged: some, but not all, of what I've written is indeed copied verbatim. Still, I don't really think that affects the quality of the answer. Also, I like your new username :-) – Jesse Madnick Apr 8 '14 at 22:34

Carmichael's Theorem: Every Fibonacci number $F_n$ has a prime factor which does not divide any earlier Fibonacci number, except for $F_1 = F_2 = 1$, $F_6 = 8$ and $F_{12} = 144$.

This is a special case of his more general result: Let $P,Q$ be nonzero integers such that $P^2 > 4Q$, and consider the Lucas sequence $D_1 = 1$; $D_2 = P$; $D_{n+2} = P \cdot D_{n+1} - Q \cdot D_n$. Then all but finitely many $D_n$ have a prime factor which does not divide $D_m$ for any $m < n$; the only possible exceptions are $D_1$, $D_2$, $D_6$ and $D_{12}$.

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It's amazing to me that the exceptions always (potentially) happen at the same members of the sequence regardless of the values of $P$ and $Q$! Is there a brief explanation of why this happens? Are there infinitely many $P,Q$ for which e.g. $D_{12}$ has no prime factors that don't divide $D_n, n\leq 11$? – Steven Stadnicki Aug 24 '12 at 20:13
    
@Steven: There is a fairly readable proof in Minoru Yabuta, A simple proof of Carmichael's theorem on primitive divisors, Fib. Quart., vol.39, pp.439-443. The reason that these terms are the only possible exceptions is that the problem for all Lucas sequences can be reduced to an analysis of the Fibonacci sequence and the Fermat sequence ($M_1 = 1$; $M_2 = 3$; $M_{n+2} = 3 M_{n+1} - 2M_n$). In slightly more detail, to each Lucas sequence $D_n(P,Q)$ we can associate another sequence $E_n(P,Q)$. (cont...) – arjafi Aug 25 '12 at 5:58
    
@Steven: (...inued) A sufficient condition that $D_n(P,Q)$ has a "primitive divisor" is that $E_n(P,Q)$ is "large enough". For each $n>2$ we can also show that either $E_n(1,-1)$ (Fibonacci) or $E_n(3,2)$ (Fermat) is the minimum for all $E_n(P,Q)$. In this way we reduce the problem to those two sequences. It does seem that the Fibonacci sequence is the only one where the 12th term has no primitive divisor. – arjafi Aug 25 '12 at 5:59

Every simply connected subregion of $\mathbb C$ is conformally equivalent to the unit disk, with the exception of $\mathbb C$ itself.

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For squarefree positive integer $d$, every imaginary quadratic field $\mathbb{Q}(\sqrt{−d})$ has class number greater than 1 unless $d$ is equal to one of the Heegner numbers: 1, 2, 3, 7, 11, 19, 43, 67, 163.

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While this is true, I don't think it is a very good example of what was being asked, because there are analogous explicit results for other class numbers, and it is known that the class number tends to infinity as $-d$ tends to $-\infty$ (so that it is known that every given class number will belong to only finitely many imaginary quadratic fields). – Gro-Tsen Jan 19 '14 at 22:28

Here's a neat result that holds for all finite groups except if the Monster shows up as a quotient:

Let $G$ be a finite group and $p,q$ primes dividing $|G|$. If $G$ contains no element of order $pq$, then either

  1. the Sylow $p$-subgroups or the Sylow $q$-subgroups are abelian, or
  2. $G/O_{\{p,q\}'}(G)$ is the Monster and $\{p,q\}=\{5,13\}$ or $\{7,13\}$.

[source]

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Mitchell's theorem: A primitive complex reflection group is either the symmetric group $S_n \subseteq GL_{n-1}(\mathbb{C})$ or one of $34$ exceptions.

Definitions: a complex reflection group is a finite subgroup $W$ of $GL_n(\mathbb{C})$ for some $n$, that is generated by reflections. A reflection is an invertible matrix with codimension one fixed space. A complex reflection group $W \subseteq GL_n(\mathbb{C})$ acting on $V=\mathbb{C}^n$ is imprimitive if there is a decomposition $V=V_1 \oplus V_2$ such that for each $w \in W$, and $i=1,2$, $w(V_i) \subseteq V_j$ for some $j$. Otherwise it is primitive.

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Was this proven before Todd-Shephard's theorem? – Mariano Suárez-Alvarez Aug 24 '12 at 20:34
    
Well, to be fair I believe what Mitchell actually proved (long before Shepard-Todd wrote down the famous list) is that a primitive CRG is either the symmetric group in its reflection rep'n, or lives in dimension at most $8$. But given this theorem, producing the list of exceptions is sort of inevitable (but still interesting). I think Shepard-Todd probably get a bit too much credit for the classification of irreducible CRGs, when it was really a community effort. – Stephen Aug 24 '12 at 22:58

How about the Big Picard theorem? http://en.wikipedia.org/wiki/Picard_theorem

If a function $f:\mathbb{C}\to \mathbb{C}$ is analytic and has an essential singularity at $z_0\in \mathbb{C}$, then in any open set containing $z_0$, $f(z)$ takes on all possible complex values, with at most one possible exception, infinitely often.

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There are several theorems in topology and geometry that hold for manifolds of all dimensions save 3 and sometimes 4. For example, of Euclidean space except $\mathbb{R}^3$ and sometimes $\mathbb{R}^4$. There are also some conjectures proved for $\mathbb{R}^n$, n ≠ 3 (resp {3,4}), but which are open questions for n = 3 ({3,4}). Wikipedia describes it thus:

[T]he cases $N = 3$ or $4$ have the richest and most difficult geometry and topology. There are, for example, geometric statements whose truth or falsity is known for all N except one or both of 3 and 4. N = 3 was the last case of the Poincaré conjecture to be proved.

Using the regular polytopes as an example:

  • There are exactly 3 regular (convex) polytopes, the n-simplexes, -cubes, and -orthoplexes which exist in all dimensions, except 3 and 4, which have several additional Platonic solids and regular polychora without higher analogues (ignoring the trivial cases of dimensions 1 and 2).
  • There are no regular non-convex polytopes except in dimensions 2, 3, and 4.

Further,

In a sense, 3- and 4-dimensional spaces are privileged. This has implications in the philosophy of physics: why did space (apparently) have 3 spatial dimensions rather than 2 or 527?

See also 3-manifold, special phenomena of 4-manifolds, low-dimensional topology.

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Map Color Theorem: For any surface $\Sigma$ with Euler characteristic $c \leq 0$, with the exception of the Klein bottle, $$\chi(\Sigma) = \frac{1}{2} (7 + \sqrt{49 - 24c}),$$ where $\chi$ is the chromatic number.

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Vinogradov's theorem that all but a finite number of odd numbers are the sum of three primes.

Not sure about the current state of Goldbach's conjecture.

A theorem at about my level of math: All primes greater than two are odd.

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There are effective versions of Vinogradov's three-primes theorem: one due to Liu and Wang ams.org/mathscinet-getitem?mr=1932763 states that every odd integer $> \exp(3100)$ is the sum of three odd primes. Of course it's likely to be true for all odd integers $\ge 9$ (and that is the case if the Generalized Riemann Hypothesis is true), but there's still some distance to go before that can be verified. – Robert Israel Aug 24 '12 at 5:25
    
I'm not sure the third one really counts. I could say "all primes other than three are not multiples of three." It's just that we don't have a special name for numbers that aren't multiples of 3. – Michael Lugo Aug 24 '12 at 21:32
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How about this then: All but one or two or three of my statements are valid answers to the OP question. – marty cohen Aug 25 '12 at 0:18
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Just the other day a proof of Goldbach's weak conjecture was announced. – lhf Jun 2 '13 at 14:18

The only consecutive integer powers are 8 and 9.

This was conjectured by Eugène Catalan in 1844 and proved by Preda Mihăilescu in 2002.

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Polynomial equations of every degree have no general solution in radicals, with the exception of degrees 1, 2, 3 and 4.

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The only spheres which admit almost complex structures are $S^2$ and $S^6$.

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+1 I love this fact! – Jesse Madnick Sep 11 '13 at 9:15

Brownian motion is transient in every dimension, with the exception of dimensions 1 and 2.

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Not sure if this fits the scope of the question, it's hardly as advanced as all the other suggestions but I find it quite elegant and surprising:

There exists no positive integer sandwiched between a perfect square and a perfect cube, with the sole exception of $26$ - equivalent to saying that the only solution of $a^2 \pm 2 = b^3$ is $(5, 3)$.

Of course this is just one among many Diophantine equations which admit only one solution, but it has a special significance when viewed from a less abstract point of view.

Another nice one:

For any prime $p$, the product of its primitive roots is congruent to $1$ modulo $p$, except for $p = 3$ for which the product is equal to $2$ (result due to Gauss).

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I liked this fact a lot two years ago. I'm 28. – Michael Lugo Aug 24 '12 at 21:32

Fermat's Last Theorem: For any positive integer $n$, except $n = 1, 2$, there is no solution of positive integers $(x, y, z)$ to the equation $$x^n + y^n = z^n.$$

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It is somewhat interesting as a result on its own (but for this question, because it succeeds for $n \ge 3$, I don't consider it is relevant), but when dividing throughout by $z^n$ you get that all the paths $s^n+c^n=1$ only have rational points $(s,c)$ for $n=1,2$. – Mark Hurd Aug 26 '12 at 3:30

The "sausage catastrophe" for finite sphere packings comes to mind:

For 1..55 spheres a linear "sausage" is the optimal packing, for higher numbers some cluster packing is optimal.

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Optimal by what measure? – Steven Stadnicki Aug 24 '12 at 20:21
    
Smallest volume of a convex "wrapping". Look at de.wikipedia.org/wiki/Theorie_der_endlichen_Kugelpackungen (sorry, no english entry). – Landei Aug 24 '12 at 21:37
    
+1 not for the $\le 55$, but for the $57, 58, 63 \text{ and } 64$ other numbers of spheres where the sausage is still optimal. – Mark Hurd Aug 28 '12 at 7:47
    
Translated from German Wikipedia: "In 1992 Pier Mario Gandini and Jörg Michael Wills could show, that starting with 56 spheres (with exception of 57, 58, 63, 64) the linear order as sausage isn't the best one, but that in this case a cluster packing is better, which is not a sausage packing. Meanwhile it was shown that for the four exceptions the sausage packing isn't optimal either. So the sausage catastrophe must occur at least at 56 spheres." – Landei Aug 28 '12 at 8:41

How about the Ax-Kochen theorem?

Every homogeneous polynomial of degree $d$ in $n$ variables with $n>d^2$ has a non-trivial zero in $\mathbb{Q}_p$ for all but finitely many $p$... and the finite set of exceptions depends on the degree $d$.

Here are some things we know about the finite exception set for various $d$: http://en.wikipedia.org/wiki/Ax-Kochen_theorem#Exceptional_primes

This is a special case of the more general fact that any first-order sentence (in the language of valued fields) which is true of all but finitely many Laurent series fields $\mathbb{F}_p((t))$ is true of all but finitely many $p$-adic fields $\mathbb{Q}_p$. Model theory gives us many more examples of the principle "true in char $0$" = "true in char $p$ for large enough $p$".

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The classification of simple finite dimensional Jordan algebras has exactly one exceptional case.

That of simple Lie algebras has five exceptions.

(All this over sensible fields, of course)

There are 3 regular polyhedra in every dimension, except in dimesions 2, 3 and 4.

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Here's a beautiful theorem of Peter J Cameron from the theory of designs:

Theorem. If symmetric $2-(v,k,\lambda)$ design $\mathscr{D}$ extends, then it is one of the following :

  1. $2-(4\lambda+3,\;\; 2\lambda+1,\;\; \lambda )$1
  2. $2-((\lambda+2)(\lambda^2+4\lambda+2), \;\;\lambda^2+3\lambda+1, \;\;\lambda)$
  3. $2-(495,39,3)$

This appeared in the 1973 paper of Prof. P J Cameron [Cam]. When it was stated, the existence of the design of parameters $2-(111, 11, 1)$2 was yet undecided. It has been now proved with an extensive computer search $[10]$ that this design does not exist.

Some (perhaps) Useful References.

[Cam] Cameron P. J., Extending Symmetric Designs, Journal of Combinatorial Theory, Series A Vol. 14, Issue 2 (Mar., 1973), pp. 215-220.

$[10]$ Lam C. W. H., Thiel L. H., Swiercz S., The Non-existence of Finite Projective Plane of Order 10 Can. J. Math., XLI (1989), pp. 1117-1123.

1 Note that these are the parameters of a Hadamard $2$-design.
2Some readers will recognise that this is a projective plane of order 10.

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P.S. This fits in here, because, if we ever have extensible, two designs, there are only finitely many that don't fit into an infinite family (but it could happen that , ... forbid, there are only finitely many). – user21436 Aug 24 '12 at 1:14

Any number of theorems about quadratic forms are true over fields where the characteristic is not equal to 2. My favorite reference for the subject, where this sense of "nothing works the same in characteristic 2" is made clear as early as the "Notes to the Reader," is Lam's Introduction to the Theory of Quadratic Forms over Fields. For an even more elementary example of this phenomenon, note that the quadratic formula only holds (or even makes sense) for fields of characteristic not equal to 2.

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A mathematician said “Who / Can quote me a theorem that's true? / For the ones that I know / Are simply not so / When the characteristic is two!” – PseudoNeo Aug 24 '12 at 19:31
    
+1. The above poem is quoted in the "Notes to the Reader" in Lam's book. – Brett Frankel Aug 28 '12 at 21:33

No free group is amenable, with the exception of $\mathbb Z$.

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No free group is abelian, with the exception of $\mathbb Z$. – j.p. Oct 31 '12 at 19:16

Every family of graphs that is closed under minors has at least one forbidden minor, with the exception of the family of all graphs.

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As stated, this is trivial, given that every graph is a minor of itself. Perhaps you meant the fact that every family of graphs that is closed under minors can be defined by a finite set of forbidden minors? – Ilmari Karonen Aug 26 '12 at 16:25

Every prime number is odd, with the exception of 2.

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All prime numbers are odd, and $2$ is the oddest of them all. – Georges Elencwajg Aug 24 '12 at 8:58
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@GeorgesElencwajg Although Hadamard-de la Vallée Poussin and Erdős–Kac taught us that they even out, in the end... – Did Aug 24 '12 at 9:10
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No prime number is divisible by $57$, apart from $57$. – PseudoNeo Aug 24 '12 at 19:25
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@PseudoNeo Funny, I thought the correct statement was that no prime number is divisible by 42, apart from 42. – Did Aug 24 '12 at 20:42
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@Thomas It is well known that in this universe 42 is everything: composite, prime, finite, infinite, and the answer to every question... (By the way, 57 could have elicited the same kind of reaction from you.) – Did Aug 25 '12 at 8:13

Other than the sphere, all closed surfaces have list chromatic number equal to chromatic number.

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The sphere $S^n$ is simply connected if and only if $n\geq 2$.

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And connected if and only if $n\geqslant1$. – Did Aug 24 '12 at 20:12

There exists an Aperiodic Tiling of $\mathbb{R}^d$ for all dimensions excepting $d=1$. For $d=1$ it is easy to show that an aperiodic set of tiles cannot exist.

Also, for $d \geq 3$ there exists an aperiodic single tile, the problem is still open in $d=2$ (A non-connected single aperiodic tile was discovered 2 years ago, but usually we ask for connected tiles).

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A graph is planar, unless it contains a copy of $K_5$ or $K_{3,3}$.

Every subgroup of $\mathbb{Q}$ is residually finite, except $\mathbb{Q}$ itself.

The only two perfect squares in the sequence $\lbrace\displaystyle\sum_{i=1}^n i^2\rbrace_n$ are $1^2$ and $70^2$.

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$A_n$ has no nontrivial normal subgroup unless $n=4$.

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