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

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{| class="wikitable" border="1" style="overflow:hidden;"
 
{| class="wikitable" border="1" style="overflow:hidden;"
 
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| style="font-size:1.2em;" | <center><m>\xymatrix{ G \ar@{-->}[dr]^(.3){\varphi'} \ar[d]_\pi \ar[r]^\varphi & H \\ G/\text{Ker}(\varphi) \ar@{.>}[ur]^(.3){\bar{\varphi} } \ar@{.>}[r]_-\theta & \text{Im}(\varphi) \ar@{^{(}->}[u]_i }</m></center>
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| 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>
 
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! Diagram of morphisms in play
 
! Diagram of morphisms in play
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** Which is of course a group homomorphism.
 
** Which is of course a group homomorphism.
 
** And has the property: {{M|1=\varphi=\bar{\varphi}\circ\pi}}
 
** And has the property: {{M|1=\varphi=\bar{\varphi}\circ\pi}}
 +
* Additionally, I take it as trivial that:
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** {{M|1=\varphi=i\circ\varphi'}}
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===Proof===
 +
* Note that {{M|1=\varphi=i\circ\varphi'}} and {{M|1=\varphi'=\theta\circ\pi}} - by substitution we see:
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** {{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.
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# {{M|\theta}} is [[surjective]].
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#* 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)
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# {{M|\theta}} is [[injective]]
 
==Notes==
 
==Notes==
 
<references group="Note"/>
 
<references group="Note"/>
 
{{Notes|Abstract Algebra|Group Theory}}
 
{{Notes|Abstract Algebra|Group Theory}}

Revision as of 17:12, 16 July 2016

Claim

Let G and H be groups, let φ:G→H be any group homomorphism, then:

  • G/Ker(φ)≅Im(φ)

Or, alternatively:

  • There exists a group isomorphism, θ:G/Ker(φ)→Im(φ) such that the following diagram commutes:
    • (so φ=i∘θ∘π) where i:Im(φ)→H is the canonical injection, i:h↦h. It is a group homomorphism.

Proof

Diagram of morphisms in play

First note:

  • We get a function, φ′:G→Im(φ) I'll call the "canonical surjection", given by φ′:g↦φ(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: φ′=θ∘π
  • 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: φ=ˉφ∘π
  • Additionally, I take it as trivial that:
    • φ=i∘φ′

Proof

  • Note that φ=i∘φ′ and φ′=θ∘π - by substitution we see:
    • φ=i∘θ∘π

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

  • I would like to do something like φ=ˉφ∘θ−1∘φ′ but I can't as θ−1 might not be a function.

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

  1. 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 θ is not surjective, then we cannot have φ′=θ∘π (as φ′ is surjective)
  2. θ is injective

Notes

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