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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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    $\begingroup$ I once saw someone proving resolutions of singularities of curves by quoting Hironaka's theorem. $\endgroup$ Oct 17, 2010 at 15:23
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    $\begingroup$ rjlipton.wordpress.com/2010/03/31/april-fool $\endgroup$ Oct 17, 2010 at 15:42
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    $\begingroup$ 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 $D$=$L$. $\endgroup$
    – JS Milne
    Oct 17, 2010 at 20:07
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    $\begingroup$ @Maxime: I have trouble believing that such a proof is actually non-circular. Surely such proofs form a step, however easy, in the classification. $\endgroup$ Oct 17, 2010 at 21:59
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    $\begingroup$ 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. $\endgroup$
    – roy smith
    Jan 29, 2011 at 6:48

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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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One can use the continuous functional calculus of a C$^*$-algebra (namely $M_N(\mathbb{C})$) to prove that a normal matrix is diagonalizable.

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    $\begingroup$ ok, but this is not really an awful sophisticated proof. this is the standard way of proofing the spectral theorem for normal operators. if one considers the special case of normal operators on finite dimensional spaces - viz, matrices - you get this. $\endgroup$ Feb 25, 2013 at 5:35
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$Forest$ is in $P$. Given a finite undirected graph $G$ one can in polynomial time decide whether the input is a forest. The class of all finite forests is a minor-closed property and by the Robertson–Seymour theorem, there are finitely many forbidden minors. We can in $O(n^3)$ time test whether $G$ contains a forbidden minor and if not, output yes.

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    $\begingroup$ Although I like the example, I'm not sure I follow your argument. For the case of forests we already know the finite set of forbidden minors: $\{C_3\}$. So Robertson-Seymour doesn't really enter the picture except via the $O(n^3)$ test, which is really a different theorem. $\endgroup$ Mar 28, 2013 at 23:33
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Kn is non-planar for n>4: it contradicts the four-color theorem.

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    $\begingroup$ To qualify as a good answer, it has to be non-circular... Are we sure this passes that test? $\endgroup$ Dec 30, 2012 at 2:42
  • $\begingroup$ @Mariano Suárez-Alvarez: I thought it's noncircular for sufficiently large n, that's why I phrased the example the way I did. It is probably circular for n=5. I am aware that this can be proved using "any subgraph of a planar graph is planar", and "K5 is nonplanar" or "Euler's theorem", all of which are preliminary results to 4-color-theorem, but it was not clear to me that this consistutes a circularity, as this is a statement with a quantifier, just a ridiculously easy one to prove. I was testing the limits of the question, in a sense. I agree it's not 100% in the spirit. $\endgroup$
    – Ron Maimon
    Jan 5, 2013 at 16:12
  • $\begingroup$ Hi @Ron, you don't know me but I was hoping to buy you a beer sometime in NYC. Shoot me an email if you'd like to say hello. (Couldn't find a better way to contact you!) $\endgroup$ Apr 17, 2013 at 16:17
  • $\begingroup$ I gave once as an easy exercise on an exam the following easy problem: If we remove $2$ edges from $K_7$, the resulting graph is not planar...One student solved using the 4 colors theorem :) $\endgroup$
    – Nick S
    Jan 8, 2014 at 23:42
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Around year 1970 a popular way to compute cohomology groups of the finite cyclic groups was by applying spectral sequences (which was quite an overkill).

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    $\begingroup$ This was popular among whom? The book by Cartan and Eilenberg, the very first textbook on the subject, already has the computation done in terms of the usual very small periodic projective resolution: after that, using anything else to compute this seems pretty weird! $\endgroup$ May 11, 2013 at 7:37
  • $\begingroup$ @Mariano, I didn't say among whom to be discreet about it. At the time I rediscovered (it sounds funny) the periodic resolutions due to the free actions of the cyclic groups on the spheres. Later I saw a paper on periodic projective resolutions by Swan (it covered more advanced material of course). $\endgroup$ May 12, 2013 at 4:12
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The Jordan curve theorem. As far as I know, the "elementary" proof is quite involved, at least with respect to the intuitive plausibility of the statement.

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  • $\begingroup$ Is that a "simple fact"? $\endgroup$ Oct 17, 2010 at 16:58
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    $\begingroup$ I think the idea of this question is to judge the simplicity of the fact by the length of the shortest possible elementary proof, not by the length of the statement. $\endgroup$
    – HJRW
    Oct 17, 2010 at 17:40
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    $\begingroup$ unknown - for suitable definitions of 'heuristic' and 'simple', yes, I do. But the key word in the question is 'disproportionate'. $\endgroup$
    – HJRW
    Oct 17, 2010 at 19:01
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    $\begingroup$ The Jordan curve theorem is not intuitive: it deals with continuous curves, and at that level of generality it is quite legitimate to expect the worst. The result is almost obvious for $C^1$ curves, of course, but there is a chasm between $C^0$ and $C^1$, and I can think of a couple of "intuitive" results like this which are not yet even proved in the $C^0$ case. See e.g. the square pegs & round holes problem quomodocumque.wordpress.com/2007/08/31/… which may be close to being solved, but has been open since 1911! $\endgroup$ Oct 17, 2010 at 22:52
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    $\begingroup$ I am not sure it is so intuitive, even in the nice $C^1$ case. For example, could you explain to a child why the results holds in the plane and not in the torus ? $\endgroup$
    – Hugh J
    Oct 18, 2010 at 22:25
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If $0\le f_n \le 1$ is a sequence of continuous functions on $[0,1]$ that converges pointwise to $0$, then $\int_0^1 f_n(t) dt $ converges to $0$. Understandable by freshman, the statement is hard to prove using only the tools of calculus but is immediate from the dominated convergence theorem.

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    $\begingroup$ I don't see this as a simple fact. To construct Lebesgue measure you usually have to prove such a statement (or something similar) anyway. $\endgroup$
    – Mark
    Jun 15, 2011 at 15:10
  • $\begingroup$ Yes, like Mark, I don't think this is in the intended spirit of the question, which is about sophisticated proofs for facts that have much easier proofs. $\endgroup$
    – Todd Trimble
    Dec 22, 2012 at 7:22
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