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 $p$. $\blacksquare$

  • 4
    $\begingroup$ 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. $\endgroup$ – DonAntonio Nov 29 '17 at 23:17
  • $\begingroup$ 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. $\endgroup$ – 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 $x\in G$.

  • 1
    $\begingroup$ (+1) Your answer is better than mine, since you explain why that proof could not work. $\endgroup$ – José Carlos Santos Nov 30 '17 at 9:51

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 $p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.