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This may seem somewhat of an easy proof, but as I'm currently working through a sample paper, I'll definition be getting all my answers checked here (if I cannot find solutions online).

Question: Suppose $G$ is cyclic, i.e. there exists $g\ \epsilon\ G$ such that $G=<g>$ (and similarly for $g^{-1}$). Show that if $|G|=ab$, then $G$ has exactly one subgroup of order $a$.

Attempt: By definition, G will contain at least one subgroup (in fact two, $G$ and $\left\{ e_G \right\}$)

...this is honestly as far as I got. From here, I thought about invoking Lagrange's Theorem, but then I ran into the problem of "What if there was more than one subgroup with the same order?". I know that for cosets, all cosets are disjoint (or equal), form a partition on $G$ and are all of the same order.

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  • $\begingroup$ Hint: Given some natural number $n$, what can you say about the subgroup generated by $g^n$? $\endgroup$ – Tobias Kildetoft Jun 25 '14 at 9:31
  • $\begingroup$ Have you seen the Fundamental Theorem of Cyclic Groups? (See: en.wikipedia.org/wiki/Subgroups_of_cyclic_groups) $\endgroup$ – Adriano Jun 25 '14 at 9:32
  • $\begingroup$ $<g^n>$? If G is infinite, then all elements of that subgroup would be in G. I guess it would be powers of those elements, so say for n=2, it would include only every second element of $G$. $\endgroup$ – Yoshi Jun 25 '14 at 9:35
  • $\begingroup$ What do you mean "if $G$ is infinite"? You have asked a question about a finite group. $\endgroup$ – Tobias Kildetoft Jun 25 '14 at 9:47
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The answer is on two points

  • existence: the order of the subgroup $\langle g^b\rangle$ is $a$
  • if $H$ is a subgroup with order $a$ so it's generated by $g^\alpha$ for some $\alpha$ and $g^{\alpha a}=e_G$ so $|G|=ab$ divides $\alpha a$ so $b$ divides $\alpha$ and then $g^\alpha\in\langle g^b\rangle$ so $H\subset \langle g^b\rangle$ and by the cardinality we have the unicity.
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In a cyclic group, there is exactly one subgroup of order $n$ for every $n\mid |G|$. It is the one generated by $g^{\frac{|G|}{n}}$ if $g$ generates $G$.

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