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 ¹

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 ²

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 ³.
Also, \( \vert K \vert \)
divides \( \vert N/P \vert \)
(Lagrange). Then Lemmarick proves: \(K\)
Is \({1}\).

Trochaic Tetrameter Tying Together ⁴

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 ⁵

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 ⁶

\( \vert X \vert \): \(1 \pmod{p}\)
Orbit \(X\) divides \(G\)’s size:
We have proved the Third.