# Cauchy's Theorem for Abelian Groups

I am trying to verify the following proposition:

Let $G$ be a finite Abelian group and let $p$ be a prime that divides the order of $G$. Then $G$ has an element of order $p$.

My proof: By Lagrange's theorem: $x^{|G|}=e$. By assumption we have $kp=|G|$ for some prime $p$. So $e=x^{|G|}=x^{kp}$. Thus $|x^k|=p$. $\blacksquare$

The book's proof uses induction and cosets -- is this necessary?

For reference, here's the book proof:

Clearly, this statement is true for the case in which $G$ has ­order 2. We prove the theorem by using the Second Principle of Mathematical Induction on $|G|$. That is, we assume that the statement is true for all Abelian groups with fewer elements than G and use this assumption to show that the statement is true for G as well. Certainly, G has elements of prime order, for if $|x| = m$ and $m = qn$, where $q$ is prime, then $|x^n| = q$. So let $x$ be an element of $G$ of some prime order $q$, say. If $q=p$, we are finished; so assume that $q \neq p$. Since every subgroup of an Abelian group is normal, we may construct the factor group $\bar{G} = G/\langle x\rangle$. Then $\bar{G}$ is Abelian and $p$ divides $|G|$, since $|\bar{G}| = |G|/q$. By induction, then, $G$ has an element — call it $y\langle x\rangle$ — of order $p$. Then, $(y\langle x\rangle)^p = y^p\langle x\rangle = \langle x\rangle$ and therefore $y^p \in \langle x\rangle$. If $y^p = e$, we are done. If not, then $y^p$ has order $q$ and $y^q$ has order $p$. $\blacksquare$

• Google "McKay's Proof of Cauchy Theorem". Increidibly simple, short and elegant, and you don't need to do the abelian and the non-abelian cases separately. – DonAntonio Nov 29 '17 at 23:17
• But your "proof" does not seem to use the fact that $p$ is a prime number? So if your proof were correct, we would have shown that for any divisor $d$ of the group order, there exists an element of order $d$ (in the group). But that is false (do you know a counterexample?). So any valid proof must use in some way the fact that the divisor considered is prime. – Jeppe Stig Nielsen Nov 30 '17 at 13:09

Your proof is incorrect. You know that $(x^k)^p=e$ (for any $x\in G$), but this does not necessarily mean $x^k$ has order $p$. All it tells you is that the order of $x^k$ divides $p$, so it is either $1$ or $p$.
In fact, your approach cannot work without some major modification. It is possible that $x^k=e$ for all $x\in G$, so there is no element of the form $x^k$ which has order $p$. For instance, if $G=(\mathbb{Z}/p\mathbb{Z})^2$, then $|G|=p^2$ so $k=p$, but $x^p=e$ for all $x\in G$.
From $x^{kp}=e$, what you can deduce is that $(x^k)^p=e$, and therefore, that $|x^k|$ divides $p$; that is, that it is equal to $p$ or equal to $1$ (that is, $x^k=e$). How do you know that it is equal to $p$?
And where did you use that $G$ is Abelian?
Here is a simple proof due to Frobenius, I believe. Let $x_{1}, x_{2}, ..., x_{n}$ be the elements of the group $G$. Let $r_{i}$ be the order of the element $x_{i}$. Furthermore, let $$Z = \big( \mathbb{Z} / r_{1} \mathbb{Z}, + \big) \times \big( \mathbb{Z} / r_{2} \mathbb{Z}, + \big) \times \ldots \times \big( \mathbb{Z} / r_{n} \mathbb{Z}, + \big)$$ Now define a map $\varphi$ from $Z$ into $G$ by $\varphi(k_{1}, k_{2}, \ldots, k_{n}) = x_{1}^{k_{1}} x_{2}^{k_{2}} \dots x_{n}^{k_{n}}$. As $G$ is abelian it is easy to see that $\varphi$ is a homomorphism. Moreover $\varphi$ is surjective as $\varphi(0,0, \ldots, 1, \ldots, 0) = x_{i}$. If $K = ker \ \varphi$, then $Z/K \cong G$ and therefore $|Z| = |K| \cdot |G|$. Since $p$ divides $|G|$, it follows that $p$ divides $|Z|$. But $|Z| = r_{1} \times r_{2} \ldots r_{n}$. Hence one of the $r_{i}$ is divisible by $p$.