ord$(h)|\max\{\text{ord}(g)|g\in G\}$ for all $h\in G$.

Question

Let $G$ be a finite abelian group and $n:=\max\{\text{ord}(g)|g\in G\}$. Now I have to proof that ord$(h)|n$ for all $h\in G$.

My idea was:

Let $g\in G$ with ord$(g)=m<n$. Then because of the euclidean divsion in $\mathbb{Z}$ one can write $n=km+r$ for $k,r\in\mathbb{Z}$ and $r<m$. So what I have to show now is $g^n = g^m = e$. From that follows $m|n$ by the definition of ord, right?

So this gives me $$g^n=g^{km+r}=g^{km}g^r=(g^m)^kg^r=e^kg^r=g^r$$

But from here on I dont know where to go next.

You're using the same letter $g$ for a few different things. This becomes somewhat confusing to read and even more confusing to write an appropriate answer trying to refer to the different $g$'s. — Arthur, May 23 '19 at 08:34
I just think the equality I mentioned is not necessarily true. — Paulo Mourão, May 23 '19 at 08:35
@PauloMourão When I show $g^n=g^m=e$ the I am done, right? Because if this holds I know that $n=km$ for a $k\in\mathbb{Z}$ and therefore $m|n$. — TwoStones, May 23 '19 at 08:36
@PauloMourão There were two "different" $g$'s just in the title alone. One was the overarching "$g\mid n$ for all $g\in G$", and the other was inside the definition of $n$. It has now been fixed. — Arthur, May 23 '19 at 08:38
@Arthur ahaha yes you're right. But I think it's just one after "My idea was" — Paulo Mourão, May 23 '19 at 08:40
@TwoStones It's true that $g^m=e$ by assumption. So it's true that $g^{km}=e$. But that only means that $g^{km+r}=eg^r=g^r$ — Paulo Mourão, May 23 '19 at 08:41
@PauloMourão Yeah I know that. But I dont know where to go from there. Or is it a dead end? — TwoStones, May 23 '19 at 08:42
Ah wait, it's written "I have to show that", sorry I didn't see that, my bad. — Paulo Mourão, May 23 '19 at 08:43
How do you use the property that $G$ is abelian? Note that it is not always true for non-abelian groups. — Groups, May 23 '19 at 08:43
@HongyiHuang Until now I didnt use it at all, youre right. But I dont really see where to use it... — TwoStones, May 23 '19 at 08:46
Duplicate of ord $(b)|\max{\text{ord}(g)|g\in G}$ for all $b\in G,$ a finite abelian group). — Bill Dubuque, Jun 13 '19 at 21:39

score 1 · Accepted Answer · answered May 23 '19 at 09:44

Since $G$ is a finite abelian group, we can write $$G=\langle g_1\rangle\times\cdots\times\langle g_k\rangle$$ as a direct product of cyclic groups, of orders $n_1,...,n_k$, respectively.

Letting $n=\text{lcm}(n_1,...,n_k)$, it follows that $g^n=1$ for all $g\in G$.

Thus, for $g\in G$, if $m=\text{ord}(g)$, then $m|n$.

Now let $h=g_1\cdots g_k$, and let $w=\text{ord}(h)$.

Since $h^n=1$, we have $w|n$.

But $h^w=1$ implies $g_i^w=1$ for all $i$, hence $n_i|w$ for all $i$, so $n|w$.

It follows that $w=n$, hence $n=\max\{\text{ord}(g)\mid g\in G\}$.

This completes the proof.

user10354138 · Answer 2 · 2019-05-23T09:32:09.877

In the other answer, you see how to use the structural theorem (which is a big sledgehammer) to crack this. There is a more elementary solution:

Lemma: Let $G$ be an abelian group. If $g,h\in G$, $\operatorname{ord}(g)=m$, $\operatorname{ord}(h)=n$, and $\min(m,n)\nmid\max(m,n)$ then $\operatorname{ord}(gh)=\operatorname{lcm}(m,n)$.

Remark: Since our $G$ is finite, repeated use of the this lemma shows lcm of all possible orders is achieved by some $g\in G$ and hence the claim follows.

Proof of Lemma: Let $$ m=p_1^{m_1}p_2^{m_2}\dots p_k^{m_k}\text{ and }n=p_1^{n_1}p_2^{n_2}\dots p_k^{n_k} $$ be the prime factorisation of $m,n$, where we allow one of $m_i,n_i$ to be zero if necessary (but not both). Then $$ \operatorname{lcm}(m,n)=p_1^{\max\{m_1,n_1\}}p_2^{\max\{m_2,n_2\}}\dots p_k^{\max\{m_k,n_k\}} $$

Certainly $\operatorname{ord}(gh)=:q\mid\operatorname{lcm}(m,n)$. Suppose, $q<\operatorname{lcm}(m,n)$ and we will derive a contradiction.

$q<\operatorname{lcm}(m,n)$ gives either $n\nmid q$ or $m\nmid q$. If both, consider $kq$ where $k$ is a product of only those prime powers missing from $m$. We still have $kq<\operatorname{lcm}(m,n)$. So swapping $g,h$ if necessary, we may assume WLOG $(gh)^r=e$, $r<\operatorname{lcm}(m,n)$ but $m\mid r$. Thus we obtain $$ e=g^rh^r=h^r $$ but this implies $r$ is a multiple of $\operatorname{ord}h=n$ and so $\operatorname{lcm}(m,n)\mid r$, contradicting our assumption $r<\operatorname{lcm}(m,n)$. QED.

ord$(h)|\max\{\text{ord}(g)|g\in G\}$ for all $h\in G$.

2 Answers2