3

Theorem (FTH) Let $f: G \rightarrow H$ be a group homomorphism and let $N$ be a normal subgroup of G which is contained in $\ker(f)$ Then there exists a unique group homomorphism $\overline{f}:G/N \rightarrow H$ such that $\overline{f}\circ v=f$. Here $v: G \rightarrow G/N$ denotes the natural epimorphism. In other words, $\overline{f}(aN)=f(a)$ for all $a \in G$. Moreover, $\operatorname{im}(f)$ and $\ker(\overline{f})= \{ aN \in G/N\mid a\in \ker(f) \}=\ker(f)/N $

Let's start with with the basics. What is $G/N$? and what is a normal subgroup of G which is contained in $\ker(f)$?. I understand what it is, but I feel like I am missing the big picture to really understand this theorem.

Zev Chonoles
  • 129,973
daniel
  • 751
  • 1
  • 9
  • 22

3 Answers3

3

HINT: Define $\bar{f}(aN) = f(a)$ for all $aN \in G/N$ and prove that this is a well-defined homomorphism and that it satisfies the given relation. And obviously this is a unique homomorphism, as $\bar{f} \circ v = f$ requires $\bar{f}(aN) = f(a)$ for all $aN \in G/N$ to hold, which is indeed the only homomorphism.

To get the intuition behind it think of a homomorphism as a grouping of the elements of one group, having a certain common property and regarding them as a single element (a representative for that property). In this case $f$ does this "collapsing" in one step, while $\bar{f} \circ v$ does it in two steps, while eventually getting the same thing.

Stefan4024
  • 35,843
  • Ok, so we have a group homomorphism $f$ between two groups G and H and we also have that there exists a normal subgroup N of G in the kernel of f. The theorem is that we can construct a homomorphism $\overline{f}$ from $G/N$ which is the quotient group to H and it will have the property that $\overline{f} \circ v=f$ where v is the natural epimorphism – daniel Aug 18 '16 at 01:46
3

A typical application of this theorem is when you are specifically interested in the group $G/N$ (or something isomorphic to it), and want to know about homomorphisms defined on it. If $G$ is a group that's easier to understand, then invoking this theorem lets you use what you know about homomorphisms out of $G$ to tell you about the homomorphisms out of $G/N$.

A simple example is the question of what homomorphisms exist between cyclic groups $\mathbb{Z} / m\mathbb{Z} \to \mathbb{Z} / n\mathbb{Z}$.

We can set $G = \mathbb{Z}$, $N = m\mathbb{Z}$, and $H = \mathbb{Z} / n\mathbb{Z}$. $G$ is very easy to understand, so we can compute there are bijections between the following sets:

  • homomorphisms $\mathbb{Z} / m\mathbb{Z}\to \mathbb{Z} / n\mathbb{Z}$
  • homomorphisms $\mathbb{Z} \to \mathbb{Z} / n \mathbb{Z}$ whose kernel contains $m \mathbb{Z}$.
  • Elements $x \in \mathbb{Z} / n \mathbb{Z}$ such that $mx = 0$

Explicitly, given a homomorphism $\varphi : \mathbb{Z} / m \mathbb{Z} \to \mathbb{Z} / n \mathbb{Z}$, the corresponding $x$ is $\varphi(\bar{1})$. Conversely, given an $x$, the corresponding homomorphism is $\bar{a} \mapsto ax$.

This example, I think, highlights a very important special case: very often, the way we define a homomorphism $G/N \to H$ is by the following recipe:

  • Define a function $\varphi: G \to H$
  • Show that it's a homomorphism
  • Show that $\varphi(N) = \{ e_H \}$

By the theorem, this recipe yields a well-defined homomorphism $G/N \to H$.

2

On equality $ker(\overline{f})=ker(f)/N$ we see the need to $N$ be contained in $ker(f)$ and, obviously, be normal in $G$. To fix ideas, this result can be understood as follows:

If $f:G\longrightarrow H$ be a group homomorphism and $N$ a normal subgroup of $G$ which is contained in $ker(f)$, then there exists a unique group homomorphism $\overline{f}:G/N\longrightarrow H$ such that the following diagram commutes $$\require{AMScd}\begin{CD}G@>{f}>>H\\@V{v}VV&@VVid_{H}V\\G/N@>>\overline{f}>H\end{CD}$$ where $v$ is the natural epimorphism and $id_{H}$ is the identity of $H$.

In other words, we can decompose $f$ in a unique composition of homomorphisms where one is $v$.

Let's prove it. We want $\overline{f}$ satisfyng $\overline{f}\circ v(x)=\overline{f}(xN)=f(x)$, so define $$\begin{matrix}\overline{f}:&G/N&\longrightarrow&H\\&xN&\longmapsto&f(x)\end{matrix}$$ We need to show that:

$1)$ $\overline{f}$ is well-defined: if $xN=yN$, then $xy^{-1}\in N\subseteq ker(f)$. Thus $$f(x)f(y)^{-1}=f(x)f(y^{-1})=f(xy^{-1})=1_{H}$$ so $f(x)=f(y)$, i.e., $\overline{f}(xN)=\overline{f}(yN)$. So we prove that $\overline{f}$ is well-defined.

$2)$ $\overline{f}$ is a group homomorphism: for all $xN,yN\in G/N$, $$\overline{f}(xN\cdot yN)=\overline{f}(xyN)=f(xy)=f(x)f(y)=\overline{f}(xN)\overline{f}(yN).$$

$3)$ $\overline{f}$ is such that the diagram above commutes: it is follow from the definition.

$4)$ $\overline{f}$ is the unique group homomorphism satisfying this property: if $g:G/N\longrightarrow H$ is such that that diagram above commutes, then for all $xN\in G/N$ we have $$g(xN)=g\circ v(x)=f(x)=id_{H}\circ f(x)=\overline{f}\circ v(x)=\overline{f}(xN).$$ Therefore $g=\overline{f}$.

The proof is done.

Rafael
  • 3,789