Group factorisation theorem
From Maths
Contents
Statement
- 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] [Note 2]
Additionally we have [ilmath]\varphi=\overline{\varphi}\circ\pi[/ilmath] (or in other terms, the diagram on the right commutes)
Proof
Notes
- ↑ The notation [ilmath]A\trianglelefteq B[/ilmath] means [ilmath]A[/ilmath] is a normal subgroup of the group [ilmath]B[/ilmath].
- ↑ This may look strange as obviously you're thinking "what if we took a different representative [ilmath]h\in[g][/ilmath] with [ilmath]h\ne g[/ilmath], then we'd have [ilmath]\varphi(h)[/ilmath] instead of [ilmath]\varphi(g)[/ilmath]!", these are actually the same, see Factor (function) for more details, I shall explain this here.
- Technically we have this: [ilmath]\bar{\varphi}:u\mapsto\varphi(\pi^{-1}(u))[/ilmath] for the definition of [ilmath]\bar{\varphi} [/ilmath]
- Note though that if [ilmath]g,h\in\pi^{-1}(u)[/ilmath] that:
- [ilmath]\pi(g)=\pi(h)=u[/ilmath] and by hypothesis we have [ilmath][\pi(x)=\pi(y)]\implies[\varphi(x)=\varphi(y)][/ilmath]
- Thus [ilmath]\varphi(g)=\varphi(h)[/ilmath]
- [ilmath]\pi(g)=\pi(h)=u[/ilmath] and by hypothesis we have [ilmath][\pi(x)=\pi(y)]\implies[\varphi(x)=\varphi(y)][/ilmath]
- So whichever representative of [ilmath][g][/ilmath] we use [ilmath]\varphi(h)[/ilmath] for [ilmath]h\in[g][/ilmath] is the same.
- Note though that if [ilmath]g,h\in\pi^{-1}(u)[/ilmath] that:
- This is actually all dealt with as a part of factor (function) not this theorem. However it is worth illustrating.
- Technically we have this: [ilmath]\bar{\varphi}:u\mapsto\varphi(\pi^{-1}(u))[/ilmath] for the definition of [ilmath]\bar{\varphi} [/ilmath]
References
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