Group factorisation theorem

Statement

[ilmath]\xymatrix{G \ar[r]^\varphi \ar[d]^\pi & H \\ G/N \ar@{.>}[ur]_{\overline{\varphi}} }[/ilmath]
Diagram

Let [ilmath]G[/ilmath] and [ilmath]H[/ilmath] be groups, and consider any [ilmath]N\trianglelefteq G[/ilmath] and [ilmath]\pi:G\rightarrow G/N[/ilmath] the canonical projection of the quotient group, let [ilmath]\varphi:G\rightarrow H[/ilmath] be any group homomorphism, then:

• If [ilmath]N\subseteq\text{Ker}(\varphi)[/ilmath] then [ilmath]\varphi[/ilmath] factors uniquely through [ilmath]\pi[/ilmath] to yield [ilmath]\bar{\varphi}:G/N\rightarrow H[/ilmath] given by [ilmath]\bar{\varphi}:[g]\mapsto\varphi(g)[/ilmath]

Additionally we have [ilmath]\varphi=\overline{\varphi}\circ\pi[/ilmath] (or in other terms, the diagram on the right commutes)

Proof

The bulk of this proof involves invoking the function factorisation theorem:

Let [ilmath]f:X\rightarrow Y[/ilmath] and [ilmath]w:X\rightarrow W[/ilmath] be functions. If [ilmath]\forall x,y\in X\big[ [w(x)=w(y)]\implies[f(x)=f(y)]\big][/ilmath] then [ilmath]f[/ilmath] "factors (uniquely, if [ilmath]w[/ilmath] is surjective) through [ilmath]w[/ilmath]" via: [ilmath]\bar{f}:W\rightarrow Y[/ilmath] given by [ilmath]\bar{f}:u\mapsto f(w^{-1}(u))[/ilmath] (which is "well defined") Which has the property that [ilmath]f=\bar{f}\circ w[/ilmath] (the diagram on the right commutes)

[ilmath]\xymatrix{X \ar[r]^w \ar[dr]_f & W \ar@{.>}[d]^{\overline{f} } \\ & Y}[/ilmath]

Applying the function factorisation theorem

Before we can apply this theorem we need to check that the objects in play are eligible (satisfy the requirements to factor) for the theorem. We set up as follows:

• Let [ilmath]G[/ilmath], [ilmath]H[/ilmath] be groups, let [ilmath]N\trianglelefteq G[/ilmath] and let [ilmath]\varphi:G\rightarrow H[/ilmath] be a group homomorphism with [ilmath]N\subseteq\text{Ker}(\varphi)[/ilmath], then:
  • [ilmath]\forall a,b\in X\big[ [\pi(a)=\pi(b)]\implies[\varphi(a)=\varphi(b)]\big][/ilmath]
• where [ilmath]\pi:G\rightarrow G/N[/ilmath] is the canonical projection of the quotient group

Proof

Let [ilmath]a,b\in X[/ilmath] be given

1. If [ilmath]\pi(a)\ne\pi(b)[/ilmath] then it doesn't matter what [ilmath]\varphi(a)[/ilmath] and [ilmath]\varphi(b)[/ilmath] are, as for logical implication if the LHS is false, it doesn't matter what the RHS for the implication to be considered true.
2. If [ilmath]\pi(a)=\pi(b)[/ilmath] then we require [ilmath]\varphi(a)=\varphi(b)[/ilmath] for the implication to be true.
   • By hypothesis we have [ilmath]\pi(a)=\pi(b)[/ilmath] so:
     1. [ilmath]\implies\pi(a)[\pi(b)]^{-1}=e[/ilmath] (here [ilmath]e[/ilmath] will denote the identity of whatever group makes sense in the context, in this case [ilmath]G/N[/ilmath] and [ilmath]\implies[/ilmath] may not be the strongest logical relation that can be used)
     • 1. [ilmath]\implies\pi(a)\pi(b^{-1})=e[/ilmath]
       2. [ilmath]\implies\pi(ab^{-1})=e[/ilmath]
     • [ilmath]\implies ab^{-1}\in\text{Ker}(\pi)[/ilmath] But! The kernel of the canonical projection of a quotient group is the normal subgroup of the quotient
     • [ilmath]\implies ab^{-1}\in N[/ilmath]
     • [ilmath]\implies ab^{-1}\in\text{Ker}(\varphi)[/ilmath] (by hypothesis, as [ilmath]N\subseteq\text{Ker}(\varphi)[/ilmath] using the implies-subset relation)
     • [ilmath]\implies \varphi(ab^{-1})=e[/ilmath] ([ilmath]e[/ilmath] is the identity of [ilmath]H[/ilmath] this time)
     • [ilmath]\implies \varphi(a)[\varphi(b)]^{-1}=e[/ilmath]
     • [ilmath]\implies \varphi(a)=\varphi(b)[/ilmath]
   • We have shown [ilmath]\pi(a)=\pi(b)\implies\varphi(a)=\varphi(b)[/ilmath] as required.