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Notation: For a group $G$, we write $G^n$ to denote the $n$-fold direct product of $G$ with itself.

Problem set up:

Consider, for a finite group $G$, and $n > 1$, the set $Q(G)_n$ of all isomorphism classes of groups of the form $\frac{G^{n}}{H_{n-1}}$, for any normal subgroup $H_{n-1}$ of $G^n$ isomorphic to $G^{n-1}$.

Note that $Q(G)_1 \subset Q(G)_2 \subset \cdots$, and so for finite groups $G$, the sequence stabilises at some $N$ - i.e. we have $Q(G)_n =Q(G)_{n+1}$ for all $n \geq N$. We call the sequence $Q(G)_1, \cdots, Q(G)_N$ the quotient series of $G$.

We recall that by the classification theorem for finite abelian groups, we can write any finite abelian group as $\bigoplus_{i = 0}^M \mathbb Z_{p_{i}^{k_i}}$ for primes $p_i$ and positive integers $k_i$.

Question: Let $G = \bigoplus_{i = 0}^M \mathbb Z_{p_{i}^{k_i}}$ be a finite abelian group. What is the quotient series of $G$?

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  • $\begingroup$ Can you give a nontrivial example? $\endgroup$ May 10, 2021 at 9:26
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    $\begingroup$ Sure, take $G = \mathbb Z_4 \times \mathbb Z_2$. Then $Q(G)_2$ is $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ and $G$ itself. $\endgroup$
    – Nate River
    May 10, 2021 at 9:48
  • $\begingroup$ You want $\oplus$ not $\otimes$ right $\endgroup$ May 10, 2021 at 11:40
  • $\begingroup$ Oh you mean in the problem statement, right! $\endgroup$
    – Nate River
    May 10, 2021 at 12:07

1 Answer 1

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I'll abbreviate $\mathbb{Z}/p^k \mathbb{Z}$ to $L_k$. Let $G = \bigoplus_{i=1}^M L_i^{a_i}$. I claim that $\bigoplus_{i=1}^N L_i^{b_i}$ is in the quotient series of $G$ if and only if the Littlewood-Richardson coefficient $$c_{(1^a_1),\ \ (2^{a_2}),\ \ \dotsc,\ \ (M^{a_M})}^{(1^{b_1} 2^{b_2} \cdots N^{b_N})}$$ is nonzero. Note the parentheses and commas: We have $M$ distinct partitions on the bottom and only one partition on the top.

You can decide whether or not this is an explicit enough answer to be useful to you.


Proof. In general, let $\alpha$, $\beta$ and $\gamma$ be three partitions. Then there is a short exact sequence $$0 \to \bigoplus L_{\alpha_i} \to \bigoplus L_{\beta_i} \to \bigoplus L_{\gamma_i} \to 0$$ if and only if $c_{\alpha \gamma}^{\beta}$ is nonzero. This is the same as asking that there be a semistandard Young tableaux of shape $\beta/\alpha$ whose rectification is a given tableaux of shape $\gamma$. See, for example, Fulton's survey Eigenvalues, invariant factors, highest weights, and Schubert calculus. The short exact sequences of abelian groups are in Section 2; the tableaux are briefly mentioned in Section 9 and can be found in more detail in many places, such as Stanley's Enumerative Combinatorics II or Fulton's Young Tableaux.

In your situation, you want to know when we have $c_{T \alpha,\ \gamma}^{(T+1) \alpha}$ nonzero for $T$ large. So we are looking for tableaux of shape $(T+1)\alpha/T \alpha$ whose rectification is a given tableaux of shape $\gamma$. But, once $T$ is large enough, $(T+1)\alpha/T \alpha$ is just a union of disconnected rectangles. Specifically, if $\alpha = 1^{a_1} 2^{a_2} \cdots M^{a_M}$, then the connected components of $(T+1)\alpha/T \alpha$ are rectangles of shape $k \times a_k$, for $1 \leq k \leq M$. Standard properties of Young tableaux then give the answer above.

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