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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.


marked as duplicate by Bill Dubuque, YuiTo Cheng, Leucippus, Lord Shark the Unknown, Shogun 2 days ago

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  • $\begingroup$ I'm sorry, but where did $g^n=g^m=e$ come from? $\endgroup$ – Paulo Mourão May 23 at 8:33
  • $\begingroup$ 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. $\endgroup$ – Arthur May 23 at 8:34
  • $\begingroup$ Is he? What are the different $g$'s? $\endgroup$ – Paulo Mourão May 23 at 8:35
  • $\begingroup$ I just think the equality I mentioned is not necessarily true. $\endgroup$ – Paulo Mourão May 23 at 8:35
  • 4

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.


Hint: Note that $G$ is a finite abelian group, $G \cong\mathbb{Z}_{n_1}\times\cdots\times\mathbb{Z}_{n_s}$ for some cyclic groups $\mathbb{Z}_{n_i}$ with $n_i\mid n_{i+1}$. So your $n$ is exactly $n_s$.

Or $G \cong G_{p_1}\times \cdots\times G_{p_m}$ for some abelian $p_i$-groups, where $p_i$ are prime numbers. It suffices to show the proposition in each abelian $p$-group.

  • $\begingroup$ I don't think he's supposed to use characterization of finitely generated abelian groups yet... This looks like an introductory exercise and that would not be something he would know in the beginning. But maybe I'm wrong. $\endgroup$ – Paulo Mourão May 23 at 8:37
  • $\begingroup$ I know that $G$ is isomorphic to a direct product of cyclic groups, but that $n_i|n_{i+1}$ is new to me $\endgroup$ – TwoStones May 23 at 8:49
  • $\begingroup$ See Corollary 6.6 in Rotman's An Introduction to the Theory of Groups.@TwoStones $\endgroup$ – Hongyi Huang May 23 at 8:56
  • $\begingroup$ Or you may check this in abelian $p$-groups and use my second factorization.@TwoStones $\endgroup$ – Hongyi Huang May 23 at 9:03

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.


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