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I calculated the number of labeled Abelian groups of order $N$ (i.e., the number of distinct, abelian group laws on a set of $N$ elements). This sequence is given by OEIS A034382, but my solution differs at $N=16$.

Please point out mistakes or confirm my solution?


Let $C_n$ be a cyclic group of order $n$, $Aut(G)$ be an automorphism set of $G$.

Tthe number of labeled Abelian groups of order $N$ is $\displaystyle{\sum \frac{N!}{\# Aut(G)}}$ where G runs representative of isomorphic equivalence.

I got

$\displaystyle{ \# Aut(C_{p^n}^k)=p^{(n-1)k^2}\prod_{j=0}^{k-1} (p^{k}-p^{j}) }$

and

$\displaystyle{ \# Aut(\prod_i C_{p^{n_i}}^{k_i}) =\prod_i \left( (p^{(n_i-1)k_i^2}\prod_{j=0}^{k_i-1} (p^{k_i}-p^{j})) ( \prod_{j\neq i} p^{\min(n_i,n_j)k_j} )^{k_i} \right) }$.

From the fundamental theorem of finite abelian groups, There are 5 groups for $N=16$: $C_{16}, C_2 \times C_8, C_4^2, C_2^2\times C_4, C_2^4$.

Therefore, the number of groups that are isomorphic to each group is:

  • $C_{16}$ ... $\displaystyle{\frac{16!}{8}}$
  • $C_2 \times C_8$ ... $\displaystyle{\frac{16!}{(1\times 2)\times (2\times 4)}}$
  • $C_4^2$ ... $\displaystyle{\frac{16!}{16\times 3\times 2}}$
  • $C_2^2\times C_4$ ... $\displaystyle{\frac{16!}{((3\times 2)\times 2^2)\times (2^2\times 2)}}$
  • $C_2^4$ ... $\displaystyle{\frac{16!}{15\times 14\times 12\times 8}}$

Sum of them is $4250979532800$. OEIS says $4248755596800$.

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    $\begingroup$ It's only off by one-twentieth of one percent – what's the big deal? $\endgroup$ Sep 12 '19 at 6:16
  • $\begingroup$ I'm sorry that this should post to Mathematics stack exchange. I'll delete this question soon. $\endgroup$
    – sugarknri
    Sep 12 '19 at 10:59
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    $\begingroup$ It may be an error in OEIS. In either case it's a good question, and I don't support the close votes. $\endgroup$ Sep 12 '19 at 12:38
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    $\begingroup$ In case it's a good question, it would benefit from being better introduced... it starts with 3 sentences which look unrelated. The notion in the title is not defined. Also the title should not be considered as the 1st line of the post: the post should be meaningful without reading the title. $\endgroup$
    – YCor
    Sep 12 '19 at 23:17
  • $\begingroup$ @YCor Thank you for the advice about the format. I change text. If the text is still not clear, It may be because of my English ability. $\endgroup$
    – sugarknri
    Sep 13 '19 at 3:35
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There is a different formula for $\#\mathrm{Aut}(\prod_i C_{p^{n_i}}^{k_i})$ given in the paper Automorphisms of Finite Abelian Groups by Hillar and Rhea: $$\#\mathrm{Aut}(\prod_{t=1}^m C_{p^{e_t}}) = \prod_{t=1}^m (p^{d_t} - p^{t-1}) p^{e_t(m-d_t) + (e_t-1)(m-c_t+1)},$$ where $1\leq e_1\leq e_2\leq \cdots\leq e_m$, and $c_t$ and $d_t$ are the minimum and maximum of the set $S_t := \{\ell\ :\ e_\ell=e_t\}$, respectively.

Below I will show that OP's formula is equivalent to the Hillar-Rhea formula.


Let $d_0:=0$. It can be seen that the $k_i$'s are the nonzero elements of the multiset $\{ d_1-d_0, d_2-d_1, \dots, d_m-d_{m-1}\}$ and the $n_i$'s are the corresponding elements of $\{e_1,e_2,\dots,e_m\}$. Define $s_0=0, s_1, \dots, s_q$ be the indices such that $k_i = d_{s_i} - d_{s_{i-1}}$ and $n_i = e_{s_i}$. Vice versa, $d_{s_i} = k_1+\dots+k_i$ and $c_{s_i} = d_{s_{i-1}}+1$.

First consider these parts of the two formulas: $$\prod_{i=1}^q \prod_{j=0}^{k_i-1} (p^{k_i} - p^j) = \prod_{i=1}^q p^{k_i(k_i-1)/2} \prod_{j=0}^{k_i-1} (p^{k_i-j} - 1)$$ and $$\prod_{t=1}^m (p^{d_t} - p^{t-1}) = p^{m(m-1)/2}\prod_{t=1}^m (p^{d_t-t+1} - 1).$$ It is easy to see that the multisets $\{ k_i - j : 0\leq j \leq k_i-1, 1\leq i\leq q \}$ and $\{ d_t - t +1\ :\ 1\leq t\leq m \}$ are the same, since the $t$-th element in the sequence $$k_1 - 0, k_1 - 1, \dots, 1, k_2 - 0, k_2 - 1, \dots, 1, \dots$$ equals $d_t-t+1$.

Now it remains to prove the equality for the powers of $p$ in the two formulas, i.e. $$\sum_{i=1}^q \bigg(k_i(k_i-1)/2 + (n_i-1)k_i^2 + \sum_{j\ne i} \min(n_i,n_j)k_ik_j\bigg) = m(m-1)/2 + \sum_{t=1}^m \big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\big).$$ In the l.h.s. we have $$\sum_{i=1}^q \sum_{j\ne i} \min(n_i,n_j)k_ik_j = 2\sum_{i=1}^q n_i k_i \sum_{j>i} k_j=2\sum_{i=1}^q n_i k_i (m-d_{s_i}).$$ In the r.h.s. we have \begin{split} \sum_{t=1}^m \big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\big) &= \sum_{i=1}^q k_i\big(e_{s_i}(m-d_{s_i}) + (e_{s_i}-1)(m-d_{s_{i-1}})\big) \\ &= \sum_{i=1}^q k_i\big(n_i(m-d_{s_i}) + (n_i-1)(m-d_{s_{i-1}})\big)\\ &=\sum_{i=1}^q k_i\big(n_i(m-d_{s_i}) + (n_i-1)(m+k_i-d_{s_i})\big)\\ &=2\sum_{i=1}^q k_i n_i(m-d_{s_i}) + \sum_{i=1}^q \big( (n_i-1)k_i^2 - (m-d_{s_i})k_i\big). \end{split} Finally, we notice that $$\sum_{i=1}^q k_i(k_i-1)/2 = m(m-1)/2 - \sum_{i=1}^q (m-d_{s_i})k_i$$ since $m=k_1+k_2+\dots+k_q$ and $m-d_{s_i} = k_{i+1}+k_{i+1}+\dots+k_q$. QED


So, we can conclude that OEIS A034382 did indeed contain an error in its 16-th term. Now it's corrected.

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