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I didn't get any response on MSE so I though I'd give this a try here (my question on MSE).

I went to an undergrad's senior honors thesis presentation a while ago. She was discussing crossing numbers and mentioned that complete graphs $K_n$ are nonplanar iff $n \geq 5$. Coincidentally? $A_n$ (the alternating group of even permutations) is a nonabelian simple group iff $n \geq 5$.

I know there are many connections between graph theory/finite geometry and group theory. It seems that many simple groups (especially sporadics) have roots in weird graphs and odd geometric objects. So this led me to wonder...

Is there a connection between $K_n$ being nonplanar and $A_n$ being nonabelian simple (when $n \geq 5$)?

...or maybe this is just a coincidence.

A colleague mentioned that once the work is done proving that $K_5$ is nonplanar, $K_n$ begin nonplanar for all $n \geq 5$ follows quickly. This too reminds me of showing $A_n$ is simple -- once you have $A_5$ is simple, a short proof extends this to all $A_n$ ($n \geq 5$).

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  • $\begingroup$ As far as I know, this is just a coincidence. Certainly, the standard proofs of these facts don't refer to each other. $\endgroup$
    – HJRW
    Commented Dec 3, 2013 at 22:10
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    $\begingroup$ For $n\ge5$, a unit sphere can be dissected into $n$ pieces which can be reassembled to form two unit spheres. As long as you're collecting properties of the number 5.... $\endgroup$
    – Gerry Myerson
    Commented Dec 3, 2013 at 22:12
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    $\begingroup$ Might just be the law of small numbers: en.wikipedia.org/wiki/Strong_Law_of_Small_Numbers $\endgroup$
    – Ian Agol
    Commented Dec 3, 2013 at 22:23
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    $\begingroup$ Christopher Simons' article, An elementary approach to the monster, gives some constructions of groups from graphs leading to simple groups. In particular, $A_n$ ($n \geq 5$) is obtained from the complete graph $K_{n-1}$. I don't know if it is related to your question. $\endgroup$
    – Seirios
    Commented Dec 3, 2013 at 22:29
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    $\begingroup$ Thanks @Seirios. I'll have to take a look at the connection between $A_n$ and $K_{n-1}$. However, I guess it looks like (from lack of other response) that I should probably just chalk my observation up to coincidence. $\endgroup$
    – Bill Cook
    Commented Dec 5, 2013 at 19:40

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