Difference between revisions of "Notes:Proof of the first group isomorphism theorem"

Latest revision as of 17:41, 16 July 2016

Let $G$ and $H$ be groups, let $\varphi:G\rightarrow H$ be any group homomorphism, then:

Or, alternatively:

There exists a group isomorphism, θ:G/Ker(φ)→Im(φ) such that the following diagram commutes:
- $\xymatrix{ G \ar[d]_\pi \ar[r]^\varphi & H \\ G/\text{Ker}(\varphi) \ar@{.>}[r]_-\theta & \text{Im}(\varphi) \ar@{^{(}->}[u]_i }$ (so $\varphi=i\circ\theta\circ\pi$ ) where $i:\text{Im}(\varphi)\rightarrow H$ is the canonical injection, $i:h\mapsto h$ . It is a group homomorphism.

Diagram of morphisms in play
$\xymatrix{ G \ar@{-->}[dr]_(.17){\varphi'} \ar[d]_\pi \ar[r]^\varphi & H \\ G/\text{Ker}(\varphi) \ar@{.>}[ur]_(.8){\bar{\varphi} } \ar@{.>}[r]_-\theta & \text{Im}(\varphi) \ar@{^{(}->}[u]_i }$

First note:

We get a function, $\varphi':G\rightarrow\text{Im}(\varphi)$ I'll call the "canonical surjection", given by $\varphi':g\mapsto\varphi(g)$ .
We can factor φ′ through π (using the group factorisation theorem) to get θ:G/Ker(φ)→Im(φ)
- Which is of course a group homomorphism.
- And has the property: $\varphi'=\theta\circ\pi$
We can factor φ through π to, to give ˉφ:G/Ker(φ)→H
- Which is of course a group homomorphism.
- And has the property: $\varphi=\bar{\varphi}\circ\pi$
Additionally, I take it as trivial that:
- $\varphi=i\circ\varphi'$

Note that φ=i∘φ′ and φ′=θ∘π - by substitution we see:
- $\varphi=i\circ\theta\circ\pi$

This shows that the diagram commutes, we only need to show that $\theta$ is a group isomorphism to finish the proof.

I would like to do something like $\varphi=\bar{\varphi}\circ\theta^{-1}\circ\varphi'$ but I can't as $\theta^{-1}$ might not be a function.

Lets try the "brute force" approach of just showing it.

θ 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 $\theta$ is not surjective, then we cannot have $\varphi'=\theta\circ\pi$ (as $\varphi'$ is surjective)
θ is injective
- Suppose θ(x)=θ(y), we wish to show that this means x=y
  - The gist is this: θ([u])=θ([v])⟹φ(u)=φ(v)⟹φ(uv−1)=e thus uv−1∈Ker(φ)
    - So $\pi(uv^{-1})\in\text{Ker}(\varphi)$ so $\pi(uv^{-1})=[e]$ (the coset that is the normal subgroup $\text{Ker}(\varphi)$ itself)
    - Thus $\pi(uv^{-1})=[e]\implies\pi(u)=[e]\pi(v)\implies[u]=[v]$
  - We have shown $\theta([u])=\theta([v])\implies[u]=[v]$

@@ Line 34: / Line 34: @@
 #** 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}}