View source for Notes:Proof of the first group isomorphism theorem

==Claim==
Let {{M|G}} and {{M|H}} be [[group|groups]], let {{M|\varphi:G\rightarrow H}} be any [[group homomorphism]], then:
* {{M|G/\text{Ker}(\varphi)\cong\text{Im}(\varphi)}}
Or, alternatively:
* There exists a [[group isomorphism]], {{M|\theta:G/\text{Ker}(\varphi)\rightarrow\text{Im}(\varphi)}} such that the following diagram commutes:
** {{M|1=\xymatrix{ G \ar[d]_\pi \ar[r]^\varphi & H \\ G/\text{Ker}(\varphi) \ar@{.>}[r]_-\theta & \text{Im}(\varphi) \ar@{^{(}->}[u]_i } }} (so {{M|1=\varphi=i\circ\theta\circ\pi}}) where {{M|i:\text{Im}(\varphi)\rightarrow H}} is the [[canonical injection]], {{M|i:h\mapsto h}}. It is a group homomorphism.
==Proof==
<div style="float:right;">
{| class="wikitable" border="1" style="overflow:hidden;"
|-
| style="font-size:1.2em;" | <center><m>\xymatrix{ G \ar@{-->}[dr]_(.17){\varphi'} \ar[d]_\pi \ar[r]^\varphi & H \\ G/\text{Ker}(\varphi) \ar@{.>}[ur]_(.8){\bar{\varphi} } \ar@{.>}[r]_-\theta & \text{Im}(\varphi) \ar@{^{(}->}[u]_i }</m></center>
|-
! Diagram of morphisms in play
|}
</div>
First note:
* We get a [[function]], {{M|\varphi':G\rightarrow\text{Im}(\varphi)}} I'll call the "canonical [[surjection]]", given by {{M|\varphi':g\mapsto\varphi(g)}}.
* We can factor {{M|\varphi'}} through {{M|\pi}} (using the [[group factorisation theorem]]) to get {{M|\theta:G/\text{Ker}(\varphi)\rightarrow\text{Im}(\varphi)}}
** Which is of course a group homomorphism.
** And has the property: {{M|1=\varphi'=\theta\circ\pi}}
* We can factor {{M|\varphi}} through {{M|\pi}} to, to give {{M|\bar{\varphi}:G/\text{Ker}(\varphi)\rightarrow H}}
** Which is of course a group homomorphism.
** And has the property: {{M|1=\varphi=\bar{\varphi}\circ\pi}}
* Additionally, I take it as trivial that:
** {{M|1=\varphi=i\circ\varphi'}}
===Proof===
* Note that {{M|1=\varphi=i\circ\varphi'}} and {{M|1=\varphi'=\theta\circ\pi}} - by substitution we see:
** {{M|1=\varphi=i\circ\theta\circ\pi }}
This shows that the diagram commutes, we only need to show that {{M|\theta}} is a [[group isomorphism]] to finish the proof.
* I would like to do something like {{M|1=\varphi=\bar{\varphi}\circ\theta^{-1}\circ\varphi'}} but I can't as {{M|\theta^{-1} }} might not be a function.
Lets try the "brute force" approach of just showing it.
# {{M|\theta}} is [[surjective]].
#* While I have failed (I think... it was a few days ago and I wasn't pleased with it) to do it, I think the following is promising:
#** Suppose {{M|\theta}} is not surjective, then we cannot have {{M|1=\varphi'=\theta\circ\pi}} (as {{M|\varphi'}} is surjective)
# {{M|\theta}} is [[injective]]
#* Suppose {{M|1=\theta(x)=\theta(y)}}, we wish to show that this means {{M|1=x=y}}
#** The gist is this: {{M|1=\theta([u])=\theta([v])\implies\varphi(u)=\varphi(v)\implies\varphi(uv^{-1})=e}} thus {{M|uv^{-1}\in\text{Ker}(\varphi)}}
#*** So {{M|1=\pi(uv^{-1})\in\text{Ker}(\varphi)}} so {{M|1=\pi(uv^{-1})=[e]}} (the coset that is the normal subgroup {{M|\text{Ker}(\varphi)}} itself)
#*** Thus {{M|1=\pi(uv^{-1})=[e]\implies\pi(u)=[e]\pi(v)\implies[u]=[v]}}
#** We have shown {{M|1=\theta([u])=\theta([v])\implies[u]=[v]}}
==Notes==
<references group="Note"/>
{{Notes|Abstract Algebra|Group Theory}}