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I will quote a question from my textbook, to prevent misinterpretation:

Let $G$ be a finite abelian group and let $m$ be the least common multiple of the orders of its elements. Prove that $G$ contains an element of order $m$.

I figured that, if $|G|=n$, then I should interpret the part with the least common multiple as $lcm(|x_1|,\dots,|x_n|)=m$, where $x_i\in G$ for $0\leq i\leq n$, thus, for all such $x_i$, $\exists a_i\in\mathbb{N}$ such that $m=|x_i|a_i$. I guess I should use the fact that $|x_i|$ divides $|G|$, so $\exists k\in \mathbb{N}$ such that $|G|=k|x_i|$ for all $x_i\in G$. I'm not really sure how to go from here, in particular how I should use the fact that $G$ is abelian.

egreg
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Tyron
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2 Answers2

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A finite abelian group can be written as a (finite) direct product of cyclic groups: $$ G=C_{m_1}\times C_{m_2}\times\dots\times C_{m_r} $$ where $C_n$ denotes a cyclic group of order $n$. Thus the order of any element in $G$ divides $\operatorname{lcm}(m_1,m_2,\dots,m_r)$. On the other hand, if $g_i$ is a generator of $C_{m_i}$, the element $$ g=(g_1,g_2,\dots,g_r) $$ has order precisely $\operatorname{lcm}(m_1,m_2,\dots,m_r)$.

Fill in the details.

egreg
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    Every finite abelian group can be written as the direct product of cyclic groups $C_{m_1}\times C_{m_2}\ldots C_{m_n}$ where $m_1 | m_2, m_2 | m_3, \ldots, m_{n-1} | m_n$. The answer is the generator of $C_{m_n}$. – Marc Bogaerts Nov 07 '16 at 10:50
  • @BogaertsMarc That's another way of seeing the same thing. – egreg Nov 07 '16 at 10:51
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Hint : in an abeilan group, for any two elements $a,\, b$ of different order, there is an element in the group, whose order is lcm of order of $ a, \,b$

jnyan
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  • Is this element $ab$? – Tyron Nov 06 '16 at 18:37
  • Yes. It has to be. – jnyan Nov 07 '16 at 02:32
  • So if it holds for two elements, it holds for an arbitrary, finite amount of elements, hence also for all of the elements in the group, completing the proof. Am I right? Also, do the elements $a$ and $b$ really have to be of different order for the sake of this argument? For say $|a|=|b|$, then $lcm(|a|,|b|)=|a|=|b|$, thus there also exists an element with order equal to $lcm(|a|,|b|)$, namely $a$ (or $b$, or $ab$). In other words, is it really necessary to distinguish between the trivial and non trivial case, or is that just more formal? Just being curious! – Tyron Nov 07 '16 at 08:10
  • Correct. If the order is same then, no issue. You could say there are more than one element of that order. But taking non trivial case, applies for every other case. If you get it why order of an element is lcm of order of two elements, then u r done. Same logic can be extended – jnyan Nov 07 '16 at 08:18
  • One more question. What about the case where $b=a^{-1}$? Since $|a|=|a^{-1}|=|b|$, $lcm(|a|,|b|)=|a|=|b|$, but then $|ab|=|e|=1$, which is only equal to $lcm(|a|,|b|)$ if $a=b=e$, am I right? Oh wait, this case is not regarded because of the assumption that $a$ and $b$ have different order! But then, how does one fill out the entire group if some combinations of elements are not allowed? I'm guessing this "filling out" is necessary since, in the question, the lcm of all elements in $G$ is regarded. – Tyron Nov 07 '16 at 08:47
  • Correct, again. – jnyan Nov 07 '16 at 08:51
  • Let us continue this discussion in chat. – Tyron Nov 07 '16 at 09:00
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    @PeldePinda, the answer you mentioned took elements of same order for a,b. I mentioned a,b of different orders. – jnyan Sep 30 '17 at 08:55