1 - Maps on Groups

Group homomorphism

let $G_{1}, G_{2}$ be groups and $f : G_{1} \to G_{2}$ be a map. $f$ is a group homomorphism if for each $x, y \in G_{1}$ , $f (x \cdot_{G_{1}} y) = f (x) \cdot_{G_{2}} f (y)$ .

furthermore, we will suppress any notation to indicate that $x y$ is a product in $G_{1}$ and $f (x) (f y)$ is a product in $G_{2}$ .

Properties of homomorphisms

For the following discussion, let $G_{1}, G_{2}$ be groups, with identities $e_{1}$ and $e_{2}$ and $f$ a homomorphism from $G_{1} t o G_{2}$ .

$f (e_{1}) = e_{2}$ (under a homomorphism, the image of the identity of the domain group, is the identity of the target group).
for each $x \in G_{1}$ , $f (x)^{- 1} \in G_{2}$ is $f (x^{- 1})$ where $x^{- 1}$ is the inverse $x$ in $G_{1}$ .
Let $G_{1}, G_{2}, G_{3}$ all be groups, and let $f$ be a homomorphism from $G_{1}$ to $G_{2}$ , $h$ a homomorphism from $G_{2}$ to $G_{3}$ , then $h \circ f$ is a homomorphism from $G_{1}$ to $G_{3}$ .(the composition of homomorphisms, is itself a homomorphism).
![[diagram-20240712 (1).svg#invert_B]]

\begin{proof}

We will use The unqiue property of the identity element of a group. Notice that $f (e_{1} e_{1}) = f (e_{1}) f (e_{1}) = f (e_{1})$ . Hence it must be that $f (e_{1}) = e_{2}$ .
Notice that for any $x \in G_{1}$ , $f (x x^{- 1}) = f (x) f (x^{- 1})$ but $x x^{- 1} = e_{1}$ . Therefore, $e_{2} = f (x) f (x^{- 1})$ Hence $f (x^{- 1})$ is the unique inverse of $f (x)$ in $G_{2}$ . In notation, $f (x)^{- 1} = f (x^{- 1})$ .
let $x, y$ be arbitrary elements of $G_{1}$ . Then, $h (f (x y)) = h (f (x) f (y)) = h (f (x)) h (f (y))$ .

\end{proof}

Group isomorphism

let $G_{1}$ , $G_{2}$ be groups. a map $f : G_{1} \to G_{2}$ is an isomorphism, if it is a bijective homomorphism. An isomorphism from $G$ onto itself is called an automorphism.

The inverse of an isomorphism is an isomorphism

if $f : G_{1} \to G_{2}$ is an isomorphism, then $f^{- 1} : G_{2} \to G_{1}$ is an isomorphism, where $G_{1}, G_{2}$ are groups.

\begin{proof}
The inverse of a bijection, is also a bijection. Hence, it is sufficient to prove that $f^{- 1}$ is a homomorphism. let $u, v$ be arbitrary elements of $G_{2}$ . Then, since $f$ is bijective, there exists unique $x, y \in G_{1}$ such that $u = f (x), v = f (y)$ . hence $u v = f (x) f (y) = f (x y)$ . Now, $f^{- 1} (u v) = f^{- 1} f (x y) = x y = f^{- 1} (u) f^{- 1} (v)$ . Hence $f^{- 1}$ is a homomorphism.

\end{proof}

The collection of all automorphisms on a group, is itself a group

Let $G$ be a group, and let $Aut (G)$ be the collection of all automorphisms on $G$ . Then this set, with map composition as the operator, forms a group.

\begin{proof}
notice that the identity map $i d (x) = x$ from $G$ to $G$ is an automorphism (it is both bijective and homomorphic). And acts as the group identity under composition. for any two maps $f, h \in Aut (G)$ . Then $h f$ is bijective, moreover due to [[#^3f5f3f]] $h f$ is a homomorphism, hence $h f \in Aut (G)$ . Associativity follows as a property of function composition, and due to [[#^e12bc0]], for each $f \in Aut (G)$ , $f^{- 1} \in Aut (G)$ : the inverse map, acts as the group inverse element.
\end{proof}

The left coset map:

left coset maps on a group

Let $G$ be a group, and $g$ an arbitrary element of $G$ . then the left coset map $f_{g} : G \to G$ is the map $f_{g} (x) = g x$ for each $x \in G$ .

left coset maps are bijective

Let $G$ be a group, and $L (G)$ be the collection of left coset maps from $G$ to $G$ . Then, every element $f_{g} \in L (G)$ is a bijection on $G$ .

\begin{proof}
Let $g$ be an arbitrary element of $G$ . We claim that $f_{g}$ is a bijection. suppose for $x, y \in G$ , $f_{g} (x) = f_{g} (y)$ . Then, $g x = g y$ hence $x = y$ , showing injectivity.
Now, suppose $y$ is an arbitrary element in $G$ . Notice that $f_{g} (g^{- 1} y) = g (g^{- 1} y) = y$ . Showing Surjectivity.

\end{proof}

the collection of left coset maps, is a group

Let $G$ be a group, then the collection of left coset maps $L (G)$ is a group under map composition. Moreover, $L (G)$ is isomorphic to $G$ .

\begin{proof}
The identity map on $G$ is $f_{e} = e x$ for all $x \in G$ , $f_{e} \in L (G)$ is the group identity as $f_{e} f_{g} = f_{g} f_{e} = f_{g}$ for any $f_{g} \in L (G)$ . Associativity follows as a property of map composition, of course, the composition of two left coset maps $f_{g}, f_{h} \in L (G)$ . is $f_{g} f_{h} = f_{g h}$ , because for all $x \in G$ $f_{g} (f_{h} (x)) = g (h x) = (g h) x = f_{g h} (x)$ . Hence, $L (G)$ is closed under composition, moreover the map $g \mapsto f_{g}$ is a homomorphism, as $g h \mapsto f_{g h} = f_{g} f_{h}$ .
For each $f_{g} \in L (G)$ the unique $f_{g}^{- 1}$ is given by $f_{g^{- 1}}$ . This is because for each $x \in G$ , $f_{g} (f_{g^{- 1}} (x)) = f_{g^{- 1}} (f_{g} (x)) = g g^{- 1} x = g^{- 1} g x = e x$ . Hence $f_{g} f_{g^{- 1}} = f_{g^{- 1}} f_{g} = f_{e}$ .
We claim that the map $g \mapsto f_{g}$ is bijective. Surjectivity is obvious. now suppose that $f_{g_{1}} = f_{g_{2}}$ then for each $x \in G$ , $g_{1} x = g_{2} x$ hence $g_{1} = g_{2}$ . Hence we produce the desired isomorphism from $G$ to $L (G$ ).

\end{proof}

two distinct left coset maps never agree on the image of a particular element.

Let $G$ be a group, and $f_{g}, f_{h} \in L (G)$ be two distinct left coset maps. Then there is no $x \in G$ such that $f_{g} (x) = f_{h} (x)$
Otherwise, this would imply that $g x = h x$ hence $g = h$ (multiplying $x^{- 1}$ on both the sides), meaning that $f_{g}$ and $f_{h}$ are identical maps.

Warning!!

A left coset map not NOT a homomorphism

Let $G$ be a group. Then a map $f_{g} \in L (G)$ if $g \neq e$ then $f_{g}$ is not a homomorphism from $G$ to $G$ .
$f_{g} (x y) = g (x y)$ , $f_{g} (x) f_{g} (y) = g x g y$ . Even if $G$ is abelian, $g x g y = g^{2} x y$ and $g^{2} = g$ only if $g = e$ . See The unique property of the identity of a group.