Notes:Group isomorphism theorems
- Note: the bulk of the information on this page comes from:
Contents
Factorisation Theorem
Let [ilmath]N[/ilmath] be a normal subgroup of a group [ilmath](G,*)[/ilmath]. Let [ilmath]\varphi:G\rightarrow H[/ilmath] be any group homomorphism (so [ilmath]H[/ilmath] is also a group), whose kernel contains [ilmath]N[/ilmath], then [ilmath]\varphi[/ilmath] factors uniquely through [ilmath]\pi:G\rightarrow G/N[/ilmath].
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This diagram commutes |
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That is to say that there exists a group homomorphism:
- [ilmath]\psi:G/N\rightarrow H[/ilmath]
such that [ilmath]\varphi=\psi\circ\pi[/ilmath]
In line with the factor (function) page, note that we require [ilmath]\forall x,y\in G[(\pi(x)=\pi(y))\implies(\varphi(x)=\varphi(y))][/ilmath] which is not immediately obvious.
Additionally, notice that the induced function, [ilmath]\psi:G/N\rightarrow H[/ilmath] is unique, this is because the canonical projection map, [ilmath]\pi:G\rightarrow G/N[/ilmath] (which [ilmath]\pi:g\mapsto gN[/ilmath] where [ilmath]gN[/ilmath] denotes a coset of [ilmath]N[/ilmath]) is surjective. This uniqueness claim is proved on the factor (function) page.
First Isomorphism Theorem
This diagram commutes |
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If [ilmath]\varphi:A\rightarrow B[/ilmath] is any group homomorphism then:
- [ilmath]A/\text{Ker}(\varphi)\cong\text{Im}(\varphi)[/ilmath]
In fact, there is a group isomorphism, [ilmath]\theta:A/\text{Ker}(\varphi)\rightarrow\text{Im}(\varphi)[/ilmath] which is unique, such that:
- [ilmath]\varphi=i\circ\theta\circ\pi[/ilmath] where [ilmath]i:\text{Im}(\varphi)\rightarrow B[/ilmath] is the inclusion map.
Notice that if we take [ilmath]\bar{\varphi}:A\rightarrow\text{Im}(\varphi)[/ilmath] being [ilmath]\bar{\varphi}:a\mapsto\varphi(a)[/ilmath] that this map has the same kernel as [ilmath]\varphi[/ilmath]. This means we get a new diagram:
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Where:
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As this diagram commutes we see:
- [ilmath]\alpha=\theta\circ i[/ilmath]
and such, there's hopefully a nice way of showing that [ilmath]\theta[/ilmath] is an isomorphism.
Second Isomorphism Theorem
Let [ilmath](A,*)[/ilmath] be a group, suppose that [ilmath]N[/ilmath] and [ilmath]N'[/ilmath] are normal subgroups of [ilmath]A[/ilmath] and that [ilmath]N'\subseteq N[/ilmath] then [ilmath]N'[/ilmath] is a normal subgroup of [ilmath]N[/ilmath], [ilmath]N/N'[/ilmath] is a normal subgroup of [ilmath]A/N[/ilmath] and:
- [math]A/N\cong\frac{A/N'}{N/N'}[/math]
In fact there is an isomorphism, [ilmath]\theta:A/N\rightarrow (A/N')/(N/N')[/ilmath] such that:
- [ilmath]\theta\circ\rho=\tau\circ\pi[/ilmath] where [ilmath]\pi[/ilmath], [ilmath]\rho[/ilmath] and [ilmath]\tau[/ilmath] are canonical projections
- Take this diagram: [ilmath]\xymatrix{ A \ar[r]^\pi \ar[d]_{\rho} & A/N' \ar[d]^{\tau} \ar@{-->}[dl]^\sigma \\ A/N \ar@{.>}[r]^-\theta & (A/N')/(N/N') }[/ilmath]
- if we first factor [ilmath]\rho[/ilmath] through [ilmath]\pi[/ilmath] to get [ilmath]\sigma[/ilmath]
- Take this diagram: [ilmath]\xymatrix{ A \ar[r]^\pi \ar[d]_{\rho} & A/N' \ar[d]^{\tau} \ar@{-->}[dl]^\sigma \\ A/N \ar@{.>}[r]^-\theta & (A/N')/(N/N') }[/ilmath]
Third Isomorphism Theorem
Let [ilmath]A[/ilmath] be a subgroup of a group, [ilmath](G,*)[/ilmath], let [ilmath]N[/ilmath] be a normal subgroup of [ilmath]G[/ilmath]. Then:
- [ilmath]AN[/ilmath] is a subgroup of [ilmath]G[/ilmath]
- [ilmath]N[/ilmath] is a normal subgroup of [ilmath]AN[/ilmath]
- [ilmath]A\cap N[/ilmath] is a normal subgroup of [ilmath]A[/ilmath] and
- [ilmath]AN/N\cong A/(A\cap N)[/ilmath]
TODO: Finish off