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.

:::

• Assuming all groups of a given order are abelian: While groups of order $p$ and $p^2$ are always abelian, groups of order $p^3$ need not be (e.g., $D_4$ and $Q_8$ are non-abelian of order 8). Never assume abelianness without proof.
• Forgetting to check semidirect product distinctness: Different homomorphisms $\phi: K \to \operatorname{Aut}(N)$ can produce isomorphic semidirect products. Always verify that two actions actually yield non-isomorphic groups.
• Misapplying Sylow’s theorems when $p$ divides the group order only once: For $|G| = pq$ with $p < q$, the condition $n_p \equiv 1 \pmod{p}$ and $n_p \mid q$ forces $n_p = 1$ when $p \nmid (q-1)$, but $n_p$ may equal $q$ when $p \mid (q-1)$.
• Confusing direct products with semidirect products: In a direct product $N \times H$, both factors are normal; in a semidirect product $N \rtimes H$, only $N$ is guaranteed normal. The notation $N \rtimes H$ always requires specifying the action.

• Prime order implies cyclic: By Lagrange’s theorem, a group of prime order $p$ has no proper subgroups, so every non-identity element generates the entire group.
• Order $p^2$ is always abelian: If $|G| = p^2$ for prime $p$, then $Z(G) \neq \{e\}$ by Theorem 6.5, and $G/Z(G)$ is cyclic, forcing $G$ to be abelian.
• Sylow theorems constrain group structure: The number of Sylow $p$-subgroups $n_p$ must divide the group order and satisfy $n_p \equiv 1 \pmod{p}$, severely limiting possible group structures.
• Non-abelian groups of order $pq$ require $p \mid (q-1)$: For $|G| = pq$ with $p < q$ primes, $G$ is cyclic unless $p$ divides $q-1$, in which case a non-abelian semidirect product exists.
• The classification table grows rapidly: While there are only 5 groups of order 8 and 5 of order 12, the number jumps to 14 groups of order 16 and 51 groups of order 32.

Problem: Classify all groups of order 6.

Solution: $|G| = 6 = 2 \times 3$. By Sylow theorems, $n_3 \mid 2$ and $n_3 \equiv 1 \pmod{3}$, so $n_3 = 1$. Thus the Sylow 3-subgroup $P_3 \cong \mathbb{Z}/3\mathbb{Z}$ is normal. Similarly, $n_2 \mid 3$ and $n_2 \equiv 1 \pmod{2}$, so $n_2 = 1$ or $3$. If $n_2 = 1$, then both Sylow subgroups are normal, giving $G \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}/6\mathbb{Z}$. If $n_2 = 3$, the action of $\mathbb{Z}/3\mathbb{Z}$ on the three Sylow 2-subgroups gives a non-abelian group, which is $S_3 \cong D_3$.

Problem: Show that the quaternion group $Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$ is the unique non-abelian group of order 8 in which every element has order dividing 4.

Solution: In $Q_8$, we have $i^2 = j^2 = k^2 = -1$ and $(-1)^2 = 1$, so every non-identity element has order 2 or 4. The element $-1$ is central, so $|Z(Q_8)| = 2$. Since $Q_8$ is non-abelian, by Proposition 16.2 it must be either $D_4$ or $Q_8$. In $D_4$, there are elements of order 2 outside the centre (reflections), but $Q_8$ has no such elements. Thus $Q_8$ is uniquely determined by this property.

Beyond order 12, the classification continues to grow in complexity:

• Order 14: Only two groups: $\mathbb{Z}/14\mathbb{Z}$ and $D_7$ (since $2 \mid (7-1)$, a non-abelian semidirect product exists).
• Order 15: Only one group: $\mathbb{Z}/15\mathbb{Z}$ (cyclic, since $3 \nmid (5-1)$ and $5 \nmid (3-1)$).
• Order 18: Five groups: two abelian ($\mathbb{Z}/18\mathbb{Z}$, $\mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$) and three non-abelian ($D_9$, $S_3 \times \mathbb{Z}/3\mathbb{Z}$, $(\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}$).
• Order 20: Five groups: $\mathbb{Z}/20\mathbb{Z}$, $\mathbb{Z}/10\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, $D_{10}$, the dicyclic group $\operatorname{Dic}_5$, and $\mathbb{Z}/5\mathbb{Z} \rtimes \mathbb{Z}/4\mathbb{Z}$.

• Cryptography: Understanding group structure is essential for elliptic curve cryptography, where the group of points on a curve must have suitable properties.
• Crystallography: The 230 space groups describe all possible crystal symmetries, built from small-order groups acting on lattices.
• Particle physics: The Standard Model is based on the gauge group $SU(3) \times SU(2) \times U(1)$, whose finite subgroups classify possible symmetry-breaking patterns.