classification of groups of order p

Let \(G\) be a group of group order \(p\) for a prime number \(p\).
Then \(G\) is isomorphic to \(\mathbb{Z}/p \mathbb{Z}\)

Let \(g \in G\) with \(g \neq 1 = e\).
Then by order of an element divides order we conclude, that

\begin{align*} \mathrm{ord}(g) = p \end{align*}

for \(g\) since \(g^1 = g \neq 0\) and order of an element divides order.

Now consider the group homomorphism

\begin{align*} f: \mathbb{Z} \rightarrow G n \mapsto g^{n} \end{align*}

(cf. Integers is the free abelian group)

Now the fundamental theorem shows that \(G \simeq \mathbb{Z}/\mathrm{ker}(f)\) and after characterizing subgroups of \(\mathbb{Z}\) as \(n \mathbb{Z}\) this shows that \(G \simeq \mathbb{Z}/p \mathbb{Z}\) (where \(n = p\) since otherwise we would have different cardinalities)

For the classification of subgroups of \(\mathbb{Z}\), we may use the euklidischer Algorithmus.