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The number of distinct transitive actions should depend on the number of orders possible for $g\in G= C_n$.

Say, take $n=12:$

Hence, the answer as per Lagrange's theorem should be: $6$, as orders of $g$ possible are: $\{1, 2, 3,4, 6, 12\}.$

jiten
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    You have the correct set of numbers, but you need to A) describe an action of all of $C_{12}$ on a set $X$ of each of those sizes, and B) explain why the cardinality of $X$ determines the transitive action uniquely up to isomorphism. – Jyrki Lahtonen Aug 02 '22 at 07:01
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    Hint: By the orbit-stabilizer theorem it is really not so much a question of orders of elements of $G$. Rather, it is about the indices of subgroups $H\le G$. The gist of orbit-stabilizer is that any transitive action is isomorphic to the action on the set of cosets $G/H$. – Jyrki Lahtonen Aug 02 '22 at 07:04
  • @JyrkiLahtonen Not read the topic till now. Next week has the topic of orbits-stabilizers. Thanks for confirmation. – jiten Aug 02 '22 at 07:09
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    There is a general result that, for any group $G$, there is a bijection between the equivalence classes of transitive actions of $G$, and the set of conjugacy classes of subgroups of $G$. – Derek Holt Aug 02 '22 at 07:41
  • @DerekHolt Thanks a lot. Link for reference, or here any question. Also, any suggestions for my answer. Actually, not sure if your last comment gives same result of $6$. – jiten Aug 02 '22 at 08:00
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    This bijection is discussed here. Yes, it gives the same result, because $C_{12}$ has exactly six subgroups. (The group is abelian, so each subgroup forms a single conjugacy class of subgroups.) – Derek Holt Aug 02 '22 at 08:10

1 Answers1

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That there's only $6$ conjugacy classes of subgroups in $C_{12}$ implies that there's only $6$ possible transitive actions. See this answer.

In general, for the cyclic group $C_n$, it's going to be $\sigma_0(n)=\sum_{d\mid n}d^0$, the number of divisors of $n$. Here $\sigma $ is the divisor function. Because a cyclic group has a characteristic subgroup of order $d$ for each divisor of $n$. Since it's characteristic, the size of its conjugacy class is one.

We're using that the number of transitive actions of $G$ is the number of conjugacy classes of subgroups of $G$, as in the first link.

calc ll
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