Prime Order Groups are Cyclic and Z_p

Corollary

If $G$ is a Group of prime order $p$ , then $G$ is a Cyclic Group. Hence $G ≅ Z_{p}$ .

Proof

Let $x \in G$ and $x \neq 1$ . 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 $G ≅ Z_{p}$ .

☐

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

Examples

Let $G = S_{3}$ where $K = ⟨ (1 2 3) ⟩ = {1, (1 2 3), (1 3 2)} ⊴ S_{3}$ . Then $\frac{S_{3}}{K}$ , using Lagrange's Divisibility Theorem of Order of Subgroups, is of order:

| \frac{S_{3}}{K} | = \frac{| S_{3} |}{| K |} = \frac{6}{3} = 2

Then using our corollary above, we get that $\frac{S_{3}}{K} ≅ \frac{Z}{2 Z}$ since $2$ is prime.

Another example. Let $G = \frac{Z}{4 Z} \times \frac{Z}{8 Z}$ is an Abelian Group. Consider using the subgroup $K = ⟨ (1, 2) ⟩ = {(0, 0), (1, 2), (2, 4), (3, 6)}$ . Then:

| \frac{\frac{Z}{4 Z} \times \frac{Z}{8 Z}}{K} | = \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:

φ : G \to \frac{Z}{8 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:

$4 (1, 0) = (0, 0)$ so then $4 \cdot \overset{―}{6} = \overset{―}{24} = \overset{―}{0}$
$8 (0, 1) = (0, 0)$ so then $8 \cdot \overset{―}{1} = \overset{―}{8} = \overset{―}{0}$
$(1, 0) + (0, 1) = (0, 1) + (1, 0)$ (so elements commute) so then $6 + 1 = 1 + 6$

Note

I've added $\overset{―}{}$ to all things that are mod 8 for this example.

This makes the general formula for $φ$ is:

φ (a, b) = a φ (1, 0) + b φ (0, 1) = a \cdot \overset{―}{6} + b \cdot \overset{―}{1} = \overset{―}{6 a + 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 ( $\overset{―}{1}$ is in the image, which generates the rest). Lastly $Ker (φ) = {(a, b) \in G | \overset{―}{6 a + b} = \overset{―}{0}}$ , so if $a = 0, 1, 2, 3$ then we can check the $b$ 's that are in the kernel:

For $a = 0$ then $b = 0$
For $a = 1$ then $\overset{―}{6 \cdot 1 + b} = \overset{―}{0} ⟺ b = \overset{―}{2}$
For $a = 2$ then we need $b = \overset{―}{4}$
For $a = 3$ then we need $\overset{―}{6 \cdot 3 + b} = \overset{―}{0} ⟺ b = \overset{―}{6}$ .
Thus $Ker (φ) = {(0, 0), (1, 2), (2, 4), (3, 6)}$ 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}{Ker (φ)} ≅ im (φ) = \frac{Z}{8 Z}