Additional Results

14.1 Cauchy”s Theorem

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 $p$ . $\blacksquare$

14.2 Worked Examples: Additional Results

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

Solution

Solution. Let $|G| = 6 = 2 \cdot 3$ . By Cauchy’s theorem, $G$ has an element $a$ of order $2$ And an element $b$ of order $3$ .

The subgroup $H = \langle b \rangle$ has index $2$ So $H \trianglelefteq G$ (Corollary 3.7). The quotient $G/H$ has order $2$ .

Since $a \notin H$ (as $|a| = 2$ and $|b| = 3$ ), every element of $G$ is either $b^k$ or $ab^k$ . The group structure is determined by $aba^{-1}$ . Since $H$ is normal, $aba^{-1} \in H$ So $aba^{-1} = b$ or $aba^{-1} = b^2$ .

Case 1: $aba^{-1} = b$ (i.e., $a$ and $b$ commute). Then $G \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \cong \mathbb{Z}/6\mathbb{Z}$ .

Case 2: $aba^{-1} = b^2 = b^{-1}$ . Then $G$ is a semidirect product with $ab = b^{-1}a$ . This is the presentation $\langle a, b \mid a^2 = b^3 = e,\ aba^{-1} = b^{-1} \rangle$ Which is $S_3$ . $\blacksquare$

Problem. Classify all groups of order $4$ .

Solution

Solution. Let $|G| = 4$ . By Lagrange, possible element orders are $1, 2, 4$ .

Case 1: $G$ has an element of order $4$ . Then $G = \langle g \rangle \cong \mathbb{Z}/4\mathbb{Z}$ .

Case 2: Every non-identity element has order $2$ . Let $a, b \in G$ with $a \neq b$ and $a, b \neq e$ . Then $G = \{e, a, b, ab\}$ (there are only $4$ elements). We have $a^2 = b^2 = (ab)^2 = e$ . From $(ab)^2 = e$ : $abab = e$ So $ba = a^{-1}b^{-1} = ab$ (since $a^{-1} = a$ and $b^{-1} = b$ ). Thus $G$ is abelian: $G \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} = V_4$ .

So there are exactly two groups of order $4$ : $\mathbb{Z}/4\mathbb{Z}$ and $V_4$ . $\blacksquare$

14.3 Simple Groups

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.

14.4 The Structure Theorem for Finitely Generated Abelian 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.

14.5 Worked Example

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:

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

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