Group factorisation theorem

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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

$\pi:G\rightarrow G/N$ the canonical projection of the quotient group, let

$\varphi:G\rightarrow H$ 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]} and $\bar{\varphi}$ is a group homomorphism.

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

Proof

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

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

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 G, H be groups, let N⊴G and let φ:G→H be a group homomorphism with N⊆Ker(φ), then:
- $\forall a,b\in X\big[ [\pi(a)=\pi(b)]\implies[\varphi(a)=\varphi(b)]\big]$
where $\pi:G\rightarrow G/N$ is the canonical projection of the quotient group

Proof

Let $a,b\in X$ be given

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

Result

We have shown that the factor (function) theorem is eligible for application here, thus we apply it and yield:

A function $\bar{\varphi}:G/N\rightarrow H$ given by $\bar{\varphi}:[u]\rightarrow\varphi(u)$ ^{[Note 3]}^{[Note 4]}
By the surjectivity of the canonical project of the quotient group (π:G→G/N) we know from the factor theorem that the induced function, ˉφ:G/N→H is the unique function that satisfies:
- $\varphi=\bar{\varphi}\circ\pi$ (there is no other $\bar{\varphi}'$ with this property that is distinct from $\bar{\varphi}$ )

Showing $\bar{\varphi}$ is a group homomorphism

We now have a function:

ˉφ:G/N→H given by ˉφ:[u]↦φ(u). We claim this is a group homomorphism. To show this we must show that:
- $\forall a,b\in G/N[\bar{\varphi}(a,b)=\bar{\varphi}(a)\bar{\varphi}(b)]$

We shall write $[u]\in G/N$ and choose $u\in[u]$ to represent the classes in what follows, however as has been mentioned in the notes several times, this isn't a problem.

Let [u],[v]∈G/N be given; as usual [uv]=[u]∗[v] where ∗ is the binary operation of the G/N group.
- Then $\bar{\varphi}([uv])=\varphi(uv)=\varphi(u)\varphi(v)$ as $\varphi$ is a group homomorphism.
- Note that ˉφ([u])ˉφ([v])=φ(u)φ(v)
  - So $\varphi(u)\varphi(v)=\bar{\varphi}([u])\bar{\varphi}([v])=\bar{\varphi}([uv])$
We have shown $\bar{\varphi}([u])\bar{\varphi}([v])=\bar{\varphi}([uv])$ as required.

Thus $\bar{\varphi}$ is a grop homomorphism from $G/N$ to $H$

Notes

Jump up ↑ The notation $A\trianglelefteq B$ means $A$ is a normal subgroup of the group $B$ .
Jump up ↑
This may look strange as obviously you're thinking "what if we took a different representative h∈[g] with h≠g, then we'd have φ(h) instead of φ(g)!", these are actually the same, see Factor (function) for more details, I shall explain this here.
- Technically we have this: ˉφ:u↦φ(π−1(u)) for the definition of ˉφ
  - Note though that if g,h∈π−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.
Jump up ↑
This is just a nicer way of writing:
- $\bar{\varphi}:g\mapsto \varphi(\pi^{-1}(g))$
Jump up ↑ Please note, again, this may be an abuse of notation, but the result is well defined, as for $a,b\in\pi^{-1}(x)$ we see $\varphi(a)=\varphi(b)$ by hypothesis.

References

Jump up ↑ Abstract Algebra - Pierre Antoine Grillet

[1] Jump up ↑ The notation $A\trianglelefteq B$ means $A$ is a normal subgroup of the group $B$ .

[3] Jump up ↑ 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.

[3] 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.

[4] 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 $\varphi(g)=\varphi(h)$

[5] $\pi(g)=\pi(h)=u$ and by hypothesis we have $[\pi(x)=\pi(y)]\implies[\varphi(x)=\varphi(y)]$
Thus $\varphi(g)=\varphi(h)$

[6] Thus $\varphi(g)=\varphi(h)$

[7] So whichever representative of $[g]$ we use $\varphi(h)$ for $h\in[g]$ is the same.

[8] This is actually all dealt with as a part of factor (function) not this theorem. However it is worth illustrating.

[4] Jump up ↑ This is just a nicer way of writing:
$\bar{\varphi}:g\mapsto \varphi(\pi^{-1}(g))$

[10] $\bar{\varphi}:g\mapsto \varphi(\pi^{-1}(g))$

[5] Jump up ↑ Please note, again, this may be an abuse of notation, but the result is well defined, as for $a,b\in\pi^{-1}(x)$ we see $\varphi(a)=\varphi(b)$ by hypothesis.

[AAPAG-2] Jump up ↑ Abstract Algebra - Pierre Antoine Grillet

[Note 1]

[1]

[Note 2]

[Note 3]

[Note 4]

Group factorisation theorem

Contents

Statement

Proof

Applying the function factorisation theorem

Proof

Result

Showing $\bar{\varphi}$ is a group homomorphism

See also

Notes

References

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Group factorisation theorem

Contents

Statement

Proof

Applying the function factorisation theorem

Proof

Result

Showing ˉφ\bar{\varphi} is a group homomorphism

See also

Notes

References

Navigation menu

Search

Showing $\bar{\varphi}$ is a group homomorphism