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> [!theorem] Theorem. ([[Cauchy's Theorem]])
> For $G$ a finite [[group]] and $p$ a [[prime number]] such that $p \b | \ |G|$, we have that $G$ contains an element of order $p$.
> [!basicexample]
> For $D_{3}$, $\tau$ is such an element.
> [!proof] Lemma 1. (Proof for when $G$ is [[cyclic group|cyclic]])
> Suppose $G$ is a [[cyclic group]] of [[order of a group|order ]] $n$; write $G=\langle x \rangle$. Let $p$ be a [[prime number]] such that $p$ [[divides]] $n$, say, $p=nk$ for some $k \in \mathbb{N}$. Then $(x^{k})^{p}=x^{kp}=x^{n}=e$, implying that $x^{k}$ has order $p$.
> [!proof] Lemma 2. (Proof for when $G$ is an [[abelian group]])
>
>
Suppose $G$ is an [[abelian group]] of order $n$. Recall that [[subgroups of abelian groups are normal]], this is important because it means we are able to [[quotient group|quotient]] by any [[subgroup]] of $G$. Proceed by induction.
The base case is immediate, for when $n=2$ we have $G \cong C_{2}$. For the inductive step, suppose that the [[Cauchy's Theorem|theorem]] holds for [[abelian group]]s of order $k < n$.
Now, with $n$ arbitrary, start by considering any $x \in G$, $x \neq e$, and let $H=\langle x \rangle$. If $p \b | \ |H|$, then since $H$ is [[cyclic group|cyclic]] it contains an element of [[order of a group|order]] $p$ by **lemma 1**; hence $G$ does too and we are done. If $p \not{\b |} \ |H |$, then we move to consider the [[quotient group|quotient]] $G / H$. [[order of quotient group is quotient of orders|Since]] $|G / H| \cdot |H|=|G|,$
we have that $p \b | \ |G / H|$ by [[Euclid's Lemma]]. So by the inductive hypothesis, $|G / H|$ has an element $\overline{y}$ of order $p$. Since the [[kernel iff normal subgroup|The universal projection homomorphism]] $\phi:G \to G / H$ is a [[surjection]] there exists $y \in G$ such that $\phi(y)=\overline{y}$. By [[group homomorphisms preserve structure|order of subgroup image under homomorphism divides order of group]] we have $|\overline{y}| \ \b | \ |y|$, i.e., $p \b |\ |y|$. So $|y|=pk$ for some $k \in \mathbb{Z}$. If the [[subgroup]] $\langle y \rangle$ equals $G$, we're done (because then $G$ is [[cyclic group|cyclic]]), so assume it has order less than $|G|$. In this case we can imply the inductive hypothesis to conclude that $\langle y \rangle$ (and thus $G$) has an element of order $p$.
> [!proof]- Proof. ([[Cauchy's Theorem]])
Let $G$ be a [[group]] and $p$ a [[prime number]] for which $p \b | \ |G|$. Consider the [[center of a group|center]] of $G$, $Z(G)$. Since $Z(G)$ is an [[abelian group]], if $p \b | |Z(G)|$ then by **lemma 2** $Z(G)$ contains an element of [[order of a group|order]] $p$ and therefore $G$ does too— so we're done.
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Thus, suppose $p \not{|} |Z(G)|$. Then, since in general the [[summation of multiples is a multiple]] $p$ must not [[divides|divide]] the size of every [[conjugate|conjugacy class]]; i.e., there exists a [[conjugate|conjugacy class]] of a non-central element $x \notin Z(G)$ whose size is not divisible by $p$. The [[class equation]] says that size is precisely $\frac{|G|}{|Z_{G}(x)|}$. Thus we can say there exists $x \notin Z(G)$ for which $p \not{|} \frac{|G|}{| Z_{G}(x)|}$. But this means $p \b | \ |Z_{G}(x)|$ [^1]. Using the induction hypothesis we may then conclude $Z_{G}(x)$ contains an element of order $p$ and therefore $G$ does as well.
[^1]: To see this, use that $p | |G|$ to write $p \not{|} \ (p\frac{k}{|Z_{G}(x)|})$. The only way this is true is for $Z_{G}(x)$ to equal $pq$ for some $q$, since that'll make the $p