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It is sometimes the case that one can produce proofs of simple facts that are of disproportionate sophistication which, however, do not involve any circularity. For example, (I think) I gave an example in this M.SE answer (the title of this question comes from Pete's comment there) If I recall correctly, another example is proving Wedderburn's theorem on the commutativity of finite division rings by computing the Brauer group of their centers.

Do you know of other examples of nuking mosquitos like this?

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I once saw someone proving resolutions of singularities of curves by quoting Hironaka's theorem. –  Richard Borcherds Oct 17 '10 at 15:23
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Brauer groups and cohomology are certainly overkill for Wedderburn's theorem: if $D$ is a finite division algebra and $L$ is a maximal subfield, then the Noether-Skolem theorem shows that the multiplicative group of $D$ is a union of conjugates of that of $L$; hence $D$=$L$. –  JS Milne Oct 17 '10 at 20:07
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@Maxime: I have trouble believing that such a proof is actually non-circular. Surely such proofs form a step, however easy, in the classification. –  Qiaochu Yuan Oct 17 '10 at 21:59
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I once convinced myself the Cantor set is non empty because it is a descending intersection of non empty closed subsets of a compact set, before noticing it contains 0. –  roy smith Jan 29 '11 at 6:48

67 Answers 67

(1) Let $G$ be a finite group. Let $H\leqslant G$ be a subgroup of index $2$. Let us prove that $H$ is normal in $G$. Let $L|K$ be a Galois extension of fields with Galois group $G$ (easily constructed via a representation of $G$ as a permutation group, taking $L$ to be a function field in suitably many variables on which $G$ acts and $K$ to be the fixed field under $G$). Let $F$ be the fixed field in $L$ under $H$. Then $F|K$ is a quadratic extension, hence normal. By the Main Theorem of Galois Theory, it follows that $H$ is normal in $G$.

(2) Let $G$ be a finite group. Let $K$ be a finite field of characteristic not dividing $|G|$. Let us prove Maschke's Theorem in this situation: $KG$ is semisimple. Given two finite dimensional $KG$-modules $X$ and $Y$, it suffices to show that $\text{Ext}^1_{KG}(X,Y) = 0$. But $\text{Ext}^1_{KG}(X,Y) = \text{H}^1(G,\text{Hom}_K(X,Y)) = 0$, since $|G|$ and $|\text{Hom}_K(X,Y)|$ are coprime.

(Well, not sure whether any of these arguments are really awfully sophisticated. It's rather breaking a butterfly on a small wheel.)

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The second one is actually a great, non-awfully sophisticated argument :) It is essentially the same as the purely algebraic proof of complete reducibility of finite dimensional modules over a fin. dim. semisimple Lie algebra---in that case, one replaces the coprimality by the action of the Casimir element, a completely parallel argument. –  Mariano Suárez-Alvarez Oct 13 '11 at 7:04

$5/2 = 2 \frac{1}{2}$ since both are the groupoid cardinality of the following action:

image

Thinking about this, it is actually quite enlightening. For more information, see the wonderful paper From Finite Sets to Feynman Diagrams by John Baez and James Dolan.

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And of course there is Fürstenberg's topological proof of the infinitude of primes. I love this because it shows that all the mathematical "plumbing" works; i.e that number theory and topology connect up as they should.

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This is a non-example. Furstenberg's proof is completely elementary. –  Darsh Ranjan Oct 17 '10 at 16:26
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To say nothing of the fact that it uses no topological content beyond words... –  BCnrd Oct 17 '10 at 16:37
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Dear trb456: there are no theorems of topology used in the argument, just topology word games, so it does not illustrate any real connection "working" between topology and number theory. This should be more widely recognized. (It is Euclid's proof in disguise, if you unravel the words.) The proofs using divergence of the harmonic series or facts about Dedekind domains are genuine connections with other areas of math to prove the result, since they use actual theorems in those areas. If Furstenberg's proof used Tychonoff or Urysohn, it would be a different story. But I agree: whatever. –  BCnrd Oct 17 '10 at 17:24
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Please no more arguing about Furstenberg's proof! –  Pete L. Clark Oct 17 '10 at 18:04
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The topology does have connections to other areas of math, namely it is the profinite topology on $\mathbb{Z}$. –  Ian Agol Oct 22 '10 at 20:36

The fundamental theorem of algebra holds because:

  1. For each degree $n$ normed polynomial $p$ over the complex numbers, there is an $n \times n$ matrix $A$ with characteristic polynomial $\pm p$.

  2. We show that $A$ has an eigenvector.

  3. We may assume that $0$ is not an eigenvalue of $A$ (otherwise $p(0)=0$), so $A \in GL_n (\mathbb{C})$.

  4. $A$ induces a self-map $f_A$ of $CP^{n-1}$, and the eigenspaces of $A$ correspond to the fixed points of $f_A$; so we need to show that $A$ has a fixed point.

  5. As $GL_n (\mathbb{C})$ is connected, $f_A$ is homotopic to the identity (this does not depend on the fundamental theorem of algebra; if $A \in GL_n (\mathbb{C})$, then $ z 1 + (1-z )A$ is invertible except for a finite number of values of $z$; and the complement of a finite set of points of the plane is path-connected (this follows, for example, from the transversality theorem).

  6. The Lefschetz number of the identity on $CP^{n-1}$ equals $n\neq 0$, thus the Lefschetz number of $f_A$ is not zero.

  7. By the Lefschetz fixed point theorem, $f_A$ has a fixed point.

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A slightly different approach is to suppose there is a finite extension $K/\mathbb C$. Then $P(K)$, the projectivisation of $K$ as a complex vector space is a compact abelian Lie group (with multiplication induced by that of the group $K^\times$). Such a thing must be a torus, so its $H^1$ is not zero. Yet it is a complex projective space, and one easily sees that its $H^1$ is zero. –  Mariano Suárez-Alvarez Dec 21 '12 at 20:04
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This is a perfectly good proof, and is it so much more sophisticated than any of the others? Many of the favorite proofs of this theorem are similarly topological. –  Ryan Reich Dec 30 '12 at 5:35
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@Ryan: yes, in a sense it is a nice proof. Lefschetz fixed point theorem is a hard result, which depends either on Poincare duality or on simplicial approximation. Most topological proofs I know are considerably more elementary (and use the topology of the complex plane, which is more obviously related to the problem than self-maps of $CP^n$). –  Johannes Ebert Jan 30 '13 at 22:54

A number of high school contest problems in number theory reduce to Mihailescu's theorem. (The only perfect powers with a difference of 1 are 8 and 9.)

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See, for instance, the following ML threads: 1) artofproblemsolving.com/Forum/… and 2) artofproblemsolving.com/Forum/viewtopic.php?f=57&t=368417 –  J. H. S. Oct 18 '10 at 5:54

There is a simple pigeonhole argument for the following fact, due to Erdős and Szekeres I believe:

In any sequence $a_1, a_2, \ldots, a_{mn+1}$ of $mn+1$ distinct integers, there must exist either an increasing subsequence of length $m+1$ or a decreasing subsequence of length $n+1$ (or both).

The "sophisticated" proof of this fact is that any Young tableau with $mn+1$ boxes must either have more than $m$ columns or more than $n$ rows, and so the result follows because the number of columns/rows corresponds to the length of the longest increasing/decreasing subsequence of the corresponding permutation under the Robinson--Schensted correspondence.

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Isn't the proof involving Young tableau the simple one? –  Will Sawin Aug 20 '12 at 3:54
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@Will: Right, it's easier than the sophisticated proof using Robinson-Schensted. –  Timothy Chow Dec 6 '12 at 22:46

Every finite integral domain is a field:

Let $D$ be a finite integral domain. Being finite, it is Artinian an Noetherian and therefore has Krull dimension zero. But $(0)$ is a prime ideal, because $D$ is a domain, therefore $(0)$ is a maximal ideal and $D$ is a field.

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This is actually the same as the elementary proof: if $x \neq 0$ then $x^n = x^m$ for some $n, m$ by finiteness, so $x^{|n - m|} = 1$ by integrality. The ideal-theoretic proof is: we get $(x^n) = (x^m)$ so $x^n = x^m y$ for some $y$, and then by integrality $x^{|n - m|} y = 1$. It turns out $y = 1$ here, but that's because in passing to ideals we gave a slightly more general proof assuming only that $D$ is artinian, which is a finiteness condition on its set of ideals and not its underlying set. –  Ryan Reich Jun 5 '11 at 15:04
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Alternatively, and with different technology: $1$ generates a field inside $D$, over which $D$ is a finite dimensional algebra. It is semisimple because the Jacobson radicial, whose elements are nilpotent, must be trivial. Wedderburn's theorem then implies $D$ is a product of matrix rings over division rings. As it is a domain, we see that $D$ must be in fact a division field. This does not need commutativity. –  Mariano Suárez-Alvarez Jun 5 '11 at 17:34

An olympiad-type question I once tried to solve was: prove that all integers $>1$ can be written as a sum of two squarefree integers$^{[1]}$. The proof I came up with (which uses at least $3$ non-trivial results!) went as follows:

We can check that it holds for $n \le 10^4$. Now, let $S$ be the set of all squarefree integers, except for the primes larger than $10^4$. Then by the fact that the Schnirelmann density of the set of squarefree integers equals $\dfrac{53}{88}$ $^{[2]}$ and some decent estimate on the prime counting function$^{[3]}$, we have that the Schnirelmann density of $S$ must be larger than $\dfrac{1}{2}$. By Mann's Theorem$^{[4]}$ we now have that every positive integer can be written as sum of at most $2$ elements of $S$. In particular, every prime number can be written as sum of $2$ elements of $S$, and every integer that is not squarefree can be written as sum of $2$ elements of $S$. All there is now left, is proving the theorem for composite squarefree numbers; $n = pq = (p_1 + p_2)q = p_1q + p_2q$, where $p$ is the smallest prime dividing $n$ and $p_1, p_2$ are squarefree integers.

$^{[1]}$ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=470&t=150908 $^{[2]}$ http://www.jstor.org/pss/2034736 $^{[3]}$ http://en.wikipedia.org/wiki/Prime-counting_function#Inequalities $^{[4]}$ http://mathworld.wolfram.com/MannsTheorem.html

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Could you derive this result from supposing Goldbach's conjecture? –  Zsbán Ambrus Feb 24 '13 at 15:06

Proving the Banach fixed point theorem for compact metric spaces using the structure of monothetic compact semigroups.

Thm. Let $X$ be a compact metric space and $f\colon X\to X$ a strict contraction, meaning $d(f(x),f(y))< d(x,y)$ for $x\neq y$. Then $f$ has a unique fixed point and for any $x_0\in X$, the iterates $f^n(x_0)$ converges to the fixed point. Pf. Contractions are clearly equicontinuous, so by the Arzelà–Ascoli theorem, the closed subsemigroup $S$ generated by $f$ is compact in the compact-open topology. Now, a monothetic compact semigroup has a unique minimal ideal $I$, which is a compact abelian group. Moreover, either $S$ is finite and $I$ consists of all sufficiently high powers of $f$ or $S$ is infinite and $I$ consists of all limit points of the sequence $f^n$. In either case, $I$ consists of strict contractions, being in the ideal generated by $f$. Thus the identity element $e$ of $I$ is a constant map, being an idempotent strict contraction. Thus $I=\{e\}$, being a group. Thus $f^n$ converges to a constant map to some point $y$. Clearly $y$ is the unique fixed point of $f$.

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The space $C[0,1]$ is not reflexive. If it was, it also had a predual. But then it would be a von Neumann algebra. However von Neumann algebras correspond to very strange topological spaces which have the property that closures of open subsets are again open. Clearly this is not the case for $[0,1]$.

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Here is a Ramsey theory proof every finite semigroup has an idempotent. Let S be a finite semigroup with finite generating set A. Choose an infinite word $a_1a_2\cdots$ over A. Color the complete graph on 0,1,2... by coloring the edge from i to j with $i\lneq j$ by the image in S of $a_{i+1}\cdots a_j$. By Ramsey's theorem there is a monochromatic clique $i\lneq j\lneq k$. This means $$a_{i+1}\cdots a_j=a_{j+1}\cdots a_k=a_{i+1}\cdots a_k$$ is an idempotent.

This proof, generalized to larger clique sizes, actually shows any infinite word contains arbitrarily long consecutive subwords mapping to the same idempotent of S, which is used in studying automata over infinite words.

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This is nice. It somehow looks like Ramsey's theorem is quite adapted to this line of reasoning, which detracts from its thermonuclearity in this context. –  Mariano Suárez-Alvarez Dec 30 '12 at 5:29

This is kind of an elementary example, but I always thought it was funny to prove that $S_3$ is isomorphic to a subgroup of $S_6$ using Cayley's theorem.

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Here is an example that I learned through MO!

The infinitude of completely split primes in a Galois extension K of Q is an easy consequence of Chebotarev's Density Theorem. A slightly simpler argument involves showing that the Dedekind Zeta Function ζK(s) has a simple pole at s = 1. However, there is a very simple arithmetic argument that accomplishes the desired task...

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Which is, of course, a disguised generalization of Euclid's proof of the infinitude of the primes. KConrad discusses such generalizations here: math.uconn.edu/~kconrad/blurbs/gradnumthy/dirichleteuclid.pdf –  Qiaochu Yuan Oct 19 '10 at 12:40

A quiver whose unoriented graph is the affine D4 Dynkin diagram is tame. Therefore the moduli space of four points on a projective line is one dimensional.

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An excellent exercise for the reader who wants to understand quiver theory: Apply this argument to all the simply laced affine Dynkin types and explain which moduli spaces you have just proved to be one dimensional. –  David Speyer Nov 1 '10 at 12:43

Every finite dimensional complex representation of a finite cyclic group decomposes into a direct sum of irreducible representations. This can be deduced from the decomposition theorem for perverse sheaves as follows: It is enough to show that the group algebra is semi simple. To check this it is enough to lift the regular representation of $\mathbb Z/n$ to $\mathbb Z=\pi^1(\mathbb C^*)$ and show that it decomposes into a direct sum of irreducible representations of $\mathbb Z$.

Consider the covering $z \mapsto z^n$ of $\mathbb C^*$ by itself.

It is easy to see, that the monodromy action on the pushforward of the constant sheaf $\mathbb C[1]$ along this map coincides with the regular representation. On the other hand since the map is small, the decomposition theorem guarantees that the pushforward decomposes into a direct sum of IC complexes. Since our map is a covering and our space is smooth these are actually irreducible local systems on $\mathbb C^*$. But irreducible local systems correspond to irreducible representation of the fundamental group.

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Arrow's theorem is a basic result in social choice theory which has several simple proofs. (For three proofs see this paper: Three Brief Proofs of Arrow's Impossibility Theorem by J. Geanakoplos)

It also has a few complicated proofs: The paper by Tang, Pingzhong and Lin, Fangzhen Computer-aided proofs of Arrow's and other impossibility theorems, Artificial Intelligence 173 (2009), no. 11, 1041–1053. Gives an inductive proof based on rather complicted inductive step and a computerized check for the base case. The paper by Yuliy Baryshnikov, Unifying impossibility theorems: a topological approach. Adv. in Appl. Math. 14 (1993), 404–415, gives a proof based on algebraic topology. My paper: A Fourier-theoretic perspective on the Condorcet paradox and Arrow's theorem. Adv. in Appl. Math. 29 (2002), 412–426, gives a fairly complicated Fourier-theoretic proof but only to a special case of the theorem.

(A complicated proof to a related theorem is by Shelah, Saharon, On the Arrow property, Adv. in Appl. Math. 34 (2005), 217–251.)

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One can also show with Fermat's last theorem that $\sqrt{2}$ is irrational - the answer of mt did $2^{1/n}$ for $n\ge 3$. Suppose that $\sqrt{2}$ is rational. Then there is a right-angled triangle with rational sides $(a,b,c)=(\sqrt{2},\sqrt{2},2)$ and area 1. Hence $1$ would be a congruent number. This contradicts Fermat's last theorem with exponent $4$.

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Every finite semigroup contains an idempotent element.

You can nuke this problem using a theorem by Ellis that every compact, semi-topological semigroup contains an idempotent (which uses Zorn's Lemma).

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Wouldn't you rather look at the power of an arbitrary element and cycles therein? –  Peter Krautzberger Jul 10 '11 at 15:24
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Not really. The minimal subsemigroup proof is much more elegant in my opinion. –  Benjamin Steinberg Aug 20 '12 at 2:44

Baryshnikov gave a topological proof of Arrow's impossibility theorem, a result for which there are well known short and elementary proofs.

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One can also prove Arrow's impossibility theorem by noting (a) that the space of ultrafilters $\beta X$ on a set $X$ is equal to its Stone-Cech compactification; and (b) every finite set is already compact. –  Terry Tao Nov 9 '10 at 3:41

Not really sure if this should count, but: From Chebyshev's proof using the central binomial coefficient that there exists some constant $C>0$ such that

$$ \pi(x) < C\frac{x}{\log x} $$

for sufficiently large $x$, and from the infinitude of primes, we get that

$$ \log x \ll x. $$

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This is quite late(and just a restatement of the regular proof in fancy terms), but I came around this while goofing off one day:

Theorem: Let $X$ a space, and $\mathscr{F}$ a sheaf of (not necessarily abelian) groups, and denote by $\pi$ the projection from the étalé space $Sp\acute{e}(\mathscr{F})$. Then $\Gamma(X,\mathscr{F})$ inject into $\mathrm{Aut}(\pi)$(taken in the category of spaces étalé over $X$).

Proof: Straightforward and not difficult(but there are a bunch of things to check).

Theorem: (Cayley's theorem) Let $G$ a finite group, then $G$ is a subgroup of a symmetric group.

Proof. Let $X$ a nonempty, connected topological space and take $\mathbb{G}$ the constant sheaf associated to $G$ on $X$. Apply previous theorem and notice that $Sp\acute{e}(\mathbb{G})$ is a globally trivial covering space, and homeomorphic(over $X$) to $\coprod_{|G|} X$, so that $G$ injects into the group of deck transformations of this covering space, which is just $\mathfrak{S}_{|G|}$!

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Liouville remarked that the fundamental theorem of algebra could be derived from his theorem that elliptic functions (doubly periodic meromorphic functions of one complex variable) must have poles. The proof goes by substituting the inverse of a polynomial as the argument of, say, Weierstrass $\wp$-function with large enough periods, and observing that it has no poles.

Of course, the proof of Liouville's theorem on elliptic functions requires the same kind of arguments used for proving the famous Liouville theorem (due to Cauchy) that bounded holomorphic functions are bounded and, apparently, already used before by Cauchy for algebraic functions.

But Liouville's observation is really more complicated than the present proof. What it simplifies, however, is the compactness argument. For elliptic functions, or for algebraic functions, one has at hand a compact Riemann surface on which some holomorphic function is bounded, hence achieves its supremum, etc. This may be the reason why the general form of Liouville theorem came only after the case of algebraic or elliptic functions.

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There is an elementary problem that goes more or less like this: you have a special telephone keyboard with nine lighted buttons (one for each number from $1$ to $9$); when pushing each button other than number $5$ (the central button) then this switches the state of the lights of the button itself and of all its surrounding buttons; pushing number $5$ only switches the state of the lights of its surrounding buttons, but not of itself. Starting with all lights off, the question asks whether we can get all lights on by pushing buttons. The obvious solution to the negative answer relies on the fact that the parity of lighted buttons at every state of the keyboard is an invariant. But there is also a sophisticated solution.

Take the set $X$ of $9$ elements and think of $\mathcal{P}(X)$ as a vector space over the field $\mathbb{Z}_2$ with the sum being the symmetric difference and the product given by $0.v=\emptyset$ and $1.v=v$. Then we can identify each state of the keyboard with a corresponding vector in this space, while pushing the button $i$ corresponds to summing a special vector $v_i$ (associated to the button) to the vector representing the state of the keyboard. Thus, we are wondering if there are some scalars $\alpha_i$ such that $\sum_{i=1}^{9} \alpha_iv_i=X$. Writing each $v_i$ and $X$ in the base of the space given by the singleton elements $1, ..., 9$, we get a system of linear equations which can be seen to have no solutions by computing the $9 \times 9$ determinant and verifying it is null.

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The determinant thing doesn't seem so odd to me, since it is, as you note, the default mechanical way of finding the solutions to the system of linear equations straightforwardly describing the problem. But I suppose "oddness" is in the eye of the beholder. –  Sridhar Ramesh Jun 19 '11 at 20:21
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Are all these references to "oddness" puns? –  Gerry Myerson Jun 20 '11 at 0:12

The Herbert Simon (Nobel Price Winner, Economics, 1978)--- Karl Egil Aubert Dispute, see

http://www.tandfonline.com/doi/abs/10.1080/00201748208601972

Aubert criticizes Simon for irrelevant use of mathematics for his "Application", but also for the fact that he uses the Brouwer fixed point theorem for a proof, when the Intermediate Value Theorem would be enough.

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The skew-field of quaternions $\mathbb H$ is isomorphic to its opposite algebra.

Indeed, by a theorem of Frobenius, division algebras over the reals are isomorphic to either $\mathbb R, \mathbb C$ or $\mathbb H$. Since $\mathbb H^\mathsf{opp}$ is again a division algebra, it must be isomorphic to one of these. There are several ways to conclude: since it is four dimensional, or since it is not commutative, or since it has more than two square roots of $-1$, etc., we conclude that the only possibility is $\mathbb H \cong \mathbb H^\mathsf{opp}$.

If you are only interested in Morita equivalence between these two algebras, you can do better: the Brauer group of $\mathbb R$ is isomorphic to $\mathbb Z_2$, and so all elements are of order $2$. This implies that the class of $\mathbb H$ coincides with its inverse, which is the class of $\mathbb H^{\mathsf{opp}}$. Thus $\mathbb H$ and $\mathbb H^\mathsf{opp}$ are Morita equivalent.

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Who in his right mind writes $\mathsf{opp}$ with two p's! –  Mariano Suárez-Alvarez Dec 30 '12 at 5:30

I think that the following proof of the fact that every subgroup of index $2$ of a given group is normal might count too. When I first came up with it (sometime during my sophomore year), I believed that I had just found the entrance to a royal road to mathematics.

Let $H\leq G$ be such that $[G:H]=2$. We'll prove that every right coset of $H$ is equal to a left coset of $H$.

Since $[G:H]=2$, $G$ is both the union of two disjoint right cosets of $H$ and the union of two disjoint left cosets of $H$. Let us suppose that $G=He \cup Hx = eH \cup yH$ where $x,y\in G\setminus H$ and $e$ denotes the identity element of $G$. According to standard lore regarding the symmetric difference of sets,

$He \cup Hx = He \triangle Hx \triangle (He \cap Hx) = He \triangle Hx \triangle \emptyset = H \triangle (Hx\triangle \emptyset) = H\triangle Hx$

and

$eH \cup yH = eH \triangle yH \triangle (eH \cap yH) = eH \triangle yH \triangle \emptyset = H \triangle (yH \triangle \emptyset) = H \triangle yH$.

Therefore, $H\triangle Hx = H\triangle yH$. Canceling $H$ on both sides of the latter equality—which is perfectly valid given that $(2^G, \triangle)$ is a group—we conclude that $Hx=yH$. Done.

If you consider that the prior argument doesn't qualify as awfully sophisticated, there is still another fancy way to derive the result in question. As a consequence of P. Hall's famous marriage theorem, M. Hall proves in Theorem 5.1.7 of his Combinatorial Theory that if $H$ is a finite index subgroup of $G$, there exists a set of elements that are simultaneously representatives for the right cosets of $H$ and the left cosets of $H$ (once he's proven the said theorem, he adds: "Simultaneous right-and-left coset representatives exist for a subgroup in a variety of other circumstances. This problem has been investigated by Ore 1."). In the case $[G:H]=2$, this implies at once that every right coset of $H$ is equal to a left coset of $H$ and we are done...

Last but not least, $[G:H]=2 \Rightarrow H \trianglelefteq G$ in the case when $|G|<\infty$ can also be seen a consequence of the well-known fact according to which any subgroup of a finite group whose index is equal to the smallest prime that divides the order of the group is of necessity a normal subgroup of the group. B. R. Gelbaum showcases in one of his books an action-free proof of this fact. He attributes both the fact and the action-free proof to Ernst G. Straus. Does any of you know on what grounds he did so? I have a Xerox copy of the relevant page in the book here. This is exactly what Gelbaum writes therein:

At some time in the early 1940s Ernst G. Straus, sitting in a group theory class, saw the proof of the ... result [i.e., $[G:H]=2 \Rightarrow H \trianglelefteq G$] ... and immediately conjectured (and proved that night): ... IF G:H [sic] IS THE SMALLEST PRIME DIVISOR P of #(G) THEN H IS A NORMAL SUBGROUP.

P.S. The Galois-theoretic proof given by Matthias Künzer is just fabulous!

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There exists transcendantal numbers because:

-- $x\mapsto \frac{1}{[{\mathbb Q}(x):{\mathbb Q}]}{\rm Tr}_{{\mathbb Q}(x)/{\mathbb Q}}x$ is a well defined, non zero, linear form from $\bar{\mathbb Q}$ to ${\mathbb Q}$.

-- The kernel of a non zero linear form form ${\mathbb R}$ to ${\mathbb Q}$ is not measurable.

-- By Solovay, every subset of ${\mathbb R}$ can be assumed to be measurable.

Conclusion: ${\mathbb R}\neq \bar{\mathbb Q}$.

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The following theorem has several essentially different proofs that need quite different levels of mathematical background, ranging from high school to graduate level. Which proof is most natural depends on who you ask, but many people (including me) will find at least some proof unnecessarily complicated.

There exists a set $ A $ that is everywhere dense on the square $ [0, 1]^2 $, but such that for any real number $ x $, the intersections $ A \cap (\{x\} \times [0, 1]) $ and $ A \cap ([0, 1] \times \{x\}) $ are both finite.

(This is a variant of a homework problem posed by Sági Gábor.)

Here's the idea of a few proofs.

  • $ A = \{(p/r, q/r) \mid p, q, r \in \mathbb{Z} \text{ and } \gcd(p,r) = \gcd(q,r) = 1 \} $ is dense because if you subdivide the square to $ 2^n $ times $ 2^n $ squares, $ A $ contains the center of each square; and has only as many points on each horizontal or vertical line as the denominator of $ x $.

  • $ A = \{(x + y\sqrt3, y - x\sqrt3) \mid x, y\in\mathbb{Q} \} $ is dense because it's a scaled rotation of $ \mathbb{Q}^2 $, but has at most one point on every horizontal or vertical line otherwise $ \sqrt3 $ would be rational.

  • Choose $ a_0, b_0, a_1, b_1 $ as four reals linear independent over rationals, this is possible because of cardinalities. $ A = \{(ma_0 + na_1, mb_0 + nb_1) \mid m, n \in \mathbb{Q}\} $ has no two points sharing coordinates because of rational independence, and $ A $ is dense because it's a non-singular affine image of $ \mathbb{Q}^2 $.

  • A is the set of a countably infinite sequence of random points independent and uniform on the square. This is almost surely dense, but almost surely has no two points that share a coordinate.

  • Choose a countable topological base of the square, then choose a point from each of its elements inductively such that you never choose a point that shares a coordinate with any point chosen previously.

  • Choose a continuum (or smaller) size topological base of the square, then choose a point from each by transfinite induction such that when you choose a point, the cardinality of points chosen previously is less than continuum, thus you can avoid sharing coordinates with those points.

  • Choose $ a, b $ as reals such that $ a, b, 1 $ are linear independent over rationals, possible because of cardinalities. Let $ A = \{((ma + nb) \bmod 1, (ma - nb) \bmod 1) \mid m, n \in \mathbb{Z}\} $. No two points share coordinates because of rational independence. Looking on the torus, A is dense somewhere on the square and the difference of any two points of A is in A so it must be dense in the origin. As A is closed to addition, it must be dense on a line passing through the origin. As it's also closed to rotation by $ \pi/2 $, it's also dense on the rotation of that line, thus, because it's closed to addition, dense everywhere.

  • Choose $ a, b $ like above. Let $ A = \{(an \bmod 1, bn \bmod 1) \mid n \in \mathbb{Z}\} $. Prove A is dense by ergodic theory and Fourier analysis.

Update: Edited the drafts of proofs to somewhat cleaner. Permuted proofs. Also fixed typo in last proof.

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Claim: $\sum\limits_{k=0}^n (-1)^k {n\choose k} = 0$ for all integers $n≥1$

Proof: Take the $n-1$-dimensional simplex $\Delta_{n-1}$. We can compute it's Euler characteristic by using simplicial homology. There are exactly $n \choose k+1$ many $k$-sub-simplexes of $\Delta_{n-1}$. Thus we get a simplicial chain complex of the form $\mathbb{Z}^{n\choose n} \to \mathbb{Z}^{n\choose n-1} \to \cdots \to \mathbb{Z}^{n\choose 2}\to\mathbb{Z}^{n\choose 1}$. So the Euler characteristic is $\chi(\Delta_{n-1}) = \sum\limits_{k=0}^{n-1} (-1)^k {n\choose k+1}=-\sum\limits_{k=1}^{n} (-1)^k {n\choose k}$
On the other hand $\Delta_{n-1}$ is contractible, and $\chi$ is homotopy-equivalence-invariant, so $\chi(\Delta_{n-1})=\chi(pt) =1$.
Putting those toghether we obtain: $0=\chi(\Delta_{n-1})-\chi(\Delta_{n-1})=1+\sum\limits_{k=1}^{n} (-1)^k {n\choose k}=\sum\limits_{k=0}^n (-1)^k {n\choose k}$

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The case of Fatou's theorem for H^2 can be proven as follows:

By Carleson's theorem the series $ \sum a_n e^{i \theta n} $ converges for almost all $\theta$ if $ \sum |a_n|^2 < \infty$. Now we can appeal to Abel's theorem to conclude that the function $ f(z)= \sum a_n z^n$ has radial limits almost everywhere on the unit circle. (I am not sure if we can get non-tangential limits this way.)

But Carleson's theorem is a much more difficult theorem than what we have proved here. (I got this example from a Hardy space course I am taking right now.)

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