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][4] 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][5]

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][6] 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][7] 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][8] 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][9] ⁶

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

[1]: {{< ref "2013-06-26-sylow-theorems" >}} [2]: http://tartarus.org/gareth/ [3]: http://mmeblair.tumblr.com/post/61532912275/carnival-of-mathematics-102-my-summation-of-other [4]: https://en.wikipedia.org/wiki/Monorhyme [5]: https://en.wikipedia.org/wiki/Limerick_%28poetry%29 [6]: https://en.wikipedia.org/wiki/Cinquain [7]: https://en.wikipedia.org/wiki/Trochaic_tetrameter [8]: https://en.wikipedia.org/wiki/Quatrain [9]: https://en.wikipedia.org/wiki/Haiku

This is not a sonnet - it is six lines too short, and is monorhymic rather than following a more varied rhyme scheme. I started out intending it to be a sonnet, but all the rhymes for "p", "G" and so forth were irresistible. "Power" is a monosyllable. ↩︎
I use a form of reverse cinquain, with syllable count 2,8,6,4,2,2,4,6,8,2. ↩︎
"Hom", of course, is short for "homomorphism". Imre Leader used it all the time, so I took it to be legitimate. ↩︎
This section is unrhymed; although Shakespeare rhymes his tetrameter, Longfellow doesn't. The strong iambic nature of English makes enjambement very natural to write when you're constrained to trochees, so I have just gone with the flow. ↩︎
Quatrains have a variety of allowable rhyme schemes, but I plumped for ABAB for the sake of variety. Yes, "N" rhymes with "one". For the purposes of scansion, pronounce each line as the first line of a limerick, with an optional weak syllable at the end if necessary. ↩︎
I know that a haiku should mention a season, etc - but that is a constraint I am willing to relax. Gareth pointed out that if "sum" and "size" were synonymous, then " |X| : 1 (mod p)/Orbit X divides G's sum/A proof of the Third" would mention the season "sum-A". ↩︎

6.2 KiB Raw Blame History

Notes on pronunciation

[Monorhymic][4] Motivation 1

Theorem One

Little [Lemmarick][5]

Introductory Interlude (to the tune of "Jerusalem")

[Cinquain][6] Claim 2

[Trochaic Tetrameter][7] Tying Together 4

Theorem Two - Quad-[quatrain][8] 5

Theorem Three - Hindmost [Haiku][9] 6

6.2 KiB

Raw Blame History

[Monorhymic][4] Motivation ¹

[Cinquain][6] Claim ²

[Trochaic Tetrameter][7] Tying Together ⁴

Theorem Two - Quad-[quatrain][8] ⁵

Theorem Three - Hindmost [Haiku][9] ⁶