15 Cosets and Usage of Lagrange's Theorem

Proof

Let $x\in G$ and $x\ne 1$. Thus $|⟨x⟩|>1$ and $|⟨x⟩|$ divides $|G|$ by Finite Groups, Order of x divides Order of G. Since $|G|$ is prime then we must have $|⟨x⟩|=|G|$. Hence $G=⟨x⟩$ is cyclic using any non-identity element as the generator. Using Cyclic Groups of the Same Order are Isomorphic then $G\cong Z_p$.

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Here we define $\phi :G\to \frac{\mathbb{Z}}{p\mathbb{Z}}$ where $g\mapsto \bar{1}$ and $g^k\to \bar{k}$ is an Isomorphism as a result.

Examples

Let $G=S_3$ where $K=⟨(123)⟩=\{1,(123),(132)\}⊴S_3$. Then $\frac{S_3}{K}$, using Lagrange's Divisibility Theorem of Order of Subgroups, is of order:
$\left|\frac{S_3}{K}\right|=\frac{|S_3|}{|K|}=\frac{6}{3}=2$
Then using our corollary above, we get that $\frac{S_3}{K}\cong \frac{\mathbb{Z}}{2\mathbb{Z}}$ since $2$ is prime.

Another example. Let $G=\frac{\mathbb{Z}}{4\mathbb{Z}}\times \frac{\mathbb{Z}}{8\mathbb{Z}}$ is an Abelian Group. Consider using the subgroup $K=⟨(1,2)⟩=\{(0,0),(1,2),(2,4),(3,6)\}$. Then:
$\left|\frac{\frac{\mathbb{Z}}{4\mathbb{Z}}\times \frac{\mathbb{Z}}{8\mathbb{Z}}}{K}\right|=\frac{32}{4}=8$
but $8$ is not prime, so sadly we can't use the theorem. But is it isomorphic to a familiar group? It was abelian since $G$ is abelian for this specific example. Thus, what if the following was an isomorphism:
$\phi :G\to \frac{\mathbb{Z}}{8\mathbb{Z}}:(1,0)\mapsto 6,(0,1)\mapsto 1$
this seems like an arbitrary thing to do! But let's roll with it and check it's first a Homomorphism (or Group Morphism). The key relations in the domain $G$ are:

1. $4(1,0)=(0,0)$ so then $4\cdot \bar{6}=\overline{24}=\bar{0}$
2. $8(0,1)=(0,0)$ so then $8\cdot \bar{1}=\bar{8}=\bar{0}$
3. $(1,0)+(0,1)=(0,1)+(1,0)$ (so elements commute) so then $6+1=1+6$

This makes the general formula for $\phi$ is:
$\phi (a,b)=a\phi (1,0)+b\phi (0,1)=a\cdot \bar{6}+b\cdot \bar{1}=\overline{6a+b}$
since any $(a,b)=a(1,0)+b(0,1)$.

We've actually got our morphism! It's also surjective since it hits all 8 outputs ($\bar{1}$ is in the image, which generates the rest). Lastly $\text{Ker}(\phi )=\{(a,b)\in G|\overline{6a+b}=\bar{0}\}$, so if $a=0,1,2,3$ then we can check the $b

Using it, we proved The First Isomorphism Theorem of Groups, The Fundamental Theorem for Group Morphisms:

In General ...

How do we handle general $\frac{G}{K}$? The answer is the observation that we have the projection morphism $\pi :G\to \frac{G}{K}$ where $\text{Ker}(\pi )=K$. But for the context of the observation, consider any morphism $\psi :\frac{G}{K}\to H$ instead. Then:

with $K\le \text{Ker}(\psi \circ \pi )$.

Say you wanted to go the other way. Suppose $\phi :G\to H$ is a Homomorphism (or Group Morphism) with $K\le \text{Ker}(\phi )$:

The claim is that the map $\psi :\frac{G}{K}\to H$ is given by $\phi =\psi \circ \pi$. There is a natural bijection:

$$\{\psi :\frac{G}{K}\to H\}\equiv \{\phi :G\to H|K\le \text{Ker}(\phi )\}$$

This will be the first instance of a "universal property of $\frac{G}{K}$". s that are in the kernel:

• For $a=0$ then $b=0$
• For $a=1$ then $\overline{6\cdot 1+b}=\bar{0}⟺b=\bar{2}$
• For $a=2$ then we need $b=\bar{4}$
• For $a=3$ then we need $\overline{6\cdot 3+b}=\bar{0}⟺b=\bar{6}$.
  Thus $\text{Ker}(\phi )=\{(0,0),(1,2),(2,4),(3,6)\}$ which was exactly $K$ (as we wanted)!

As a result, then using the The First Isomorphism Theorem of Groups, The Fundamental Theorem for Group Morphisms then:
$\frac{G}{K}=\frac{G}{\text{Ker}(\phi )}\cong \text{im}(\phi )=\frac{\mathbb{Z}}{8\mathbb{Z}}$