Difference between revisions of "Notes:Proof of the first group isomorphism theorem"
From Maths
(Saving work) |
m (Saving work) |
||
| Line 9: | Line 9: | ||
{| class="wikitable" border="1" style="overflow:hidden;" | {| class="wikitable" border="1" style="overflow:hidden;" | ||
|- | |- | ||
| − | | style="font-size:1.2em;" | <center><m>\xymatrix{ G \ar@{-->}[dr] | + | | 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> |
|- | |- | ||
! Diagram of morphisms in play | ! Diagram of morphisms in play | ||
| Line 22: | Line 22: | ||
** 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: | ||
| + | ** {{M|1=\varphi=i\circ\varphi'}} | ||
| + | ===Proof=== | ||
| + | * Note that {{M|1=\varphi=i\circ\varphi'}} and {{M|1=\varphi'=\theta\circ\pi}} - by substitution we see: | ||
| + | ** {{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. | ||
| + | # {{M|\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 {{M|\theta}} is not surjective, then we cannot have {{M|1=\varphi'=\theta\circ\pi}} (as {{M|\varphi'}} is surjective) | ||
| + | # {{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 [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
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.
- [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)
- 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:
- [ilmath]\theta[/ilmath] is injective
