Difference between revisions of "Group factorisation theorem"

Revision as of 23:17, 15 July 2016 (view source) Alec (Talk | contribs) (Saving work)		Revision as of 00:38, 16 July 2016 (view source) Alec (Talk | contribs) m (→‎Proof: Saving work) Newer edit →
Line 19:		Line 19:
	==Proof==		==Proof==
		+	The bulk of this proof involves invoking the [[function factorisation theorem]]:
		+	<center>
		+	{| class="wikitable" border="1" style="width:80%;"
		+	|-
		+	| Let {{M|f:X\rightarrow Y}} and {{M|w:X\rightarrow W}} be [[function|functions]].
		+	* If {{M|1=\forall x,y\in X\big[ [w(x)=w(y)]\implies[f(x)=f(y)]\big]}} then {{M|f}} "factors (uniquely, if {{M|w}} is [[surjective]]) through {{M|w}}" via:
		+	** {{M|\bar{f}:W\rightarrow Y}} given by {{M|1=\bar{f}:u\mapsto f(w^{-1}(u))}} (which is "well defined")
		+	* Which has the property that {{M|1=f=\bar{f}\circ w}} (the diagram on the right commutes)
		+	| style="overflow:hidden;" | <m>\xymatrix{X \ar[r]^w \ar[dr]_f & W \ar@{.>}[d]^{\overline{f} } \\ & Y}</m>
		+	|}</center>
		+	===Applying the function factorisation theorem===
		+	Before we can apply this theorem we need to check that the objects in play are eligible (satisfy the requirements to factor) for the theorem. We set up as follows:
		+	*Let {{M|G}}, {{M|H}} be [[group|groups]], let {{M|N\trianglelefteq G}} and let {{M|\varphi:G\rightarrow H}} be a [[group homomorphism]] with {{M|N\subseteq\text{Ker}(\varphi)}}, then:
		+	** {{M|1=\forall a,b\in X\big[ [\pi(a)=\pi(b)]\implies[\varphi(a)=\varphi(b)]\big]}}
		+	* where {{M|\pi:G\rightarrow G/N}} is the [[canonical projection of the quotient group]]
		+	====Proof====
		+	Let {{M|a,b\in X}} be given
		+	# If {{M|\pi(a)\ne\pi(b)}} then it doesn't matter what {{M|\varphi(a)}} and {{M|\varphi(b)}} are, as for [[logical implication]] if the LHS is false, it doesn't matter what the RHS for the implication to be considered true.
		+	# If {{M|1=\pi(a)=\pi(b)}} then we require {{M|1=\varphi(a)=\varphi(b)}} for the [[implies|implication]] to be true.
		+	#* By hypothesis we have {{M|1=\pi(a)=\pi(b)}} so:
		+	#*# {{M|1=\implies\pi(a)[\pi(b)]^{-1}=e}} (here {{M|e}} will denote the identity of whatever group makes sense in the context, in this case {{M|G/N}} and {{M|\implies}} may not be the strongest logical relation that can be used)
		+	#**# {{M|1=\implies\pi(a)\pi(b^{-1})=e}}
		+	#**# {{M|1=\implies\pi(ab^{-1})=e}}
		+	#** {{M|1=\implies ab^{-1}\in\text{Ker}(\pi)}} But! [[The kernel of the canonical projection of a quotient group is the normal subgroup of the quotient]]
		+	#** {{M|1=\implies ab^{-1}\in N}}
		+	#** {{M|1=\implies ab^{-1}\in\text{Ker}(\varphi)}} (by hypothesis, as {{M|N\subseteq\text{Ker}(\varphi)}} using the [[implies-subset relation]])
		+	#** {{M|1=\implies \varphi(ab^{-1})=e}} ({{M|e}} is the identity of {{M|H}} this time)
		+	#** {{M|1=\implies \varphi(a)[\varphi(b)]^{-1}=e}}
		+	#** {{M|1=\implies \varphi(a)=\varphi(b)}}
		+	#* We have shown {{M|1=\pi(a)=\pi(b)\implies\varphi(a)=\varphi(b)}} as required.
		+
	==Notes==		==Notes==
	<references group="Note"/>		<references group="Note"/>

---

Revision as of 00:38, 16 July 2016

Statement

Let $G$ and $H$ be groups, and consider any $N\trianglelefteq G$ ^{[Note 1]} and $\pi:G\rightarrow G/N$ the canonical projection of the quotient group, let $\varphi:G\rightarrow H$ be any group homomorphism, then^[1]:

• If $N\subseteq\text{Ker}(\varphi)$ then $\varphi$ factors uniquely through $\pi$ to yield $\bar{\varphi}:G/N\rightarrow H$ given by $\bar{\varphi}:[g]\mapsto\varphi(g)$ ^{[Note 2]}

Additionally we have $\varphi=\overline{\varphi}\circ\pi$ (or in other terms, the diagram on the right commutes)

Proof

The bulk of this proof involves invoking the function factorisation theorem:

Let $f:X\rightarrow Y$ and $w:X\rightarrow W$ be functions. If $\forall x,y\in X\big[ [w(x)=w(y)]\implies[f(x)=f(y)]\big]$ then $f$ "factors (uniquely, if $w$ is surjective) through $w$" via: $\bar{f}:W\rightarrow Y$ given by $\bar{f}:u\mapsto f(w^{-1}(u))$ (which is "well defined") Which has the property that $f=\bar{f}\circ w$ (the diagram on the right commutes)

$\xymatrix{X \ar[r]^w \ar[dr]_f & W \ar@{.>}[d]^{\overline{f} } \\ & Y}$

Applying the function factorisation theorem

Before we can apply this theorem we need to check that the objects in play are eligible (satisfy the requirements to factor) for the theorem. We set up as follows:

• Let $G$, $H$ be groups, let $N\trianglelefteq G$ and let $\varphi:G\rightarrow H$ be a group homomorphism with $N\subseteq\text{Ker}(\varphi)$, then:
  • $\forall a,b\in X\big[ [\pi(a)=\pi(b)]\implies[\varphi(a)=\varphi(b)]\big]$
• where $\pi:G\rightarrow G/N$ is the canonical projection of the quotient group

Proof

Let $a,b\in X$ be given

1. If $\pi(a)\ne\pi(b)$ then it doesn't matter what $\varphi(a)$ and $\varphi(b)$ are, as for logical implication if the LHS is false, it doesn't matter what the RHS for the implication to be considered true.
2. If $\pi(a)=\pi(b)$ then we require $\varphi(a)=\varphi(b)$ for the implication to be true.
   • By hypothesis we have $\pi(a)=\pi(b)$ so:
     1. $\implies\pi(a)[\pi(b)]^{-1}=e$ (here $e$ will denote the identity of whatever group makes sense in the context, in this case $G/N$ and $\implies$ may not be the strongest logical relation that can be used)
     • 1. $\implies\pi(a)\pi(b^{-1})=e$
       2. $\implies\pi(ab^{-1})=e$
     • $\implies ab^{-1}\in\text{Ker}(\pi)$ But! The kernel of the canonical projection of a quotient group is the normal subgroup of the quotient
     • $\implies ab^{-1}\in N$
     • $\implies ab^{-1}\in\text{Ker}(\varphi)$ (by hypothesis, as $N\subseteq\text{Ker}(\varphi)$ using the implies-subset relation)
     • $\implies \varphi(ab^{-1})=e$ ($e$ is the identity of $H$ this time)
     • $\implies \varphi(a)[\varphi(b)]^{-1}=e$
     • $\implies \varphi(a)=\varphi(b)$
   • We have shown $\pi(a)=\pi(b)\implies\varphi(a)=\varphi(b)$ as required.

Notes

1. Jump up ↑ The notation $A\trianglelefteq B$ means $A$ is a normal subgroup of the group $B$.
2. Jump up ↑ This may look strange as obviously you're thinking "what if we took a different representative $h\in[g]$ with $h\ne g$, then we'd have $\varphi(h)$ instead of $\varphi(g)$!", these are actually the same, see Factor (function) for more details, I shall explain this here.
   • Technically we have this: $\bar{\varphi}:u\mapsto\varphi(\pi^{-1}(u))$ for the definition of $\bar{\varphi}$
     • Note though that if $g,h\in\pi^{-1}(u)$ that:
       • $\pi(g)=\pi(h)=u$ and by hypothesis we have $[\pi(x)=\pi(y)]\implies[\varphi(x)=\varphi(y)]$
         • Thus $\varphi(g)=\varphi(h)$
     • So whichever representative of $[g]$ we use $\varphi(h)$ for $h\in[g]$ is the same.
   • This is actually all dealt with as a part of factor (function) not this theorem. However it is worth illustrating.

References

1. Jump up ↑ Abstract Algebra - Pierre Antoine Grillet

v • d • e   Group theory Overview of the basic and important objects of group theory - a branch of abstract algebra Primitives semigroup $\supset$ monoid $\supset$ group $\supset$ Abelian group (AKA: commutative group), group action Morphisms Group homomorphism, Group isomorphism Subobjects subgroup, normal subgroup, cosets Important theorems First isomorphism theorem, second isomorphism theorem, third isomorphism theorem (overview of the group isomorphism theorems) Important groups Symmetric group (Permutation group), Alternating group, Dihedral group, Quaternion group, Klein-4 group