Cauchy’s Theorem for Finite Abelian Groups

Note that this article assumes that the reader is comfortable with normal subgroups and quotient or factor groups.

Today’s article will be relatively short. However, I hope that it is both enjoyable and useful to you! Today, we will prove Cauchy’s theorem for finite Abelian groups. In fact, this version of Cauchy’s theorem can be generalized for all finite groups; however, we will not prove that case today. I find this proof to be really fun and clever. You jump between a group and a quotient group to prove the statement! This particular proof is the first one that I learned that used this tactic and, therefore, holds a special place in my heart.

A Lovely Proof

Theorem (Cauchy’s Theorem ): Let $G$ be a finite, Abelian group with $\#G = n.$ Then, for all prime numbers $p$ such that $p\mid n,$ there is some $x\in G$ such that $|x| = p.$

Remark 1: We will be using the notation: $\#G = n$ to mean that the order of $G$ is equal to $n.$ In other sources, you will see the notation $|G| = n.$ However, in the proof we are interested in divisors of the order of groups and to avoid $p\mid |G|,$ I like to write $p \mid \#G$ better.

Remark 2: We will also be using the standard notation $|x| = k$ to mean that the order of $x$ equals $k.$ That is $k$ is the smallest positive integer such that $q^k =e.$ Where $e\in G$ is the identity element.

Example: Consider an Abelian group of order 15. Then, there is some element with order 3 and some other element with order 5. A concrete example is $G = \Z/15\Z.$ In this case, $3\in Z/15\Z$ has order 5, and $5\in Z/15\Z$ has order 3. I know, it’s tricky to keep track!

$\mathrm{in}\;\;\; Z/15\Z \\\;\;\;\;\; |3| = 5 \\ \;\;\;\;\;|5| = 3.$

As mentioned, the proof will jump between using $G$ and $G/N$ for some normal subgroup $N \trianglelefteq G$ . Note that the symbol $“ \trianglelefteq“$ means “is a normal subgroup of”.

Proof: (Click in the Discovery)

We will proceed by strong induction on $\#G =n.$

Base Case: $\#G = 1.$ In this case, Cauchy’s Theorem is trivially true. That is, there are no primes that divide $\#G.$

Induction Hypothesis: Assume that Cauchy’s theorem is true for all groups with order less than $n.$

Induction Step: Let $\#G = n$ and let $p\mid n.$ We claim that there is at least one element in $G$ with prime order. We cannot yet guarantee that the prime is equal to $p,$ but we can guarantee that there is always an element with its order equal to a prime. Indeed, let $g\in G$ have order $m.$ It follows that $m = qk$ for some prime $q$ and some $k\in \N.$ Next, let $x = g^k$ and observe that

x^q = (g^k)^q = g^{m} = e

which implies that the order of $x = g^k$ is less than or equal to $q.$ However, if the order was $t<q,$ then ${kt}<m$ and

g^{kt} = (g^k)^t = x^t = e

contradicting that the order of $g$ is equal to $m.$ Furthermore, we know that $q\mid n$ by Lagrange’s theorem.

Now that we know there is an element $x\in G$ with prime order $q,$ we might get lucky and have $p=q,$ in which case we’d be done! So, let’s suppose that $p\neq q$ and consider the cyclic group generated by $x.$ We know that $\langle x\rangle \trianglelefteq G$ is a normal subgroup since all subgroups of Abelian groups are normal. Therefore, the quotient group $G/\langle x\rangle$ is indeed a group with a well-defined binary operation. And, we also know by Lagrange’s theorem that the order of $G/\langle x\rangle$ is $n/q.$

Observe that $p\mid (n/q)$ and thus we have $p\mid \#(G/\langle x\rangle).$ Furthermore, we know that $n/q<n$ and therefore we can use our induction hypothesis on $G/\langle x\rangle.$ (How clever is that?) Thus, there is some coset $y\langle x\rangle \in G/\langle x\rangle$ with order $p.$ I.e., $(y\langle x\rangle )^p = \langle {x}\rangle .$ We will use this coset to find an element in $G$ with order $p.$

Let $|y| = \ell.$ It follows that $(y\langle{x}\rangle)^{\ell} = \langle{x}\rangle.$ Therefore, $p\mid \ell$ since we know that the order of $y\langle x\rangle \in G/\langle x\rangle$ is equal to $p.$ This means there is some $t\in \N$ such that $\ell = pt$ and it follows that $(y^t)^p = e.$ We claim that the order of $y^t$ is $p,$ that is, $|y^t| = p.$ If not, then the order of $y^t$ must be less than $p.$ Suppose $|y^t|=s<p.$ Then we have $(y^t)^s = e$ where $st<pt=\ell$ a contradiction. Thus, $y^t\in G$ has order $p$ as desired.

$\square$

General Cauchy

Theorem (Cauchy’s Theorem): Let $G$ be a finite group with $\#G = n.$ Then, for all prime numbers $p$ such that $p\mid n,$ there is some $x\in G$ such that $|x| = p.$

The proof of Cauchy’s theorem usually uses what is known as the class equation. And this is something that I would like to dedicate its own article to. So for today, we will have to be satisfied with the case where $G$ is Abelian. In my humble opinion, the proof given above is WAY better than the proof of the full version of Cauchy. And, in fact, the proof for the full version of Cauchy’s theorem uses the finite Abelian version in a crucial way! So what we covered today is one step toward understanding the proof of Cauchy’s full theorem.

I hope that you had some fun today, and maybe even learned something useful! In either case, let me know in the comments where I could have done better in my explanation of the proof!

Be Kind. Be Curious. Be Compassionate. Be Creative.

And Have Fun!

Footnotes:

A Kick in the Discovery