# Semiregular subgroups of automorphism group of cayley graphs

Let $\Gamma$ be a Cayley graph over group $K$ and $H$ be a semiregular subgroup of $Aut(\Gamma)$ with two orbits. Then $|K|=2|H|$. Is there any other relation between $H$ and $K$ in general? What about special cases?

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In order to conclude that $|K|=2|H|" you should add WITH TWO ORBITS at the end of your first sentence. Now that your question has changed my earlier answer is not applicable and I have deleted it. – Nick Gill Nov 11 '12 at 16:17 add comment ## 2 Answers In general, you can't say mauch. Take$\Gamma$to be the complete graph$K_n$. Then$\Gamma$is a Cayley graph for any group$K$of order$n$, and any group$H$of order$n/2$acts semiregularly on$\Gamma$with two orbits. - Ok. Thanks so much. Do you think may be any relation between this two subgroups in special cases? For example for (non-complete)Cayley graphs of order twice a prime? – majid arezoomand Nov 20 '12 at 13:10 add comment EDIT: I assumed that$H$was a subgroup of$K$, which was not in the question. No, there are no other relations: given a group$K$, a subgroup$H$of index$2$and a Cayley graph$\Gamma$on$K$, it is always the case that$H$is a semiregular subgroup of$Aut(\Gamma)$with two orbits. In fact, there is nothing special about$2$here. If$\Gamma$is a Cayley graph on$K$, then$K$acts regularly on the vertices of$\Gamma$. In particular, any subgroup$H$of$K$will act semiregularly on the vertices, and the orbits of$H$will simply be the cosets of$H$in$K$. In particular, the number of orbits is the index of$H$in$K$. This all follows directly from the definitions and, as such, not really research level. - As you mentioned, if$|K:H|\geq 3$then$H$has at least 3 orbits and so 2 is neccesary here. – majid arezoomand Nov 11 '12 at 19:00 There was a typo in the first part of my answer, which is now fixed. Again, I want to emphasise that there is almost no content in my answer and that this is all seen easily from the relevant from the definitions. – verret Nov 11 '12 at 22:33 Majid didn't say that$H$is a subgroup of$K$, only that$H$is a subgroup of Aut($\Gamma$). The case where$H$is not a subgroup of$K\$ is the interesting case. –  Brendan McKay Nov 11 '12 at 23:36
You're right, I missed that. I'll edit. While I agree that this more general situation is more interesting, it's not really clear what is the question exactly. –  verret Nov 12 '12 at 0:13
I think that the question is clear. I want any relation between these two subgroups. –  majid arezoomand Nov 15 '12 at 18:21