11-1100-Pgadey-Lect6 - Revision history

Pgadey at 20:22, 6 October 2011

2011-10-06T20:22:35Z

← Older revision		Revision as of 16:22, 6 October 2011
Line 36:		Line 36:
	<math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math>		<math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math>

	The last group <math>Q</math> is the famous ''unit quaternion'' group (See [http://en.wikipedia.org/wiki/Quaternion_group]).		The last group <math>Q</math> is the famous ''unit quaternion'' group (See [http://en.wikipedia.org/wiki/Quaternion_group], also [http://en.wikipedia.org/wiki/History_of_quaternions]).
	; Theorem		; Theorem
	: Any <math>p</math>-group has a non-trivial centre.		: Any <math>p</math>-group has a non-trivial centre.

Pgadey at 20:19, 6 October 2011

2011-10-06T20:19:58Z

← Older revision		Revision as of 16:19, 6 October 2011
Line 36:		Line 36:
	<math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math>		<math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math>

	The last group <math>Q</math> is the famous ''unit quaternion'' group (See [http://en.wikipedia.org/wiki/Quaternion_group~~:WP~~]).		The last group <math>Q</math> is the famous ''unit quaternion'' group (See [http://en.wikipedia.org/wiki/Quaternion_group]).
	; Theorem		; Theorem
	: Any <math>p</math>-group has a non-trivial centre.		: Any <math>p</math>-group has a non-trivial centre.

Pgadey at 20:19, 6 October 2011

2011-10-06T20:19:24Z

← Older revision		Revision as of 16:19, 6 October 2011
Line 36:		Line 36:
	<math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math>		<math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math>

	The last group <math>Q</math> is the famous ''unit ~~quaternions~~'' -- ~~They~~ ~~need a better description here~~.		The last group <math>Q</math> is the famous ''unit quaternion'' group (See [http://en.wikipedia.org/wiki/Quaternion_group:WP]).

	; Theorem		; Theorem
	: Any <math>p</math>-group has a non-trivial centre.		: Any <math>p</math>-group has a non-trivial centre.
Line 67:		Line 66:
	* <math>\|G\| \not\equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>.		* <math>\|G\| \not\equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>.

	If <math>\|Z(G)\| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>\|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>\|Stab(x_i)\|</math>. We have that <math>\|Stab(x_i)\| < \|G\|</math~~> <span style="color:green">[Why happens here?]</span~~> We then have that <math>p^k \leq Stab(x_i) < \|G\|</math> and by induction there is <math>\|P\| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.		If <math>\|Z(G)\| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>\|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>\|Stab(x_i)\|</math>. We have that <math>\|Stab(x_i)\| < \|G\|</math> We then have that <math>p^k \leq Stab(x_i) < \|G\|</math> and by induction there is <math>\|P\| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.

	WIf <math>\|Z(G)\| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>\|<x>\| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>\|P'\| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> \|\pi^{-1}(P')\| = p(p^{k-1}) = p^k </math>.		WIf <math>\|Z(G)\| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>\|<x>\| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>\|P'\| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> \|\pi^{-1}(P')\| = p(p^{k-1}) = p^k </math>.

Pgadey at 20:16, 6 October 2011

2011-10-06T20:16:29Z

← Older revision		Revision as of 16:16, 6 October 2011
Line 4:		Line 4:

	;Theorem		;Theorem
	Theorem
	: If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.		: If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.

Line 11:		Line 10:

	; Stabilizer of a point		; Stabilizer of a point
	: We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$.		: We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of <math>x</math>.

	''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.		''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.
Line 64:		Line 63:
	: <math>Syl_p(G) \neq \emptyset</math>		: <math>Syl_p(G) \neq \emptyset</math>

	We proceed by induction on the order of <math>p</math>. Assume the claim holds for all groups of order less than $\|G\|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>\|G\| \equiv 0\mod\ p</math> we have either:		We proceed by induction on the order of <math>p</math>. Assume the claim holds for all groups of order less than <math>\|G\|</math>. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>\|G\| \equiv 0\mod\ p</math> we have either:
	* <math>\|G\| \equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\mod\ p</math>.		* <math>\|G\| \equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\mod\ p</math>.
	* <math>\|G\| \not\equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>.		* <math>\|G\| \not\equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>.

Pgadey at 20:15, 6 October 2011

2011-10-06T20:15:28Z

← Older revision		Revision as of 16:15, 6 October 2011
Line 64:		Line 64:
	: <math>Syl_p(G) \neq \emptyset</math>		: <math>Syl_p(G) \neq \emptyset</math>

	We proceed by induction on the order of <<p>>. Assume the claim holds for all groups of order less than $\|G\|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>\|G\| \equiv 0\mod\ p</math> we have either:		We proceed by induction on the order of <math>p</math>. Assume the claim holds for all groups of order less than $\|G\|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>\|G\| \equiv 0\mod\ p</math> we have either:
	* <math>\|G\| \equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\mod\ p</math>.		* <math>\|G\| \equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\mod\ p</math>.
	* <math>\|G\| \not\equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>.		* <math>\|G\| \not\equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>.
Line 82:		Line 82:
	: If <math>x \in G</math> has <math>\|<x>\| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.		: If <math>x \in G</math> has <math>\|<x>\| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.

	<span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]		<span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]</span>

	We show the first statement. We have that <math>\|P / P \cap H\| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>\|PH\| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.		We show the first statement. We have that <math>\|P / P \cap H\| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>\|PH\| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.

Pgadey at 20:14, 6 October 2011

2011-10-06T20:14:37Z

← Older revision		Revision as of 16:14, 6 October 2011
Line 64:		Line 64:
	: <math>Syl_p(G) \neq \emptyset</math>		: <math>Syl_p(G) \neq \emptyset</math>

	We proceed by induction on the ~~oder~~ of $p$. Assume the claim holds for all groups of order less than $\|G\|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>\|G\| \equiv 0\mod\ p</math> we have either:		We proceed by induction on the order of <<p>>. Assume the claim holds for all groups of order less than $\|G\|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>\|G\| \equiv 0\mod\ p</math> we have either:
	* <math>\|G\| \equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\mod\ p</math>.		* <math>\|G\| \equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\mod\ p</math>.
	* <math>\|G\| \not\equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>.		* <math>\|G\| \not\equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>.

Pgadey at 20:13, 6 October 2011

2011-10-06T20:13:57Z

← Older revision		Revision as of 16:13, 6 October 2011
Line 73:		Line 73:

	; Sylow 2		; Sylow 2
	: Every Sylow <math>p</math>-subgroup of <math>G> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.		: Every Sylow <math>p</math>-subgroup of <math>G</math> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.

	; Sylow 3		; Sylow 3

Pgadey at 20:12, 6 October 2011

2011-10-06T20:12:43Z

← Older revision		Revision as of 16:12, 6 October 2011
Line 4:		Line 4:

	;Theorem		;Theorem
			Theorem
	: If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.		: If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.

Line 45:		Line 46:
	Observe that <math>\|Gx_i\|\| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that		Observe that <math>\|Gx_i\|\| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that
	<math> \|G\| = \|Z(G)\| + \sum_{\|Gx_i\| > 1} \frac{\|G\|}{\|Stab(x)\|} </math>		<math> \|G\| = \|Z(G)\| + \sum_{\|Gx_i\| > 1} \frac{\|G\|}{\|Stab(x)\|} </math>
	The formula above is called ''"the class formula"''. We have that <math>\|G\| / \|Stab(x)\| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>\|G\| \equiv 0\ \mod\ p</math> and <math>\sum_{\|Gx_i\| > 1} \frac{\|G\|}{\|Stab(x_i)\|} \equiv 0\ \mod\ p</math>. It follows that <math>\|Z(G)\| \equiv 0\ \mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq \|Z(G)\|</math> and thus <math>p \leq \|Z(G)\|</math>.		The formula above is called ''"the class formula"''. We have that <math>\|G\| / \|Stab(x)\| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>\|G\| \equiv 0\mod\ p</math> and <math>\sum_{\|Gx_i\| > 1} \frac{\|G\|}{\|Stab(x_i)\|} \equiv 0\mod\ p</math>. It follows that <math>\|Z(G)\| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq \|Z(G)\|</math> and thus <math>p \leq \|Z(G)\|</math>.

	==Sylow==		==Sylow==
Line 58:		Line 59:

	; Sylow set		; Sylow set
	: If <math>\|G\| = p^k m</math> for <math>m \not\equiv 0\ \mod\ p</math> then <math>Syl_p(G) = \{P \leq G : \|P\| = p^k</math>.		: If <math>\|G\| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : \|P\| = p^k</math>.

	; Sylow I		; Sylow I
	: <math>Syl_p(G) \neq \emptyset</math>		: <math>Syl_p(G) \neq \emptyset</math>

	We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $\|G\|$. <span style="color:green">[Dror: "Stare at the class equation.]</span> Since <math>\|G\| \equiv 0\ \mod\ p</math> we have either:		We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $\|G\|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>\|G\| \equiv 0\mod\ p</math> we have either:
	* <math>\|G\| \equiv 0\ \mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\ \mod\ p</math>.		* <math>\|G\| \equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\mod\ p</math>.
	* <math>\|G\| \not\equiv 0\ \mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\ \mod\ p</math>.		* <math>\|G\| \not\equiv 0\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>.

	If <math>\|Z(G)\| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>\|G\|/\|Stab(x_i)\| \not\equiv 0\ \mod\ p</math>. Thus <math>p^k</math> divides <math>\|Stab(x_i)\|</math>. We have that <math>\|Stab(x_i)\| < \|G\|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < \|G\|</math> and by induction there is <math>\|P\| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.		If <math>\|Z(G)\| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>\|G\|/\|Stab(x_i)\| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>\|Stab(x_i)\|</math>. We have that <math>\|Stab(x_i)\| < \|G\|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < \|G\|</math> and by induction there is <math>\|P\| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.

	WIf <math>\|Z(G)\| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>\|<x>\| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>\|P'\| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> \|\pi^{-1}(P')\| = p(p^{k-1}) = p^k </math>.		WIf <math>\|Z(G)\| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>\|<x>\| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>\|P'\| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> \|\pi^{-1}(P')\| = p(p^{k-1}) = p^k </math>.
Line 75:		Line 76:

	; Sylow 3		; Sylow 3
	: Let <math>n_p(G) = \|Syl_p(G)\|</math>. We have <math>n_p \equiv 0\ \mod\ \|G\|</math> and <math>n_p \equiv 1\ \mod\ p</math>.		: Let <math>n_p(G) = \|Syl_p(G)\|</math>. We have <math>n_p \equiv 0\mod\ \|G\|</math> and <math>n_p \equiv 1\mod\ p</math>.

	; A Nearly Tautological Lemma		; A Nearly Tautological Lemma
Line 89:		Line 90:
	==Groups of Order 15==		==Groups of Order 15==

	If <math>\|G\| = 15</math> then <math>n_3 \equiv 0\ \mod\ 15</math> and <math>n_3 \equiv 1\ \mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\ \mod\ 15</math> and <math>n_5 \equiv 1\ \mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.		If <math>\|G\| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.

Pgadey at 20:11, 6 October 2011

2011-10-06T20:11:27Z

← Older revision		Revision as of 16:11, 6 October 2011
Line 1:		Line 1:
			==Theory of Transitive <math>G</math>-sets==
	; Theorem		; Theorem
	: Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"		: Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"
Line 46:		Line 47:
	The formula above is called ''"the class formula"''. We have that <math>\|G\| / \|Stab(x)\| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>\|G\| \equiv 0\ \mod\ p</math> and <math>\sum_{\|Gx_i\| > 1} \frac{\|G\|}{\|Stab(x_i)\|} \equiv 0\ \mod\ p</math>. It follows that <math>\|Z(G)\| \equiv 0\ \mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq \|Z(G)\|</math> and thus <math>p \leq \|Z(G)\|</math>.		The formula above is called ''"the class formula"''. We have that <math>\|G\| / \|Stab(x)\| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>\|G\| \equiv 0\ \mod\ p</math> and <math>\sum_{\|Gx_i\| > 1} \frac{\|G\|}{\|Stab(x_i)\|} \equiv 0\ \mod\ p</math>. It follows that <math>\|Z(G)\| \equiv 0\ \mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq \|Z(G)\|</math> and thus <math>p \leq \|Z(G)\|</math>.

			==Sylow==
	SYLOW

	A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.		A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.
Line 64:		Line 65:
	We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $\|G\|$. <span style="color:green">[Dror: "Stare at the class equation.]</span> Since <math>\|G\| \equiv 0\ \mod\ p</math> we have either:		We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $\|G\|$. <span style="color:green">[Dror: "Stare at the class equation.]</span> Since <math>\|G\| \equiv 0\ \mod\ p</math> we have either:
	* <math>\|G\| \equiv 0\ \mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\ \mod\ p</math>.		* <math>\|G\| \equiv 0\ \mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \equiv 0\ \mod\ p</math>.
	* <math>\|G\| \equiv 0\ ~~\not~~\mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\ \mod\ p</math>.		* <math>\|G\| \not\equiv 0\ \mod\ p</math> and <math>\sum \|G\|/\|Stab(x_i)\| \not\equiv 0\ \mod\ p</math>.

	If <math>\|Z(G)\| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>\|G\|/\|Stab(x_i)\| \not\equiv 0\ \mod\ p</math>. Thus <math>p^k</math> divides <math>\|Stab(x_i)\|</math>. We have that <math>\|Stab(x_i)\| < \|G\|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < \|G\|</math> and by induction there is <math>\|P\| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.		If <math>\|Z(G)\| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>\|G\|/\|Stab(x_i)\| \not\equiv 0\ \mod\ p</math>. Thus <math>p^k</math> divides <math>\|Stab(x_i)\|</math>. We have that <math>\|Stab(x_i)\| < \|G\|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < \|G\|</math> and by induction there is <math>\|P\| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.
Line 86:		Line 87:


	~~GROUPS~~ OF ~~ORDER~~ 15		==Groups of Order 15==

	If <math>\|G\| = 15</math> then <math>n_3 \equiv 0\ \mod\ 15</math> and <math>n_3 \equiv 1\ \mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\ \mod\ 15</math> and <math>n_5 \equiv 1\ \mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.		If <math>\|G\| = 15</math> then <math>n_3 \equiv 0\ \mod\ 15</math> and <math>n_3 \equiv 1\ \mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\ \mod\ 15</math> and <math>n_5 \equiv 1\ \mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.

Pgadey at 20:09, 6 October 2011

2011-10-06T20:09:29Z

← Older revision		Revision as of 16:09, 6 October 2011
Line 77:		Line 77:

	; A Nearly Tautological Lemma		; A Nearly Tautological Lemma
	: If <math>P \in Syl_p(G)</math> and <math>H \~~lea~~ N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.		: If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.
	: If <math>x \in G</math> has <math>\|<x>\| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.		: If <math>x \in G</math> has <math>\|<x>\| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.