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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.

Related: https://math.stackexchange.com/questions/1315302/group-action-so-that-every-subgroup-is-a-stabilizer

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  • $\begingroup$ 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? $\endgroup$ – Theo Johnson-Freyd May 1 at 23:38
  • $\begingroup$ @TheoJohnsonFreyd: No, I don't think so. Already for the cyclic group there is a non-trivial formula. See the edit to the question. $\endgroup$ – orgesleka May 2 at 4:57
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    $\begingroup$ $|X^g| = \sum_{H : g \in H} [G:H]$ $\endgroup$ – Sean Eberhard May 2 at 10:15
  • $\begingroup$ @SeanEberhard: Thank you for your comment. That looks quite interesting. How did you derive it? $\endgroup$ – orgesleka May 2 at 10:22
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    $\begingroup$ @oregsleka: SeanEberhard's formula can be derived as follows. Fix $H\le G$ and let $H_1,\dots H_m$ be the distinct $G$-conjugates of $H$, and say that $g\in H_i\iff 1\le i\le k$. Then $|\{x\,|\,g^x\in H_i\}|$, which is independent of $i$, equals $(k/m)|G|$. Let $X_H=\cup_{i=1}^m G/H_i$. Then $|H||X_H^g|=\sum_{i=1}^m|\{x\,|\,g^x\in H_i\}|=m(k/m)|G|=k|G|=\sum_{1\le i\le m,\ g\in H_i} |G|$. Now divide by $|H|$ and sum over a set of representatives of conjugacy classes of subgroups of $G$. $\endgroup$ – Richard Lyons May 4 at 18:09

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