View source for Group factorisation theorem

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==Statement==
<div style="float:right;margin:0.2em;">
{| class="wikitable" border="1" style="overflow:hidden;"
|-
| style="width:7.5em;" | <center><m>\xymatrix{G \ar[r]^\varphi \ar[d]^\pi & H \\ G/N \ar@{.>}[ur]_{\overline{\varphi}} }</m></center>
|-
! Diagram
|}
</div> Let {{M|G}} and {{M|H}} be [[groups]], and consider any {{M|N\trianglelefteq G}} <ref group="Note">The notation {{M|A\trianglelefteq B}} means {{M|A}} is a [[normal subgroup]] of the [[group]] {{M|B}}.</ref> and {{M|\pi:G\rightarrow G/N}} the [[canonical projection of the quotient group]], let {{M|\varphi:G\rightarrow H}} be any [[group homomorphism]], then{{rAAPAG}}:
* ''If'' {{M|N\subseteq\text{Ker}(\varphi)}} ''then'' {{M|\varphi}} [[factor (function)|factors]] ''uniquely'' through {{M|\pi}} to yield {{M|\bar{\varphi}:G/N\rightarrow H}} given by {{M|\bar{\varphi}:[g]\mapsto\varphi(g)}} <ref group="Note">This may look strange as obviously you're thinking "what if we took a different representative {{M|h\in[g]}} with {{M|h\ne g}}, then we'd have {{M|\varphi(h)}} instead of {{M|\varphi(g)}}!", these are actually the same, see [[Factor (function)]] for more details, I shall explain this here.
* Technically we have this: {{M|\bar{\varphi}:u\mapsto\varphi(\pi^{-1}(u))}} for the definition of {{M|\bar{\varphi} }}
** Note though that if {{M|g,h\in\pi^{-1}(u)}} that:
*** {{M|1=\pi(g)=\pi(h)=u}} and by hypothesis we have {{M|1=[\pi(x)=\pi(y)]\implies[\varphi(x)=\varphi(y)]}}
**** Thus {{M|1=\varphi(g)=\varphi(h)}} 
** So whichever representative of {{M|[g]}} we use {{M|\varphi(h)}} for {{M|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.</ref> and {{M|\bar{\varphi} }} is a [[group homomorphism]].
Additionally we have {{M|1=\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]]:
<center>
{| class="wikitable" border="1" style="width:80%;"
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| 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.
====Result====
We have shown that the [[factor (function)]] theorem is eligible for application here, thus we apply it and yield:
* A [[function]] {{M|\bar{\varphi}:G/N\rightarrow H}} given by {{M|\bar{\varphi}:[u]\rightarrow\varphi(u)}}<ref group="Note">This is just a nicer way of writing:
* {{M|\bar{\varphi}:g\mapsto \varphi(\pi^{-1}(g))}}</ref><ref group="Note">Please note, again, this may be an abuse of notation, but the result is well defined, as for {{M|a,b\in\pi^{-1}(x)}} we see {{M|1=\varphi(a)=\varphi(b)}} by hypothesis.</ref>
* By the [[surjective|surjectivity]] of the [[canonical project of the quotient group]] ({{M|\pi:G\rightarrow G/N}}) we know from the [[factor (function)|factor theorem]] that the induced function, {{M|\bar{\varphi}:G/N\rightarrow H}} is the unique function that satisfies:
** {{M|1=\varphi=\bar{\varphi}\circ\pi}} (there is no other {{M|\bar{\varphi}'}} with this property that is distinct from {{M|\bar{\varphi} }})
===Showing {{M|\bar{\varphi} }} is a group homomorphism===
We now have a function:
* {{M|\bar\varphi:G/N\rightarrow H}} given by {{M|\bar\varphi:[u]\mapsto\varphi(u)}}. We claim this is a [[group homomorphism]]. To show this we must show that:
** {{M|1=\forall a,b\in G/N[\bar{\varphi}(a,b)=\bar{\varphi}(a)\bar{\varphi}(b)]}}
We shall write {{M|[u]\in G/N}} and choose {{M|u\in[u]}} to represent the classes in what follows, however as has been mentioned in the notes several times, this isn't a problem.
* Let {{M|[u],[v]\in G/N}} be given; as usual {{M|1=[uv]=[u]*[v]}} where {{M|*}} is the binary operation of the {{M|G/N}} group.
** Then {{M|1=\bar{\varphi}([uv])=\varphi(uv)=\varphi(u)\varphi(v)}} as {{M|\varphi}} is a group homomorphism.
** Note that {{M|1=\bar{\varphi}([u])\bar{\varphi}([v])=\varphi(u)\varphi(v)}}
*** So {{M|1=\varphi(u)\varphi(v)=\bar{\varphi}([u])\bar{\varphi}([v])=\bar{\varphi}([uv])}}
* We have shown {{M|1=\bar{\varphi}([u])\bar{\varphi}([v])=\bar{\varphi}([uv])}} as required.
Thus {{M|\bar{\varphi} }} is a grop homomorphism from {{M|G/N}} to {{M|H}}
==See also==
* [[Group homomorphism theorem]] ({{AKA}}: [[First group isomorphism theorem]])
* [[Function factorisation theorem]]
==Notes==
<references group="Note"/>
==References==
<references/>
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{{Theorem Of|Abstract Algebra|Group Theory}}