Notes:Proof of the first group isomorphism theorem

Claim

Let [ilmath]G[/ilmath] and [ilmath]H[/ilmath] be groups, let [ilmath]\varphi:G\rightarrow H[/ilmath] be any group homomorphism, then:

• [ilmath]G/\text{Ker}(\varphi)\cong\text{Im}(\varphi)[/ilmath]

Or, alternatively:

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

Proof

[ilmath]\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 }[/ilmath]
Diagram of morphisms in play

First note:

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

Proof

• Note that [ilmath]\varphi=i\circ\varphi'[/ilmath] and [ilmath]\varphi'=\theta\circ\pi[/ilmath] - by substitution we see:
  • [ilmath]\varphi=i\circ\theta\circ\pi[/ilmath]

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

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

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

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