Syyyyylooooow

by math_explorer, Jan 5, 2013, 8:33 AM

Group actions on sets are effective but incredibly annoying to write.

Abstract algebra notation is nuts. We have $Gx$ (orbit of $x$ under group action $G$ ) and $G_x$ (stabilizer/isotropy subgroup of $x$ under group action $G$ ) and $^Gx$ (conjugacy class of $x$ in group $G$ ) and $X_G$ and $X^G$ and goodness knows what else.

Anyway, I'm going to use $\text{Stab}(x)$ instead of $G_x$ for the stabilizer subgroup and $\text{Cl}(x)$ instead of $^Gx$ for the conjugacy classes, and I can't find any reasonable literate notation for the orbit of $x$ but that's okay because it wouldn't help the inexplicable summation notation.

We keep using this silly counting argument which I thought was something like Burnside's lemma but actually is a lot simpler, just partitioning the set into orbits and slapping the orbit-stabilizer theorem on.

If $G$ is the group and $S$ is the set then

$|S| = \sum_{Gx} |Gx| = \sum_{Gx} \frac{|G|}{|\text{Stab}(x)|}$

where in the sum, $x$ ranges over a set of representatives of the orbits, i.e. exactly one $x$ is taken from each orbit, i.e. $Gx$ is equal to each orbit exactly once. Anyway, that sum is a terrible abuse of notation so I'll just leave it out and explain things in English.

So this proof is randomly cobbled together from the book and professor-given hints.

Below, let $p$ denote some fixed prime.

-- Cauchy's (not him again!) lemma --
Let $G$ be an abelian subgroup. Then if $p \mid |G|$ then $G$ has a subgroup of order $p$ .

Proof: Induct on the order of $G$ .

If we have an element $a$ with order $pr$ for some positive integer $r$ then $\langle a^r \rangle$ (the subgroup generated by $a^r$ ) works.

Otherwise, we can at least pick an element $a \neq 1$ with order not divisible by $p$ . As $G$ is abelian, we can take the factor group $G/\langle a \rangle$ , whose order is divisible by $p$ . By induction, this factor group contains a subgroup with order $p$ ; taking a generator of it, we obtain a coset $b\langle a\rangle$ with order $p$ , where $b \in G$ . Then $b$ has order divisible by $p$ , because if $b^s = 1$ then $(b\langle a\rangle)^s = \langle a\rangle$ , the identity of the factor group, and $p \mid s$ . As before, let $b$ 's order be $pr$ and take $\langle b^r \rangle$ .

(Alternatively: this is entirely trivial if you know the structure of all finitely generated abelian groups, so whatever.)

-- end Cauchy's lemma --

-- Sylow I --

Let $p$ be a prime and let $k$ be a positive integer. If $p^k \mid |G|$ then there exists a subgroup $H$ of $G$ with order $p^k$ .

Proof:

Consider the group action $G$ acting via conjugation ( $g: x \mapsto gxg^{-1}$ ) on the set $G$ itself. See? I told you it was confusing.

Then

$|G| = \sum |\text{Cl}(x)| = \sum \frac{|G|}{|C(x)|}$

$x$ ranges over a set of representatives of the conjugacy classes, or orbits under conjugation.

Consider the center $C(G)$ . It's just the set of $x$ for which $|C(x)| = |G|$ , so $\frac{|G|}{|C(x)|} = 1$ and the conjugacy class of $x$ contains only $x$ itself. We separate them from the summation like so:

$|G| = |C(G)| + \sum \frac{|G|}{|C(x)|}$

where $x$ ranges over a set of representatives of the conjugacy classes with at least two elements.

I.1. If $|C(G)|$ is divisible by $p$ :

Of course, $C(G)$ is abelian. So, by Cauchy's Lemma above, there exists a subgroup $L < C(G)$ with order $p$ . As $L$ is a subgroup of the center, it commutes with every element of $G$ , so it's normal. Then we can compute the factor group $G/L$ , whose order is divisible by $p^{k-1}$ . Thus by induction $G/L$ contains a subgroup $H$ with order $p^{k-1}$ . The union of all the cosets in $H$ is a subgroup with order $p^k$ , so we're done.

I.2. If $|C(G)|$ is not divisible by $p$ :

Since $|G|$ is divisible by $p$ , there would have to be another term on the RHS of the equation not divisible by $p$ . Thus, for some $x \in G$ , the value $\frac{|G|}{|C(x)|}$ is not divisible by $p$ . This is only possible if $p^k \mid |C(x)|$ . But $C(x) \lneqq G$ (if the equality sign held $x$ should have been moved out into $C(G)$ ), hence by the induction hypothesis on $C(x)$ we have a subgroup $L < C(X) < G$ with $|L| = p^k$ and we're done.

-- end Sylow I --

-- Sylow II --

Let $G$ be a finite group; suppose $p^k \parallel |G|$ (i.e. $p^k \mid |G|, p^{k+1} \nmid |G|$ ). Consider the collection $\mathcal{S} = \{H < G \mid |H| = p^k \}$ of all subgroups of $G$ with order $p^k$ (which are called Sylow subgroups).

Then:
(a) for any two Sylow subgropus $H_1, H_2 \in \mathcal{S}$ , they are conjugate, i.e. $\exists x \in G$ such that $xH_1x^{-1} = H_2$
(b) $|\mathcal{S}|$ is a divisor of $\frac{|G|}{p^k}$ , the index of any Sylow subgroup
(c) $|\mathcal{S}| \equiv 1 \bmod p$
(d) for any subgroup $L < G$ with order $p^r$ for some nonnegative integer $r$ , there is some Sylow subgroup $H \in \mathcal{S}$ with $L < H$ .

Proof.

First, we prove this statement: (d') given a fixed Sylow subgroup $H$ , for any subgroup $L < G$ with order $p^r$ for some nonnegative integer $r$ , there exists a conjugation $xHx^{-1}$ of $H$ such that $L < xHx^{-1}$ .

Consider the group action $L$ acting via left translaion ( $\ell: gH \mapsto \ell gH$ ) on the set of left cosets of $H$ , which we'll denote $\mathcal{T} = \{gH \mid g \in G\}$ .

Again:
$|\mathcal{T}| = \sum \frac{|L|}{|\text{Stab}(gH)|}$

where $gH$ ranges over a set of representatives of the orbits in $\mathcal{T}$ .

Now, we certainly know that $|\mathcal{T}| = \frac{|G|}{|H| = p^k}$ is not divisible by $p$ . Hence again there's a term on the RHS that's not divisible by $p$ . As $|L| = p^r$ this is only possible if, for some coset $gH$ , we have $\text{Stab}(gH) = L$ .

Thus, that coset $gH$ satisfies: for any $\ell \in L$ , we have $\ell gH = gH$ , i.e. $\ell \in gHg^{-1}$ . Then $L \subset gHg^{-1}$ , and as both are subgroups of $G$ we're done.

With this we have proved (d), since (by Sylow I) at least one $H$ exists, and the $gHg^{-1}$ produced is a subgroup with the same order as $H$ and is thus another Sylow subgroup. It also implies (a), just by setting $L$ to be another Sylow group.

Now we prove (c). Fix a Sylow subgroup $H_1$ and consider the group action $H_1$ acting via conjugation ( $x: H \mapsto xHx^{-1}$ ) on the set of Sylow subgroups $\mathcal{S}$ . Then

$|\mathcal{S}| = \sum \frac{|H_1|}{|\text{Stab}(H_j)|}$

where $H_j$ ranges over a set of representatives of the orbits in $\mathcal{S}$ .

If $H_j = H_1$ it's obvious that every element of $H_1$ is in the stabilizer, hence the summand is $1$ .

On the other hand, if $H_j \neq H_1$ , we claim that $\text{Stab}(H_j) \neq H_1$ . Suppose otherwise; then $xH_jx^{-1} = H_j$ for any $x \in H_1$ , so $H_j$ is a normal subgroup of the subgroup generated by these two Sylow subgroups, $M = \langle H_1, H_j \rangle$ .

Consider the factor group $M/H_j$ ; we claim that its order is also a power of $p$ . This is because if we take the intersection of each coset with $H_1$ , we actually get a factor group of $H_1$ . (I think it's the "first isomorphism theorem" or something.) That is, $M/H_j \cong H_1/(H_1 \cap H_j)$ so its order is a divisor of $|H_1| = p^k$ .

Thus $|M| = p^k|M/H_j|$ is also a power of $p$ , and $|M| > |H_1| = p^k$ . But as $M$ is a subgroup of $G$ , we know $|M|$ (a higher power of $p$ than $p^k$ ) divides $|G|$ , contradicting our selection of $k$ .

Therefore, $\text{Stab}(H_j) \neq H_1$ and therefore $\frac{|H_1|}{|\text{Stab}(H_j)|}$ is greater than 1. As $|H_1| = p^k$ , the result is divisible by $p$ . Therefore, all terms vanish $\bmod p$ except for the orbit of $H_1$ whose summand is $1$ , so $|\mathcal{S}| \equiv 1 \bmod p$ .

Now, (b) is easy. Consider the group action $G$ acting via conjugation ( $g: H \mapsto gHg^{-1}$ ) on the set of Sylow subgroups $\mathcal{S}$ . Since there's only one orbit,

$|\mathcal{S}| = \frac{|G|}{|\text{Stab}(H)|}$

from which it's obvious $|G|$ is divisible by $|\mathcal{S}|$ . And since $|\mathcal{S}|$ is not divisible by $p$ , we know that $|G|/p^k$ is divisible by $|\mathcal{S}|$ too.

This post has been edited 1 time. Last edited by math_explorer, Aug 23, 2013, 7:25 AM

abstract algebra

Cauchy

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Syyyyylooooow

by math_explorer, Jan 5, 2013, 8:33 AM

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