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

Latest revision as of 17:41, 16 July 2016

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

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: $\varphi=\bar{\varphi}\circ\pi$
• Additionally, I take it as trivial that:
  • $\varphi=i\circ\varphi'$

Proof

• Note that $\varphi=i\circ\varphi'$ and $\varphi'=\theta\circ\pi$ - 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.

1. $\theta$ 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)
2. $\theta$ is injective
   • Suppose $\theta(x)=\theta(y)$, we wish to show that this means $x=y$
     • The gist is this: $\theta([u])=\theta([v])\implies\varphi(u)=\varphi(v)\implies\varphi(uv^{-1})=e$ thus $uv^{-1}\in\text{Ker}(\varphi)$
       • 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 $\theta([u])=\theta([v])\implies[u]=[v]$

Notes