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]^(.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 }[/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]