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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.

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  • $\begingroup$ Just out of curiosity, what is your textbook? $\endgroup$ – Airymouse Nov 6 '16 at 16:53
  • $\begingroup$ @Airymouse "Groups and Symmetry" by M. A. Armstrong, I'm not sure which edition (can't find it in the book) but after some research my guess is its the second. $\endgroup$ – Tyron Nov 6 '16 at 18:10
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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$

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  • $\begingroup$ Is this element $ab$? $\endgroup$ – Tyron Nov 6 '16 at 18:37
  • $\begingroup$ Yes. It has to be. $\endgroup$ – jnyan Nov 7 '16 at 2:32
  • $\begingroup$ 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! $\endgroup$ – Tyron Nov 7 '16 at 8:10
  • $\begingroup$ 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 $\endgroup$ – jnyan Nov 7 '16 at 8:18
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    $\begingroup$ @PeldePinda, the answer you mentioned took elements of same order for a,b. I mentioned a,b of different orders. $\endgroup$ – jnyan Sep 30 '17 at 8:55
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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.

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    $\begingroup$ 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 $C_{m_n}$. $\endgroup$ – Marc Bogaerts Nov 7 '16 at 10:50
  • $\begingroup$ @BogaertsMarc That's another way of seeing the same thing. $\endgroup$ – egreg Nov 7 '16 at 10:51

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