I imagine this is a duplicate, but I can't find it. In the first chapter on groups, Aluffi shows that if $g$ is a group element of finite order, then $$|g^n|=\frac {|g|}{\gcd(n,|g|)}.$$ In an exercise, he asks the reader to show that if $g$ is an element of maximal finite order in an abelian group and $h$ is any element of finite order, then $|h|\mid|g|$. I proved this using the following hopefully-valid lemma:


If $g$ has finite order and $m\mid |g|$ then there is an element with order $m$.


$$|g^{|g|/m}|=\frac{|g|}{|g|/m}=m$$ $\square$

If $g$ is of maximal finite order and $h$ is of finite order, then $|g|$ and $\frac{|h|}{\gcd(|g|,|h|)}$ are relatively prime, so by a previous exercise, $$|gh|=|g|\frac{|h|}{\gcd(|g|,|h|)}.$$ By maximality, $\gcd(|g|,|h|)=|h|$ so $|h|\mid|g|$.

Is this correct? The hint Aluffi gives for the exercise involves powers of primes and such and thus strikes me as more complicated.


I see that Prove that for any element $b$, $|b|$ divides $|a|$ (order of $b$ divides order of $a$). addresses a similar question, but the solutions all seem quite different from mine and don't answer the question of whether mine is valid.

  • $\begingroup$ Aluffi's hint might not be as complicated as you thought. See here. $\endgroup$ – EuYu Nov 25 '13 at 22:14

In my opinion, the usual definition of the order as the smallest positive natural number for which the corresponding power becomes trivial is conceptually wrong. It should be a characterization, not a definition. The correct definition is the following: Let $G$ be a group, $g \in G$. Then $\{z \in \mathbb{Z} : g^z = 1\}$ is obviously a subgroup of $\mathbb{Z}$. Hence it has the form $d \mathbb{Z}$ for some unique $d \geq 0$. We call $d$ the order of $g$ (and yes, this includes $d=0$, but for silly reasons one usually defines $d=\infty$ then). Hence, we have the following defining property of $d$:

For all $n \in \mathbb{Z}$ we have $g^n = 1$ iff $d\mid n$.

This is how one can really work with orders without pain. For example we can prove the desired formula as follows: Let $d$ be the order of $g$, and $n \in \mathbb{N}$.

$$(g^n)^m = 1\iff g^{n \cdot m}=1 \iff d\mid n \cdot m \iff d/\gcd(d,n) \mid n/\gcd(d,n) \cdot m \iff \text{(since $d/\gcd(d,n)$ and $n/\gcd(d,n)$ are coprime) } d/\gcd(d,n)\mid m.$$

This shows that $d/\gcd(d,n)$ satisfies the defining property of the order of $g^n$.

No need to prove lemmas or deal with prime powers.

  • $\begingroup$ I fully agree with your opinion. Not only $\{z\in\mathbb{Z}:g^z=1\}$ is a subgroup, but it's the kernel of a homomorphism, so we can apply the homomorphism theorem to count the elements of $\langle g\rangle$. $\endgroup$ – egreg Nov 26 '13 at 0:23
  • $\begingroup$ Yes, you are right, one immediately gets $\langle g \rangle \cong \mathbb{Z}/d\mathbb{Z}$. $\endgroup$ – Martin Brandenburg Nov 26 '13 at 15:19

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