4 - Examples of Groups

Notice that except the identity permutation $k_0$ , every other permutation, can be written as the composition of two independent swaps. Let $(i,j)$ denote the permutation $i\mapsto j,j\mapsto i$ And leaves everything unchanged. Then, $k_1=(1,3)(2,4)$ , $k_2=(1,2)(3,4)$ and, $k_3=(1,4)(2,3)$.
Furthermore, notice that $k_1^2=k_2^2=k_3^2=k_0$. and $k_1{k}_2=k_2{k}_1=k_3$ and so on...

The set $K_4=\{e,a,b,c\}$ with $a^2=b^2=c^2=e$ and $ab=ba=c$, $ac=ca=b$, $bc=cb=a$ as the products is called the Klein-Four group.

Below, we depict two necklaces, one on a rigid triangular wire, and one on a rigid square wire. We can imagine distinctly colored beads on each vertex. The question then is, how can we manipulate the necklace, so that when you place it on the table and take a picture, it looks different in the arrangement of beads?
call the "do nothing" transformation $e=r_0$. Then, call the counter-clockwise rotation by $2\pi /n$ (for a regular $n$-gon), $r_1$.
Then every rotation of the necklace is given by $r_0,r_1,\dots ,r_{n-1}$. Where $r_n=r_0$.

Now notice that you've got a "front" of the necklace and "back" of the necklace. And to get to the back of the necklace from the front, you would reflect it about an axis, say $S_0$ call that $s_0$.
notice that two reflections about the same axis is equivalent to "doing nothing". $s_i^2=r_0$.

And by observing this picture more, one can work out all the other products. Two reflections should be a rotation, because you come back to the "front" of the necklace, and so on...

A rotation, then a reflection leaves you in the "back" of the necklace, so it should be equivalent to a reflection about some axis...
![[diagram-20240712 (4).svg#invert_B]]

The set $D_n=\{r_0,r_1,\dots r_{n-1},s_0,s_1,\dots s_{n-1}\}$ with the following products:

• $r_i{r}_j=r_{i+1}$
• $s_i{s}_j=r_{i-j}$
• $s_i{r}_j=s_{i-j}$
• $r_i{s}_j=s_{i+j}$

The set $\mathbb{R}$ with addition $+$ forms an abelian Group.
The set $\mathbb{R}-0$ with multiplication $\cdot$ forms an abelian group as well.

The map $x\mapsto e^x$ is the desired isomorphism from $(\mathbb{R},+)$ to $(\mathbb{R}-0,\cdot )$

In the complex numbers $\mathbb{C}$, the unit circle $S^1=\{e^{i\theta }:\theta \in \mathbb{R}\}$ is an abelian group with complex multiplication.

$\theta \mapsto e^{i\theta }$ is the desired homomorphism from $(\mathbb{R},+)$ to $S^1$. if $\theta$ is restricted to $[0,2\pi )$, then the given map is an isomorphism.