Sep 1, 2014
1 note

The Sylow Theorems, 2:

Theorem 3. Let \(G\) finite. Then

a) If \(H\) is a \(p\)-subgroup of \(G\), then the inclusion of \(H\) into \(G\) factors through a Sylow \(p\)-subgroup.

b) All \(p\)-Sylow subgroups are conjugate.

c) The number of \(p\)-Sylow subgroups of \(G\) is \(\equiv 1 \mod p\).

Proof of a). Let \(G\) act on the conjugate class of a \(p\)-Sylow subgroup by conjugation; the stabilizer of a subgroup contains that subgroup and so has index (and hence the conjugacy class of that subgroup is of size) coprime to \(p\); by the previous lemma, we then have a fixpoint, say K, under the action of conjugation by \(H\). Then \(H\) is contained in the normalizer of \(K\). Examining the canonical isomorphism

\[HK / K \simeq H / H \cap K\] we see that \(H \subseteq K\) since otherwise the quantities \([HK / K : K],[H:H \cap K]\) would be nonzero, hence the order of \(HK\) a power of \(p\) larger than \(|K|\), a contradiction.

Proof of b). Take \(H\) to be a \(p\)-Sylow subgroup. Then \(H\) acts on the set of all \(p\)-Sylow subgroups by conjugation. Suppose we have \(K\) in an orbit disjoint from the conjugacy class of \(H\) (which contains \(H\)). Then the orbit of \(K\) under conjugation by \(H\) has a fixpoint, so that \(H = K\), a contradiction.

Proof of c). Let \(H\) be a \(p\)-Sylow subgroup of \(G\); then \(H\) acts on the set of all \(p\)-Sylow subgroups of \(G\) by conjugation; we see that \(H\) is the only fixpoint of this action, and the result follows from the \(p\)-group fixed point theorem. \(_\square\)

Aug 31, 2014
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The p-group fixed point theorem

An easy but surprisingly useful result.

Theorem. Let \(G \to \mathrm{Sym}(X)\) be the action of a \(p\)-group \(G\) on finite \(X\). Then the fixpoints of the action are \( \equiv |X| \mod p\). In particular, if \(p \not{|} |X|\) then \(|X|\) has a fixed point, if \(|X|\) has 1 fixed point, then \(|X| \equiv 1 \mod p\), and if \(p | |X|\), then it has \(0 \mod p\) fixed points.

Proof. Let \(S\) be the fixpoints of \(|X|\). By orbit-stabilizer,

\[|X| = \sum \text{ orbits } = |S| + \sum [G : \mathrm{Stab}_x]\] where the final final sum is taken over representatives of distinct nontrivial orbits; that sum vanishes mod \(p\), giving the result. The other statements follow.

Aug 31, 2014
1 note

The Sylow Theorems, 1:

(These are my notes of Lang’s development of these theorems.)

Lemma 1. (Cauchy’s theorem for abelian groups.) Let \(G\) be finite and abelian, and let \(p | |G|\). Then \(G\) has an element (and hence a cylic subgroup) of order \(p\).

Proof. Recall the exponent of a finite group \(G\) is the least integer \(n\) such that \(\forall g \in G, g^n = e_G\). One sees this is equivalent to it being \(\mathrm{lcm}\{\mathrm{ord}(g) : g \in G\}\). Let \(n\) be the exponent of \(G\) and suppose \(p | n\). With Cauchy’s theorem we see \(p | |G| \iff p | n\); \(|G|\) admits a decomposition into prime factors, i.e. \(\prod_{p \in \mathbb{Z} : p | |G|} p^{a_p}\), hence \(|G|\) divides \(n^{\max \{a_p\}_{p | |G|}}\).

(Alternately, without Cauchy’s theorem, since we’re trying to prove the easier case for abelian groups, let \(x\) be any nontrivial element of \(G\); then \(\langle x \rangle\) has exponent that divides \(n\) and the exponent of \(G / \langle x \rangle\) divides \(n\); then by induction \(|G / \langle x \rangle|\) divides \(n\) and so does \(|G|\), since \(|G| = [G:\langle x \rangle][\langle x \rangle : 1]\).)

Now, let \(p\) divide \(|G|\). Since \(p | n\) then there exists \(g \in G\) so that \(g^{ps} = e_G\), so \(g^s\) has order \(p\), as desired. \(_\square\)

Now, this extends easily to a proof of the existence of \(p\)-Sylow groups in the abelian case:

Theorem 1. Let \(G\) be a finite abelian group and \(p\) a prime dividing the order of \(G\). Then there exists a \(p\)-Sylow subgroup of \(G\).

Let \(m\) be the exponent of \(p\) in the prime factorization of \(|G|\). By applying the above lemma \(m\) times one obtains the sequence of quotient groups

\[G \to G / \langle x_1 \rangle \to (G / \langle x_1 \rangle) / \langle x_2 \rangle \to \dots \text{ (m times). }\]

where each \(x_n\) has order \(p\) in its ambient group. Pulling the identity of \(( \dots (G / \langle x_1) \rangle) / \dots ) / \langle x_m \rangle\) back along the sequence yields a  \(p\)-Sylow subgroup of \(G\), as desired.

Modifying this proof a little further yields a proof of the existence of \(p\)-Sylow subgroups for the general case; since the important step is quotienting by a cyclic group, then as one might suspect, we try to pass our process for obtaining cyclic groups of order \(p\) to the next-best thing for an abelian group, i.e. the center of the group.

Theorem 2. Let \(G\) be a finite abelian group and \(p\) a prime dividing the order of \(G\). Then there exists a \(p\)-Sylow subgroup of \(G\).

We induct, i.e. assume the statement is proven for all groups of order less than \(G\)’s. We may discard the cases where \(G\) is prime or where \(G\) has a subgroup whose index is coprime with \(p\), since in that case any \(p\)-Sylow subgroups of those subgroups would also be \(p\)-Sylow subgroups in \(G\).

Letting \(G\) act on itself by conjugation, we see the fixpoints of this action correspond precisely to elements of the center \(Z\) of \(G\). Since \(p\) divides the order of \(G\), then by the class formula

\[|G| = [Z:1] + \sum [G:\mathrm{Stab}_x]\] (where the \(x\) are taken from disjoint nontrivial orbits) \(p\) divides \((Z:1)\) and hence \(Z\) is nontrivial.

Now, let \(a\) have order \(p\) in \(Z\); then \(\langle a \rangle\) is normal in \(G\) since it’s in \(Z\). Then by the induction hypothesis \(G / \langle a \rangle\) has a \(p\)-Sylow subgroup of order \(p^{n-1}\) where \(p^n\) maximal in the factorization of \(|G|\). Taking the preimage of this subgroup along the projection \(G \to G / \langle a \rangle\) gives a \(p\)-Sylow subgroup of \(G\), as desired. \(_\square\)

Aug 28, 2014
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Another random linear algebra problem, 2:

If \(X \hookrightarrow V\) as vector spaces, then denote by \(V^*\) \(\mathrm{Hom}_{\text{VectF}} (V,F)\) and by \(X^0\) the annihilator of the submodule \(X\) i.e. \(\{f \in V^* : X \hookrightarrow V = X \hookrightarrow \ker f \hookrightarrow V\}\).

Now, let \(T : V \to W\) a linear map between finite-dimensional vector spaces. Recall the transpose of \(T\) is the map \(T^t : W^* \to V^*\) by \(f \to f \circ T\). Show that

a) \(\mathrm{im}(T)^0 = \ker(T^t)\)


b) \(\dim \mathrm{im}(T) = \dim \mathrm{im}(T^t)\).

Proof of a). Let \(g \in W^*\) such that \(\mathrm{im}T \subseteq \ker g\). Since the category of \(\mathbf{R}\)-modules is abelian, and the assumption tells us precisely that the inclusion of \(\mathrm{im}T \hookrightarrow W\) factors through \(\ker g\), we write \(V \overset{g \circ T}{\to} R\) as

\[V \overset{\pi_T}{\to} \mathrm{im} T \hookrightarrow \ker g \hookrightarrow W \overset{g}{\to} \mathbf{R}\]

and since the right half of that chain of maps is zero, so is the entire chain of maps, i.e. \(\mathrm{im}(T)^0 \subseteq \ker(T^t)\).

Conversely, \(g \in \ker T^t \implies g \circ T = 0 \implies \mathrm{im} T \hookrightarrow W\) factors through \(\ker T\) by the universal property of kernels, and we are done.

b) Equivalently, we want \(\mathrm{im} T \simeq \mathrm{im} T^t\) as vector spaces. From the above it’ll suffice to show \((\mathrm{im} T)^* \simeq W^* / \mathrm{im} (T)^0\). Since we’re in finite dimension and vector spaces are free, this is equivalent to showing \(\mathrm{im} (T)^0 \simeq W^* / (\mathrm{im} T)^* \). Let \(\{x_1, \dots x_n\}\) be a basis of \(\mathrm{im} T\) in \(W\); extend it to a basis \(\{x_1, \dots x_n, w_1, \dots w_m\}\) of \(W\), and define the canonical map \(c:W \overset{\sim}{\to} W^*\) by \(b_i \mapsto \delta_i\) for each basis element \(b_i\).

Now, quotienting by subspaces corresponds precisely to making the basis of that subspace vanish. I claim \(\{w_1, \dots w_m\}\) is a basis for \(\mathrm{im}(T)^0\). Indeed, every functional in \(\mathrm{im}(T)^0\) has a representation \(\sum_{i \leq n + m} a_i \delta_i\) but by definition \(a_i\) is zero for every \(a_i\) corresponding to a \(c(x_i)\), which gives the isomorphism by identification. \(_\square\)

Aug 28, 2014
0 notes

Another random linear algebra problem:

Let \(V\) be a vector space over \(\mathbf{C}\) and let \(v_1, \dots v_n\) be eigenvectors of an operator \(T\) with distinct eigenvalues. Show that \(v_1, \dots v_n\) are linearly independent.

Clearly for any pair \((v_i,v_j) : i \neq j, v_i \not \in \mathrm{span}(v_j)\) since \(v_i = a v_j \implies \lambda_i v_i = a \lambda_j v_j \implies v_i = a \lambda_j / \lambda_i v_j\) but since the \(\lambda\)s are distinct, \(v_i \neq v_i\).

Now we make the following observation: if \(v_2, \dots v_m\) are linearly independent eigenvectors with distinct eigenvalues, then \(v_1\) another eigenvector with eigenvalue distinct from the others can’t lie in \(\mathrm{span}(v_2, \dots v_m)\), since

\[v_1 \in \mathrm{span}(v_2, \dots v_m) \implies v_1 = \sum_{2 \leq i \leq m} a_iv_i \implies Tv_i = \sum_{2 \leq i \leq m} a_i \lambda_i v_i \implies \sum_{2 \leq i \leq m} (a_i \lambda_i / \lambda_1 - a_i) v_i = 0 \implies\] (by independence of \(v_2, \dots v_m\)) \(\lambda_i / \lambda_1 = 1\) for each \(i\), a contradiction.

Then choose any pair of distinct eigenvectors \((v_i, v_j)\); the claim then follows by induction. \(_\square\)

Aug 28, 2014
26 notes


Hyperbolic paraboloid



Hyperbolic paraboloid

Aug 26, 2014
0 notes

One can

embed \(\mathcal{P}(X)\) (viewed as a poset category) contravariantly into \(\mathcal{P}(\mathrm{Hom}(X,X))\) by \(s \mapsto \{f : f(x) = x, \forall x \in s\}\); this is also a set-valuedpresheaf over \(X\) viewed as a topological space with the discrete topology.

Aug 26, 2014
1 note

For a friend:

We’ll show \(ab = \gcd(a,b) \cdot \mathrm{lcm}(a,b)\).

Let \(\rho(z)\) denote the set of distinct primes that divide \(z\). By the fundamental theorem of arithmetic every integer \(z\) admits a unique decomposition into prime factors, i.e. \(z = \prod_{p \in \rho(z)} p^{z_p}\).

One sees that

\[\mathrm{lcm}(a,b) = \prod_{p \in \rho(a) \cup \rho(b)} p^{\max(a_p,b_p)}\]

and that dually, \[\gcd(a,b) = \prod_{p \in \rho(a) \cap \rho(b)} p^{\min(a_p,b_p)}\]

and upon inspection, one easily sees that

\(\gcd(a,b) \cdot \mathrm{lcm}(a,b)\) divides \(ab\) and vice versa.

Aug 25, 2014
12 notes




Need ideas for an expository talk in a nonspecialized graduate seminar.

Audience background: math grad students, most of them are numerical, math bio or analysis/PDEs. They might’ve heard the word “category”.

Time limit: 50 mins

Objective: talk about something cool for an hour

Real objective: expanding CV

Ricci flows? 

ugh, yeah, that’s something i need to learn anyway

on the other hand - ugh, geometry

presheafs and sheafs? they have a concise definition as a contravariant functor,

also you can make a bunch of toy models with finite sets because discrete topology

Aug 25, 2014
0 notes

Enumerating set-maps in Mathematica:

SetMaps[n_, m_] :=
Select[Subsets[SetProduct[Range[n], Range[m]]],
Length[#] == n && (Sort[First /@ #] == Range[n]) &]

Is there a faster way to do this (i.e. something besides applying an indicator function comprising some predicates to the power set of \(n \times m\)?)

edit: (where SetProduct[x_,y_]:=Tuples[{x,y}])
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