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This is an elaboration on Stanley's reference to Exercise 5.13 in EC2. In these two blog posts you can find proofs using groupoid cardinality of the following results. For a finitely generated group $G$ let $s_n(G)$ denote the number of subgroups of index $n$ and let $c_n(G)$ denote the number of conjugacy classes of subgroups of index $n$.

Exercise 5.13a: $$\sum_{n \ge 0} \frac{|\text{Hom}(G, S_n)|}{n!} z^n = \exp \left( \sum_{n \ge 1} s_n(G) \frac{z^n}{n} \right).$$

Exercise 5.13c: $$\sum_{n \ge 0} \frac{|\text{Hom}(\mathbb{Z} \times G, S_n)|}{n!} z^n = \prod_{n \ge 1} \frac{1}{(1 - z^n)^{c_n(G)}}.$$

Taking logarithms on both sides of Theorem 2 and applying Theorem 1 gives

$$s_n(\mathbb{Z} \times G) = \sum_{d \mid n} d c_d(G)$$

which, when applied to $G = \mathbb{Z}^k$, gives the second recurrence in Carl-Fredrik's answer (which is essentially the content of Exercise 5.13d).

(Not to toot my own horn excessively and with all respect to Stanley, I believe my proof of Exercise 5.13c is better motivated than the argument that appears in EC2.)

This is an elaboration on Stanley's reference to Exercise 5.13 in EC2. In these two blog posts you can find proofs using groupoid cardinality of the following results. For a finitely generated group $G$ let $s_n(G)$ denote the number of subgroups of index $n$ and let $c_n(G)$ denote the number of conjugacy classes of subgroups of index $n$.

Exercise 5.13a: $$\sum_{n \ge 0} \frac{|\text{Hom}(G, S_n)|}{n!} z^n = \exp \left( \sum_{n \ge 1} s_n(G) \frac{z^n}{n} \right).$$

Exercise 5.13c: $$\sum_{n \ge 0} \frac{|\text{Hom}(\mathbb{Z} \times G, S_n)|}{n!} z^n = \prod_{n \ge 1} \frac{1}{(1 - z^n)^{c_n(G)}}.$$

Taking logarithms on both sides of 5.13c and applying 5.13a gives

$$s_n(\mathbb{Z} \times G) = \sum_{d \mid n} d c_d(G)$$

which, when applied to $G = \mathbb{Z}^k$, gives the second recurrence in Carl-Fredrik's answer (which is essentially the content of Exercise 5.13d).

(Not to toot my own horn excessively and with all respect to Stanley, I believe my proof of Exercise 5.13c is better motivated than the argument that appears in EC2.)

This is an elaboration on Stanley's reference to Exercise 5.13 in EC2. In these two blog posts you can find proofs using groupoid cardinality of the following results. For a finitely generated group $G$ let $s_n(G)$ denote the number of subgroups of index $n$ and let $c_n(G)$ denote the number of conjugacy classes of subgroups of index $n$.

Exercise 5.13a: $$\sum_{n \ge 0} \frac{|\text{Hom}(G, S_n)|}{n!} z^n = \exp \left( \sum_{n \ge 1} s_n(G) \frac{z^n}{n} \right).$$

Exercise 5.13c: $$\sum_{n \ge 0} \frac{|\text{Hom}(\mathbb{Z} \times G, S_n)|}{n!} z^n = \prod_{n \ge 1} \frac{1}{(1 - z^n)^{c_n(G)}}.$$

Taking logarithms on both sides of Theorem 2 and applying Theorem 1 gives

$$s_n(\mathbb{Z} \times G) = \sum_{d \mid n} d c_d(G)$$

which, when applied to $G = \mathbb{Z}^k$, gives the second recurrence in Carl-Fredrik's answer (which is essentially the content of Exercise 5.13d).

(Not to toot my own horn excessively and with all respect to Stanley, I believe my proof of Exercise 5.13b is better motivated than the argument that appears in EC2.)

This is an elaboration on Stanley's reference to Exercise 5.13 in EC2. In these two blog posts you can find proofs using groupoid cardinality of the following results. For a finitely generated group $G$ let $s_n(G)$ denote the number of subgroups of index $n$ and let $c_n(G)$ denote the number of conjugacy classes of subgroups of index $n$.

Exercise 5.13a: $$\sum_{n \ge 0} \frac{|\text{Hom}(G, S_n)|}{n!} z^n = \exp \left( \sum_{n \ge 1} s_n(G) \frac{z^n}{n} \right).$$

Exercise 5.13c: $$\sum_{n \ge 0} \frac{|\text{Hom}(\mathbb{Z} \times G, S_n)|}{n!} z^n = \prod_{n \ge 1} \frac{1}{(1 - z^n)^{c_n(G)}}.$$

Taking logarithms on both sides of Theorem 2 and applying Theorem 1 gives

$$s_n(\mathbb{Z} \times G) = \sum_{d \mid n} d c_d(G)$$

which, when applied to $G = \mathbb{Z}^k$, gives the second recurrence in Carl-Fredrik's answer (which is essentially the content of Exercise 5.13d).

(Not to toot my own horn excessively and with all respect to Stanley, I believe my proof of Exercise 5.13c is better motivated than the argument that appears in EC2.)

This is an elaboration on Stanley's reference to Exercise 5.13 in EC2. In these two blog posts you can find proofs using groupoid cardinality of the following results. For a finitely generated group $G$ let $s_n(G)$ denote the number of subgroups of index $n$ and let $c_n(G)$ denote the number of conjugacy classes of subgroups of index $n$.

Exercise 5.13a: $$\sum_{n \ge 0} \frac{|\text{Hom}(G, S_n)|}{n!} z^n = \exp \left( \sum_{n \ge 1} s_n(G) \frac{z^n}{n} \right).$$

Exercise 5.13c: $$\sum_{n \ge 0} \frac{|\text{Hom}(\mathbb{Z} \times G, S_n)|}{n!} z^n = \prod_{n \ge 1} \frac{1}{(1 - z^n)^{c_n(G)}}.$$

Taking logarithms on both sides of Theorem 2 and applying Theorem 1 gives

$$s_n(\mathbb{Z} \times G) = \sum_{d \mid n} d c_d(G)$$

which, when applied to $G = \mathbb{Z}^k$, gives the second recurrence in Carl-Fredrik's answer (which is essentially the content of Exercise 5.13d).

This is an elaboration on Stanley's reference to Exercise 5.13 in EC2. In these two blog posts you can find proofs using groupoid cardinality of the following results. For a finitely generated group $G$ let $s_n(G)$ denote the number of subgroups of index $n$ and let $c_n(G)$ denote the number of conjugacy classes of subgroups of index $n$.

Exercise 5.13a: $$\sum_{n \ge 0} \frac{|\text{Hom}(G, S_n)|}{n!} z^n = \exp \left( \sum_{n \ge 1} s_n(G) \frac{z^n}{n} \right).$$

Exercise 5.13c: $$\sum_{n \ge 0} \frac{|\text{Hom}(\mathbb{Z} \times G, S_n)|}{n!} z^n = \prod_{n \ge 1} \frac{1}{(1 - z^n)^{c_n(G)}}.$$

Taking logarithms on both sides of Theorem 2 and applying Theorem 1 gives

$$s_n(\mathbb{Z} \times G) = \sum_{d \mid n} d c_d(G)$$

which, when applied to $G = \mathbb{Z}^k$, gives the second recurrence in Carl-Fredrik's answer (which is essentially the content of Exercise 5.13d).

(Not to toot my own horn excessively and with all respect to Stanley, I believe my proof of Exercise 5.13b is better motivated than the argument that appears in EC2.)