Additional Results

Theorem 14.1 (Cauchy’s Theorem). If $p$ is a prime dividing $|G|$Then $G$ has an element of Order $p$.

Proof. Consider the set $X = \{(g_1, g_2, \ldots, g_p) \in G^p : g_1 g_2 \cdots g_p = e\}$. $|X| = |G|^{p-1}$ (choose $g_1, \ldots, g_{p-1}$ freely; $g_p$ is determined). The cyclic group $\mathbb{Z}/p\mathbb{Z}$ acts on $X$ by cyclic permutation. Orbits have size $1$ or $p$. An orbit has size $1$ precisely when $(g, g, \ldots, g) \in X$I.e., $g^p = e$. Since $|X| = |G|^{p-1}$ is divisible by $p$ (as $p$ divides $|G|$), the number of fixed points Is congruent to $0 \pmod{p}$. The element $(e, e, \ldots, e)$ is a fixed point, so there exists At least $p - 1$ other fixed points, giving a non-identity element with $g^p = e$. Since $p$ is Prime, $g$ has order $p$. $\blacksquare$

Problem. Use Cauchy’s theorem to show that every group of order $6$ is isomorphic to either $\mathbb{Z}/6\mathbb{Z}$ or $S_3$.

Problem. Classify all groups of order $4$.

A group $G$ is simple if its only normal subgroups are $\{e\}$ and $G$.

Proposition 14.2. $A_n$ is simple for all $n \geq 5$.

This is a key result in the classification of finite simple groups, which states that every finite Simple group is either cyclic of prime order, an alternating group $A_n$ ($n \geq 5$), a group of Lie type, or one of 26 sporadic groups.

Theorem 14.4. Every finitely generated abelian group $G$ is isomorphic to a direct product of Cyclic groups:

$G \cong \mathbb{Z}^r \times \mathbb{Z}/p_1^{k_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_m^{k_m}\mathbb{Z}$

Where $r \geq 0$ is the rank and $p_i^{k_i}$ are powers of (not necessarily distinct) primes. The integers $r, k_1, \ldots, k_m$ are uniquely determined.

Problem. Classify all abelian groups of order 72.

Solution. Since $72 = 2^3 \cdot 3^2$Every abelian group of order 72 is a direct product of an Abelian group of order $2^3$ and one of order $3^2$.

For order $2^3$: the partitions of 3 give (3), (2,1), (1,1,1), corresponding to $\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

For order $3^2$: the partitions of 2 give (2), (1,1), corresponding to $\mathbb{Z}/9\mathbb{Z}$ $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.

Taking all products, the six abelian groups of order 72 are:

1. $\mathbb{Z}/72\mathbb{Z} \cong \mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z}$
2. $\mathbb{Z}/8\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$
3. $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z}$
4. $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$
5. $(\mathbb{Z}/2\mathbb{Z})^3 \times \mathbb{Z}/9\mathbb{Z}$
6. $(\mathbb{Z}/2\mathbb{Z})^3 \times (\mathbb{Z}/3\mathbb{Z})^2$ $\blacksquare$

• Applying Cauchy’s theorem backwards: $|G|$ divisible by $p$ does not imply $G$ has a normal subgroup of order $p$; only a subgroup.
• Confusing Cauchy’s theorem with Sylow’s theorems: Cauchy gives existence of a single element, while Sylow gives existence of subgroups of maximal prime-power order.
• Forgetting that the Structure Theorem requires the group to be both finitely generated and abelian.
• Assuming the classification of finite simple groups applies to infinite groups.
• Mixing up the partitions of exponents when applying the Structure Theorem (e.g., $\mathbb{Z}/8\mathbb{Z}$ vs $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ are both order 8 but non-isomorphic).