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Quaternion group $Q_8$ has though $6$ elements of order $4$, but there are three sets ($(i, -i), (j, -j), (k, -k)$) having seperate subgroups, with identity and $-1$ common only. Apart from that have one element of order $2$, and $e$. But, how to use that information, is unclear.

It is not a cyclic group, hence cannot simply state based on orders of elements.

jiten
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  • See also this question for context. – Dietrich Burde Aug 02 '22 at 11:11
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    For this you really need the correspondence with conjugacy classes of subgroups that I mentioned in your earlier post. $Q_8$ has six such classes, one each of order $1,2,8$ and three of order $4$. The corresponding transitive action is on the cosets of subgroup in the conjugacy class. – Derek Holt Aug 02 '22 at 11:12
  • @DietrichBurde Is the context of a small paragraph really important? – jiten Aug 02 '22 at 11:13
  • @DerekHolt Need more details, even to work on own. – jiten Aug 02 '22 at 11:14
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    The context is important, as Derek says, since this is not your first question on this topic, and one should not repeat everything again. So it is helpful to have links. You can provide more links, which are better suited, of course. – Dietrich Burde Aug 02 '22 at 11:14

1 Answers1

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Hint Just like in your previous post about transitive actions of $C_{12}$, the general result (that @Derek Holt clued us in on) is that there are as many inequivalent actions as conjugacy classes of subgroups (of, in this case, $Q_8$).

The aforementioned also gave us a headstart in the comments that in this case there are $6$ conjugacy classes of subgroups.

The subgroups are: $, \{1\},\langle-1\rangle, \langle i\rangle, \langle j\rangle, \langle k\rangle $ and $Q_8$.

Automorphisms preserve the order of subgroups, so the only question is if the $3$ subgroups of order $4$ are conjugate. Since they each have index $2$, they're all normal. So not conjugate. That's where the number $6$ comes from (each subgroup is its own conjugacy class).

Here's some terminology: the quaternions are Hamiltonian. That's they're a non-abelian group such that every subgroup is normal.

In this problem, that tells us the number of transitive actions is equal to the number of subgroups.

calc ll
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  • Please help me with my last post at: https://math.stackexchange.com/q/4505323/424260. – jiten Aug 03 '22 at 12:45
  • Have thoroughly updated the linked post to the one listed in last comment. Kindly help in answer finding by any way you like. – jiten Aug 03 '22 at 21:27