This is a collection of poems which together prove the Sylow theorems.

Notes on pronunciation

Pronounce \(\vert P \vert\) as “mod P”, \(a/b\) or \(\dfrac{a}{b}\) as “a on b”, and \(=\) as “equals”.

\(a^b\) for positive integer \(b\) is pronounced “a to the b”.

\(g^{-1}\) is pronounced “gee inverse”.

“Sylow” is pronounced “see-lov”, for the purposes of these poems.

\(p\) and \(P\) and \(n_p\) are different entities, so they’re allowed to rhyme.

Monorhymic Motivation

Suppose we have a finite group called \(G\).

This group has size \(m\) times a power of \(p\).

We choose \(m\) to have coprimality:

the power of \(p\)’s the biggest we can see.

Then One: a subgroup of that size do we

assert exists. And Two: such subgroups be

all conjugate. And \(m\)’s nought mod \(n_p\),

while \(n_p = 1 \pmod{p}\); that’s Three.

Theorem One

Little Lemmarick

Subtitle: “The size of the normaliser \(N\) of a maximal \(p\)-subgroup \(P\) has \(N/P\) coprime to \(p\)”

There was a \(p\)-subgroup of \(G\)

(by Cauchy). The largest was \(P\).

Let \(N\) normalise,

Take \(\dfrac{N}{P}\)’s size,

Suppose that it’s zero mod \(p\).

Now \(\dfrac{N}{P}\) also has some

p-subgroup (by Cauchy); take one.

Take it un-projected,

\(P\)’s most big? Corrected!

We’ve found one sized \(p \vert P \vert\): done.

Introductory Interlude (to the tune of “Jerusalem”)

Subtitle: “\(\{P\}\) is an orbit of size \(1\) under the conjugation action of \(P\) on the set of \(G\)-conjugates of \(P\)”

Let \(X\) be \(P\)’s orbit under \(G\)

Acting by conjuga-ti-on.

Mod \(G\) o’er \(N\)’s the size of \(X\)

The Orbit/Stabiliser’s done.

And in its turn, \(P\) acts on \(X\)

By conjugating, as before,

Then \(P\) is certainly all alone:

Its orbit is itself, no more.

Let \(gPg^{-1}\) be alone,

\(P\) stabilises it, and hence

\(pgPg^{-1}p^{-1}\)

Is \(gPg^{-1}\) - from whence

We conjugate by \(g^{-1}\):

\(g^{-1}Pg\) fixes \(P\).

\(g^{-1}Pg\) is in \(N\),

so \(\pi\) applies. From this, we’ll see:

Cinquain Claim

Subtitle: “\(\{P\}\) is the only orbit of size \(1\)”

A claim:

\(\pi(g^{-1}Pg)\) is \({1}\).

Call it \(K\). If false, \(p\)

divides \(\vert K \vert\),

as \(\pi\)

a hom .

Also, \(\vert K \vert\)

divides \(\vert N/P \vert\)

(Lagrange). Then Lemmarick proves: \(K\)

Is \({1}\).

Trochaic Tetrameter Tying Together

Subtitle: “\(\{P\}\) is Sylow, since \(G/N\) has size coprime to \(p\)”

\(\pi\) has kernel \(P\) - but also

\(K\) is \({1}\), so lies inside it.

\(P\) contains \(g^{-1}Pg\);

Both have size \(p^a\). So since they’re finite, they’re the same set.

Any set alone in orbit

must be \(P\). The class equation

Tells us \(\vert G \vert / \vert N \vert\) is

Just precisely \(1 \pmod{p}\). Then

\(\vert G \vert / \vert P \vert\) is not a

multiple of \(p\) because it’s

\(\vert \dfrac{N}{P} \vert\) multiplied by

\(\dfrac{ \vert G \vert }{ \vert N \vert }\) and \(p\) can’t

possibly divide those two. So

Maximal the power of \(p\) is:

\(P\)’s a Sylow \(p\)-subgroup.

Theorem Two - Quad-quatrain

A Sylow \(p\)-subgroup let \(Q\) be:

a subgroup, size \(p^a\).

Because it’s the same size as was \(P\),

it acts on \(X\) in the same way.

Mod \(p\), we have \(\vert X \vert\) is \(1\) -

the orbits of \(Q\) will divide it;

Now invoke the class equation:

an orbit, size \(1\), lies inside it.

We dub this one \(gPg^{-1}\),

then \(g^{-1}Qg\)’s in \(N\).

Projection works just as well in verse:

\(\pi(g^{-1}Qg)\) is \({1}\).

The previous poem’s our saviour:

\(g^{-1}Qg\) is in \(P\).

The Pigeonhole tells its behaviour:

that \(P\) is \(g^{-1}Qg\).

Theorem Three - Hindmost Haiku

\(\vert X \vert\): \(1 \pmod{p}\)

Orbit \(X\) divides \(G\)’s size:

We have proved the Third.