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Let $G = \langle a \rangle$ be a cyclic group of order $n$. Show that for every divisor $d$ of $n$, there exists a subgroup $G$ whose order is $d$.

If $d \mid n$ then there exists $m \in \mathbb{Z}$ such that $n = md$. Since $a^n = 1_G$, we have $a^{md} = (a^m)^d = 1_G $. How to proceed from here? We have that $a^m$ is an element of order $d$, but do I still have to prove that there is a subgroup $G$ with arbitrary elements of the form $a^m$?

Bill Dubuque
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St Vincent
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2 Answers2

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Hint: $a^{\frac{n}{d}}$ has order $d$.

Nicky Hekster
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  • $ a^n = a^{n(d/d)} = (a^{n/d})^d $ right? Thanks for the hint – St Vincent Jan 30 '15 at 08:36
  • Yes, but you also have to show that if $(a^{\frac{n}{d}})^e=1$, then $d \mid e$. – Nicky Hekster Jan 30 '15 at 08:54
  • Isn't $e $ equal to $1$? You used both – St Vincent Jan 30 '15 at 09:11
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    $e$ is here just a natural number - not 2.7182818 ... :-)). In this case the order of $a^{\frac{n}{d}}$. You have to check if $d|e$ and $e|d$, whence $e=d$. In group theory the identity element is denoted as $1$ (that is what I use mostly), $e$, $1_G$. – Nicky Hekster Jan 30 '15 at 09:25
  • So we have to show that $ d $ is a natural number essentially? – St Vincent Jan 30 '15 at 09:29
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    No, no, that is not necessary. I put $e$ as the order of $a^{\frac{n}{d}}$, so $e$ is by definition the smallest natural number such that $(a^{\frac{n}{d}})^e=1$. You already showed that $(a^{\frac{n}{d}})^d=1$, and this implies $e \mid d$ (can you see why?). Conversely, $(a^{\frac{n}{d}})^e=1$, but since $n$ is the order of $a$, we get $n \mid \frac{ne}{d}$, say $\frac{ne}{d}=fn$. This implies $e=df$, whence $d \mid e$. But we already had $e \mid d$, so $d=e$ is the order of $a^{\frac{n}{d}}$. Hope this clarifies it. – Nicky Hekster Jan 30 '15 at 20:17
  • This makes it clear, thank you – St Vincent Jan 30 '15 at 23:05
  • @StVincent the claim in the Hint is also proved here, slightly more concisely. – Bill Dubuque Mar 21 '24 at 03:04
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Just consider the cyclic subgroup generated by $a^m$.

Anurag A
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