Cauchy's Theorem for Groups

Theorem 1 (Cauchy's Theorem for Groups): Let $G$ be a finite group of order $n$. If $p$ is prime such that $p \mid n$ then there exists an element $g \in G$ such that the order of $g$ is $p$.

• Proof: Let $P(n)$ be the statement that if $G$ is a finite group of order $n$ and if $p$ is a prime such that $p \mid n$ then $G$ contains an element of order $p$.

• Base Step: $P(1)$ is vacuously true, since if $|G| = 1$, no prime $p$ divides $1$. Observe that $P(2)$ is also true, since if $|G| = 2$ then $G \cong \mathbb{Z}/2\mathbb{Z}$. So if $p$ is a prime such that $p \mid 2$ then $p = 2$. Certainly, $\mathbb{Z}/2\mathbb{Z}$ has an element of order $2$, namely $1 \in \mathbb{Z}/2\mathbb{Z}$ since $2 \cdot 1 = 2 \equiv 0 \pmod 2$.

• Induction Step: Suppose that for some $n > 1$, $P(1)$, $P(2)$, …, $P(n-1)$ are all true. We want to show that $P(n)$ is true. Let $G$ be a group of order $n$ and suppose that $p$ is a prime such that $p \mid n$.

• Consider the class equation for $G$:

(1)

\begin{align} \quad n = |G| = |Z(G)| + \sum [G:C_G(g)] \end{align}

• Where the sum runs over one representative for each nontrivial conjugacy class. We break the proof up into cases.

• Case 1: Suppose that $p$ divides each term in $\displaystyle{\sum [G:C_G(g)]}$. Then $\displaystyle{p \mid \sum [G:C_G(g)]}$. Since $p \mid n$ and $\displaystyle{p \mid \sum [G:C_G(g)]}$ we have by the class equation that $p \mid |Z(G)|$.

• Case 1.1: Suppose that $Z(G) \neq G$. Then $|Z(G)| < |G| = n$. By the induction hypothesis, since $p$ is a prime such that $p \mid |Z(G)|$ there exists an element $g \in Z(G)$ such that $\mathrm{ord}(g) = p$. But then $g \in G$ is such that $\mathrm{ord}(g) = p$. So $P(n)$ is true.

• Case 1.2: Suppose that $Z(G) = G$. Then $G$ is an abelian group. Let $g \in G$ with $g \neq e$ (where $e$ denotes the identity element). Consider the subgroup generated by $g$, $\langle g \rangle$. Note that $\langle g \rangle$ is a normal subgroup of $G$ since $G$ is abelian.

• Case 1.2.1: Suppose that $G = \langle g \rangle$. Then $G$ is a cyclic group of order $n$. Since $p \mid n$, we see that $g^{n/p} \in G$ with $\mathrm{ord}(g^{n/p}) = p$. So $P(n)$ is true.

• Case 1.2.2: Suppose that $G \neq \langle g \rangle$. Then $|\langle g \rangle | < |G| = n$.

• Case 1.2.2.1: Suppose that $p \mid |\langle g \rangle| < n$. Then by the induction hypothesis there exists an $h \in \langle g \rangle$ with $\mathrm{ord}(h) = p$. But then $h \in G$ is such that $\mathrm{ord}(h) = p$, so $P(n)$ is true.

• Case 1.2.2.2: Suppose that $p \nmid |\langle g \rangle| < n$. By Lagrange's Theorem we have that:

(2)

\begin{align} \quad |G| &= |\langle g \rangle| [G : \langle g \rangle] \\ &= |\langle g \rangle| | G/\langle g \rangle| \end{align}

• Since $p \mid |G|$ and $p \nmid |\langle g \rangle|$ we conclude that $p \mid |G/\langle g \rangle|$ (This is simply the theorem that states that if $p$ is a prime and $p \mid ab$ then $p$ divides at least one of $a$ or $b$). Since $g \neq e$ we see that $|\langle g \rangle| > 1$. Thus:

(3)

\begin{align} \quad |G/\langle g \rangle| = |G|/|\langle g \rangle| < n \end{align}

• Since $p \mid |G/\langle g \rangle|$ and $|G/\langle g \rangle| < n$, by the induction hypothesis there exists an element of $G/\langle g \rangle$ of order $p$. Let $a \langle g \rangle \in G/\langle g \rangle$ be such that $\mathrm{ord}(a \langle g \rangle) = p$. So by definition, $p$ is the smallest positive integer such that:

(4)

\begin{align} \quad (a \langle g \rangle)^p = \langle g \rangle \end{align}

• Thus $a^p \in \langle g \rangle$. Let $k = |\langle g \rangle|$. Let $c = a^k$. We claim that $c$ has order $p$. Note that since $a^p \in \langle g \rangle$ and $k = \langle g \rangle$ we have that:

(5)

\begin{align} \quad c^p = (a^k)^p = a^{kp} = a^{pk} = (a^p)^k = (a^p)^{|\langle g \rangle|} = e \end{align}

• So $\mathrm{ord}(c) \leq p$. Note that $c \neq e$, otherwise if $c = e$ then $a^k = e$. Then $(a\langle g \rangle)^k = \langle g \rangle$. Since $\mathrm{ord}(a \langle g \rangle) = p$ we see that $p \mid k$. This cannot happen since by assumption, $p \nmid |\langle g \rangle|$. So indeed, $c \neq e$.

• So if $\mathrm{ord}(c) = j$ then $j \neq 1$ and $j \mid p$, i.e., $j = p$. Thus $c \in G$ is such that $\mathrm{ord}(c) = p$. So $P(n)$ is true.

• Case 2: Suppose that $p$ does not divide some term in the sum $\sum [G:C_G(g)]$. Then there exists a $g \in Z(G)$ such that $p \nmid [G:C_G(g)]$. By Lagrange's theorem we have that:

(6)

\begin{align} \quad n = |G| = |C_G(g)|[G:C_G(g)] \end{align}

• Since $p \mid n$ and $p \nmid [G:C_G(g)]$ we must have that $p \mid |C_G(g)|$. Since $g \not \in Z(G)$ we have that $[G:C_G(g)] \neq 1$. Thus $|C_G(g)| < |G| = n$. Since $p \mid |C_G(g)|$ and since $|C_G(g)| < n$, we have by the induction hypothesis that there exists an $h \in C_G(g)$ such that $\mathrm{ord}(h) = p$. But then $h \in G$ is such that $\mathrm{ord}(h) = p$. So $P(n)$ is true.

• Conclusion: By the principle of mathematical induction, $P(n)$ is true for all $n \in \mathbb{N}$, i.e., if $G$ is a group of order $n$ and $p \mid n$ then $G$ contains an element of order $p$. $\blacksquare$