Notes:Proof of the first group isomorphism theorem

Claim

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

• $G/\text{Ker}(\varphi)\cong\text{Im}(\varphi)$

Or, alternatively:

• There exists a group isomorphism, $\theta:G/\text{Ker}(\varphi)\rightarrow\text{Im}(\varphi)$ 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.

Proof

$\xymatrix{ G \ar@{-->}[dr]^(.3){\varphi'} \ar[d]_\pi \ar[r]^\varphi & H \\ G/\text{Ker}(\varphi) \ar@{.>}[ur]^(.3){\bar{\varphi} } \ar@{.>}[r]_-\theta & \text{Im}(\varphi) \ar@{^{(}->}[u]_i }$
Diagram of morphisms in play

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 $\varphi'$ through $\pi$ (using the group factorisation theorem) to get $\theta:G/\text{Ker}(\varphi)\rightarrow\text{Im}(\varphi)$
  • Which is of course a group homomorphism.
  • And has the property: $\varphi'=\theta\circ\pi$
• We can factor $\varphi$ through $\pi$ to, to give $\bar{\varphi}:G/\text{Ker}(\varphi)\rightarrow H$
  • Which is of course a group homomorphism.
  • And has the property: $\varphi=\bar{\varphi}\circ\pi$