3 - Fibers, Kernel, Conjugation, Centre

Fibers, Images

note: This term is inherited from category theory, where it carries more meaning. But in our case, it's just a fancy word for the inverse image of set over a map.

let $G_{1}, G_{2}$ be groups and $f : G_{1} \to G_{2}$ be a map, then the fibers of $H_{2} \subset G_{2}$ over $f$ is the set $f^{- 1} (H_{2}) = {x \in G_{1} : f (x) \in H_{2}}$ .

The image of a set $H_{1} \subset G_{1}$ is the set $f (H_{1}) = {f (x) : x \in H_{1}}$ .

The kernel of a map between groups

let $G_{1}, G_{2}$ be groups and $f : G_{1} \to G_{2}$ be a map. Then the kernel of $f$ , $K e r (f)$ is the fibres of the identity $e_{2} \in G_{2}$ . that is, $K e r (f) = f^{- 1} ({e_{2}}) = {x \in G_{1} : f (x) = e_{2}}$ .

conjugation, Normalizer

let $G$ be a group, and $S$ a subset of $G$ . Let $g$ be any particular element of $G$ . Then the conjugation of $S$ with $g$ is the set $g S g^{- 1} = {g s g^{- 1} : s \in S}$ .

The normalizer of $S$ , $N (S)$ is the set of all $g \in G$ for which the conjugation of $S$ with $g$ is equal to $S$ itself. That is, $N (S) = {g \in G : g S g^{- 1} = S}$ .

The kernel of a homomorphism is a subgroup of the domain group

Let $G_{1}, G_{2}$ be groups and $f : G_{1} \to G_{2}$ a homomorphism. Then, $K e r (f) ⩽ G_{1}$ .

\begin{proof}
We will use properties of homomorphisms for the following argument. Notice that $f (e_{1}) = e_{2}$ hence $e_{1} \in K e r (f)$ .

Now suppose $x, y \in K e r (f)$ then $f (x) = f (y) = e_{2}$ . Hence $f (x) f (y) = e_{2} e_{2} = e_{2}$ (see: the unique property of the identity element of a group). But, $f$ is a homomorphism. Hence, $f (x y) = e_{2}$ . Therefore, $x y \in K e r (f)$ .

Now, suppose $x \in K e r (f)$ , then consider the equation $f (x x^{- 1}) = f (e_{1}) = e_{2}$ . Since $f$ is a homomorphism, we obtain $f (x) f (x^{- 1}) = e_{2}$ . But, $f (x) = e_{2}$ . Therefore, $f (x^{- 1}) = e_{2}$ hence $x^{- 1} \in K e r (f)$ .
\end{proof}

The image of the domain group is a subgroup of the target under a homomorphism

Let $G_{1}, G_{2}$ be groups and $f : G_{1} \to G_{2}$ a homomorphism. Then, $f (G_{1}) ⩽ G_{2}$ .

\begin{proof}
First, notice that $f (G_{1})$ is non empty, as $e_{1} \in G_{1}$ , hence $e_{2} \in f (G_{1})$ .
Now, suppose $f (x), f (y) \in f (G_{1})$ . Then, $f (x) f (y) = f (x y) \in f (G_{1})$ as $x y \in G_{1}$ . Similarly, if $f (x) \in G_{1}$ , then $f (x)^{- 1} = f (x^{- 1}) \in f (G_{1})$ as $x^{- 1} \in G_{1}$ . See: properties of homomorphisms .
\end{proof}

The conjugation of a subgroup

Let $G$ be a group, and $H$ be a subgroup of $G$ . then for any element $g \in G$ , $g H g^{- 1} ⩽ G$ .

\begin{proof}
Since $e \in H$ , $g e g^{- 1} = e \in g H g^{- 1}$ .

Now, suppose $u, v \in g H g^{- 1}$ . Then for some $x, y \in H$ , $u = g x g^{- 1}$ and $v = g y g^{- 1}$ . Therefore, $u v = g (x y) g^{- 1}$ , hence $u v \in g H g^{- 1}$ .

Now, suppose $t \in g H g^{- 1}$ then $t = g h g^{- 1}$ for some $h \in H$ . $t^{- 1} = g h^{- 1} g^{- 1}$ (The reader is asked to pause and carry out the product $t t^{- 1}$ and $t^{- 1} t$ , should any doubt arise).
Finally apply necessary and sufficent conditions for Subgroups.

\end{proof}

The normalizer of a subgroup

Let $G$ be a group, and $H$ a subgroup of $G$ , then, $H ⩽ N (H) ⩽ G$ .

\begin{proof}
We will begin by showing that $N (H)$ is a subgroup of $G$ :

$N (H) = {g \in G : g H g^{- 1} = H}$ . Clearly, $N (H)$ is non empty as $e H e = H$ . ( $e = e^{- 1}$ ).

Suppose $g, h \in N (H)$ . then $g H g^{- 1} = h H h^{- 1} = H$ . Notice that $(g h) H (h^{- 1} g^{- 1}) = g (h H h^{- 1}) g^{- 1} = g (H) g^{- 1} = H$ . Therefore, $g h \in N (H)$ .

Now suppose that $g \in N (H)$ , then $g H g^{- 1} = H$ . But $e H e = H$ . Hence, $g^{- 1} (g H g^{- 1}) g = H$ , therefore, $g^{- 1} H g = H$ . Hence $g^{- 1} \in N (H)$ .

Finally, apply necessary and sufficent conditions for Subgroups.

Now, we will show that $H$ is a subgroup of $N (H)$ .

Of course $H$ is a group, so it is sufficient to show that $H \subset N (H)$ . For any $h \in H$ , $h H h^{- 1} = (h H) h^{- 1}$ but $h H$ is just the image of the left coset map on $H$ given by $f_{h} (x) = h x$ for all $x \in H$ . That is, $h H$ is just notation for $f_{h} (H)$ . But left coset maps are bijective. Hence, $h H = H$ .

Therefore, we have $h H h^{- 1} = (h H) h^{- 1} = H h^{- 1}$ . Using a similar argument, $H h^{- 1} = g_{h^{- 1}} (H)$ Where $g_{h^{- 1}}$ is the right coset map induced by $h^{- 1}$ . Analogously, right coset maps are bijective too. hence $H h^{- 1} = H$ .

Finially, this gives us $h H h^{- 1} = H$ . Therefore, any $h \in H$ is also in $N (H)$ .

\end{proof}

The center of a group

The Center of a group $G$ , denoted by $Z (G)$ , is the subset of $G$ whose all elements commute with the entirety of $G$ . That is $Z (G) = {g \in G : g x = x g, \forall x \in G}$ .

The center of a group is a subgroup

Let $G$ be a group. Then, $Z (G) ⩽ G$ .

\begin{proof} Since $e \in Z (G)$ , $Z (G)$ is not empty. Let $g, h \in Z (G)$ . then for all $x \in G$ , $g x = x g$ and $h x = x h$ . But, $h x, x h \in G$ . Hence $(g h) x = (g x) h = x (g h)$ Hence $g h \in Z (G)$ .
Now, if $x \in G$ , and $g \in Z (G)$ , then $x^{- 1} \in G$ . Hence, $g x^{- 1} = x^{- 1} g$ hence $x g^{- 1} = g^{- 1} x$ . hence $g^{- 1} \in Z (G)$ .

\end{proof}