Group factorisation theorem

Statement

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

Let

$G$ and

$H$ be groups, and consider any

$N\trianglelefteq G$ ^{[Note 1]} and

$\varphi:G\rightarrow H$ be any group homomorphism, then^[1]:

If $N\subseteq\text{Ker}(\varphi)$ then $\varphi$ factors uniquely through $\pi$ to yield $\bar{\varphi}:G/N\rightarrow H$ given by $\bar{\varphi}:[g]\mapsto\varphi(g)$ ^{[Note 2]}

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

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