Classification of Groups of Small Order

The following table summarizes the classification of groups of small order:

Proposition 16.1. Every group of prime order is cyclic, and every group of order $p^2$ (where $p$ Is prime) is abelian (hence isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$).

Proof. For prime order, see Section 3.3. For order $p^2$See Corollary 6.6. $\blacksquare$

Proposition 16.2. There are exactly five groups of order $8$.

Proof. The abelian groups of order $8$ are classified by the structure theorem: $\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, $(\mathbb{Z}/2\mathbb{Z})^3$.

For non-abelian groups of order $8$: by Theorem 6.5, $|Z(G)| \geq 2$. If $|Z(G)| = 4$ Then $G/Z(G)$ has order $2$ and is cyclic, making $G$ abelian (contradiction). So $|Z(G)| = 2$ and $G/Z(G) \cong V_4$.

If every non-central element has order $2$: $G$ is generated by three involutions Commuting with each other and with $Z(G)$Giving $(\mathbb{Z}/2\mathbb{Z})^3$ (abelian). So some non-central element has order $4$Say $a$ with $a^4 = e$ and $a^2 \in Z(G)$.

Pick $b \notin \langle a, Z(G) \rangle$. Then $bab^{-1} = a$ or $bab^{-1} = a^{-1}$. If $bab^{-1} = a$: $G$ is abelian, contradiction. If $bab^{-1} = a^{-1}$: we get $D_4$ when $b^2 = e$ and $Q_8$ when $b^2 = a^2$. These are the only two non-abelian groups of order $8$. $\blacksquare$

Proposition 16.3. There are exactly five groups of order $12$.

Proof sketch. $|G| = 12 = 2^2 \cdot 3$.

$n_3 \equiv 1 \pmod{3}$ and $n_3$ divides $4$So $n_3 = 1$ or $4$.

$n_3 = 1$: The Sylow $3$-subgroup $P_3 \cong \mathbb{Z}/3\mathbb{Z}$ is normal. $G$ is a semidirect product $\mathbb{Z}/3\mathbb{Z} \rtimes K$ where $K$ is a Sylow $2$-subgroup ($\mathbb{Z}/4\mathbb{Z}$ or $V_4$). Computing the possible actions gives: $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z} \cong \mathbb{Z}/12\mathbb{Z}$ $\mathbb{Z}/3\mathbb{Z} \times V_4 \cong \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ $\mathbb{Z}/3\mathbb{Z} \rtimes V_4 \cong D_6$And $\mathbb{Z}/3\mathbb{Z} \rtimes \mathbb{Z}/4\mathbb{Z}$ (the dicyclic group of order $12$).

$n_3 = 4$: The Sylow $3$-subgroup is not normal. There are four Sylow $3$-subgroups, Contributing $4 \cdot 2 = 8$ elements of order $3$. The remaining $4$ elements (plus $e$) form The unique Sylow $2$-subgroup, which must be $V_4$ (since $D_4$ has order $8$ and $\mathbb{Z}/4\mathbb{Z}$ Has no element of order $2$ besides its unique subgroup… Actually, the argument is more subtle). This gives $A_4$.

Total: five groups of order $12$. $\blacksquare$

:::caution Common Pitfall The number of groups grows rapidly with the order. While there are exactly $5$ groups of order $8$ There are $14$ groups of order $16$ and $267$ groups of order $64$. Classification by hand is only Feasible for small orders. For prime-squared orders, the abelian classification is straightforward, But non-abelian cases require careful analysis of possible semidirect products.

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