# A question concerning some group action

Let $$G$$ be a finite group. Consider the set $$X = \bigcup_{H \le G} G/H$$ which is a disjoint union of left cosets of subgroups $$H$$ of $$G$$. Then $$G$$ acts on $$X$$ by left multiplication, and the number $$|X/G|$$ of orbits is the number of subgroups of $$G$$. I want to apply Burnsides Lemma in this situation $$|X/G| = \frac{1}{|G|} \sum_{g \in G} |X^g|$$ where $$X^g = \{ x \in X | g \cdot x = x\}$$, to maybe get a "formula" for the number of subgroups of $$G$$. For this I need to "compute" $$|X^g|$$. Is there any other nice description of this quantity? Thanks for your help!

We have $$|X^g| = |\{g'H \in X| g\cdot g' \cdot H = g' \cdot H\}|$$, but how to proceed?

Edit The reason I suspect such a formula can be computed is the group $$G=C_n$$, for which we have:

$$\tau(n)=\frac{1}{n}\sum_{k=0}^{n-1}\sigma(\gcd(n,k))$$

Also, using the Lagarias inequality, one can show that an upper bound on $$\tau(n)$$ is equivalent to RH.

• Won't your "formula" end up being of the form (number of subgroups of G) = (sum over subgroups of G)(something depending on the subgroup)? Won't that (something depending on the subgroup) end up being identically the number 1? – Theo Johnson-Freyd May 1 at 23:38
• @TheoJohnsonFreyd: No, I don't think so. Already for the cyclic group there is a non-trivial formula. See the edit to the question. – orgesleka May 2 at 4:57
• $|X^g| = \sum_{H : g \in H} [G:H]$ – Sean Eberhard May 2 at 10:15
• @SeanEberhard: Thank you for your comment. That looks quite interesting. How did you derive it? – orgesleka May 2 at 10:22
• @orgesleka Well, perhaps you will find something. But note that your formula for $\tau(n)$ is not of the type that you ask. Rather, your sum ranges over elements of $G$, not subgroups of $G$. – Theo Johnson-Freyd May 3 at 17:18