Cauchy's theorem on subgroup existence

Cauchy’s the­o­rem states that if \(G\) is a finite group and \(p\) is a prime di­vid­ing \(|G|\) the or­der of \(G\), then \(G\) has a sub­group of or­der \(p\). Such a sub­group is nec­es­sar­ily cyclic (proof).

Proof

The proof in­volves ba­si­cally a sin­gle magic idea: from thin air, we pluck the defi­ni­tion of the fol­low­ing set.

Let

$$X = \{ (x_1, x_2, \dots, x_p) : x_1 x_2 \dots x_p = e \}$$
the col­lec­tion of \(p\)-tu­ples of el­e­ments of the group such that the group op­er­a­tion ap­plied to the tu­ple yields the iden­tity. Ob­serve that \(X\) is not empty, be­cause it con­tains the tu­ple \((e, e, \dots, e)\).

Now, the cyclic group \(C_p\) of or­der \(p\) acts on \(X\) as fol­lows:

$$(h, (x_1, \dots, x_p)) \mapsto (x_2, x_3, \dots, x_p, x_1)$$
where \(h\) is the gen­er­a­tor of \(C_p\). So a gen­eral el­e­ment \(h^i\) acts on \(X\) by send­ing \((x_1, \dots, x_p)\) to \((x_{i+1}, x_{i+2} , \dots, x_p, x_1, \dots, x_i)\).

This is in­deed a group ac­tion (ex­er­cise).

  • It cer­tainly out­puts el­e­ments of \(X\), be­cause if \(x_1 x_2 \dots x_p = e\), then

    $$x_{i+1} x_{i+2} \dots x_p x_1 \dots x_i = (x_1 \dots x_i)^{-1} (x_1 \dots x_p) (x_1 \dots x_i) = (x_1 \dots x_i)^{-1} e (x_1 \dots x_i) = e$$

  • The iden­tity acts triv­ially on the set: since ro­tat­ing a tu­ple round by \(0\) places is the same as not per­mut­ing it at all.

  • \((h^i h^j)(x_1, x_2, \dots, x_p) = h^i(h^j(x_1, x_2, \dots, x_p))\) be­cause the left-hand side has performed \(h^{i+j}\) which ro­tates by \(i+j\) places, while the right-hand side has ro­tated by first \(j\) and then \(i\) places and hence \(i+j\) in to­tal. <div><div>

Now, fix \(\bar{x} = (x_1, \dots, x_p) \in X\).

By the Or­bit-Sta­bil­iser the­o­rem, the or­bit \(\mathrm{Orb}_{C_p}(\bar{x})\) of \(\bar{x}\) di­vides \(|C_p| = p\), so (since \(p\) is prime) it is ei­ther \(1\) or \(p\) for ev­ery \(\bar{x} \in X\).

Now, what is the size of the set \(X\)?

It is \(|G|^{p-1}\).

In­deed, a sin­gle \(p\)-tu­ple in \(X\) is speci­fied pre­cisely by its first \(p\) el­e­ments; then the fi­nal el­e­ment is con­strained to be \(x_p = (x_1 \dots x_{p-1})^{-1}\). <div><div>

Also, the or­bits of \(C_p\) act­ing on \(X\) par­ti­tion \(X\) (proof). Since \(p\) di­vides \(|G|\), we must have \(p\) di­vid­ing \(|G|^{p-1} = |X|\). There­fore since \(|\mathrm{Orb}_{C_p}((e, e, \dots, e))| = 1\), there must be at least \(p-1\) other or­bits of size \(1\), be­cause each or­bit has size \(p\) or \(1\): if we had fewer than \(p-1\) other or­bits of size \(1\), then there would be at least \(1\) but strictly fewer than \(p\) or­bits of size \(1\), and all the re­main­ing or­bits would have to be of size \(p\), con­tra­dict­ing that \(p \mid |X|\). pic­ture of class equation

Hence there is in­deed an­other or­bit of size \(1\); say it is the sin­gle­ton \(\{ \bar{x} \}\) where \(\bar{x} = (x_1, \dots, x_p)\).

Now \(C_p\) acts by cy­cling \(\bar{x}\) round, and we know that do­ing so does not change \(\bar{x}\), so it must be the case that all the \(x_i\) are equal; hence \((x, x, \dots, x) \in X\) and so \(x^p = e\) by defi­ni­tion of \(X\).

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  • Group

    The alge­braic struc­ture that cap­tures sym­me­try, re­la­tion­ships be­tween trans­for­ma­tions, and part of what mul­ti­pli­ca­tion and ad­di­tion have in com­mon.