Surjection moment method

In a paper by M.M.Wood (Theorem 3.1), we find the following theorem:

Theorem (Wood). Let 

• $(X_{n})_{n \geq 1}$ and $(Y_{n})_{n \geq 1}$ be sequences of random finitely generated abelian groups and
• $\mathcal{A}_{m}$ be the set of isomorphism classes of finite abelian groups with exponent dividing $m \in \mathbb{Z}_{\geq 1}$ (i.e., $mG = 0$ for all $G \in \mathcal{A}_{m}$).

Suppose that for every $G \in \mathcal{A}_{m},$ the limit  $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)).$$ exists and $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) \leq |\wedge^{2}G|.$$

Then for every $H \in \mathcal{A}_{m},$ the limit $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists, and $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)).$$

If we also have $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(Y_{n}, G))$$ for all $G \in \mathcal{A}_{m},$ then $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) = \lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H).$$

Remarks. 

1. Note that it is implicit in the statement of the theorem that if we denote by $\mathcal{S}$ the probability space of finitely generated abelian groups for $X_{n},$ then for any $H \in \mathcal{A}_{m},$ the subset $$\{X \in \mathcal{S} : X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H\}$$ is measurable, and similarly for $Y_{n}.$
2. It is also implicit in the statement that the expected value $\mathbb{E}(\#\mathrm{Sur}(X_{n}, G))$ exists for every $n \geq 1,$ and similarly for $Y_{n}.$
3. In the references we follow, it is not mentioned that the abelian groups appearing as representatives of classes in $\mathcal{A}_{m}$ are of finite size. It seems that the proof of Theorem requires it, so we assume this. (This does not affect the main results in the references.)

Rest of this posting. We cover the proof of the above theorem, which we learn from another paper by Wood (Theorem 8.3).

Step 1. We first assume that for every $H \in \mathcal{A}_{m}$ the limit $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists and deduce $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z})\#\mathrm{Sur}(H, G) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G))$$ from this assumption.

Claim. For every $G \in \mathcal{A}_{m},$ there exists $G' \in \mathcal{A}_{m}$ such that the infinite sum $$\sum_{H \in \mathcal{A}_{m}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')}$$ converges.

Proof of Claim. If $m = 1,$ then the only isomorphism class in $\mathcal{A}_{m}$ is that of the trivial group, so the claim immediately holds. Hence, assume that $m > 1.$ Consider the prime factorization $m = p_{1}^{e_{1}} \cdots p_{k}^{e_{k}},$ and write $$H = H_{1} \oplus \cdots \oplus H_{k}$$ with $p_{i}^{e_{i}}H_{i} = 0$ for $H \in \mathcal{A}_{m},$ which implies that $$\sum_{H \in \mathcal{A}_{m}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')} = \sum_{H_{1} \in \mathcal{A}_{p_{1}^{e_{1}}}}\frac{\#\mathrm{Hom}(H_{1}, G_{1})}{\#\mathrm{Hom}(H_{1}, G'_{1})} \cdots \sum_{H_{k} \in \mathcal{A}_{p_{k}^{e_{k}}}}\frac{\#\mathrm{Hom}(H_{k}, G_{k})}{\#\mathrm{Hom}(H_{k}, G'_{k})}$$ for any $G' \in \mathcal{A}_{m}.$ This reduces the problem to the case where $m = p^{e}$ where $p$ is a prime and $e \in \mathbb{Z}_{\geq 1}.$ Without loss of generality, assume that $G$ is nontrivial. In this case, we have $$G \simeq \mathbb{Z}/p^{\lambda_{1}}\mathbb{Z} \times \mathbb{Z}/p^{\lambda_{2}}\mathbb{Z} \cdots \times \mathbb{Z}/p^{\lambda_{l}}\mathbb{Z}$$ with $e \geq \lambda_{1} \geq \cdots \geq \lambda_{l} \geq 1,$ so if we let $\lambda = (\lambda_{1}, \dots, \lambda_{l})$ be the corresponding partition, then we choose $$G' = \mathbb{Z}/p^{\mu_{1}}\mathbb{Z} \times \mathbb{Z}/p^{\mu_{2}}\mathbb{Z} \cdots \times \mathbb{Z}/p^{\mu_{l'}}\mathbb{Z},$$ where $\mu_{i} = 2\lambda'_{i} + 1$ for $1 \leq i \leq e.$ (Here, we have $\lambda'_{i} = 0$ for $\lambda_{1} < i \leq e.$) By using a standard formula that counts the number of group maps from a finite abelian $p$-groups to another (Lemma 7.1 of Wood), we can compute that $$\sum_{H \in \mathcal{A}_{p^{e}}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')} = \sum_{c_{1} \geq \cdots \geq c_{e} \geq 0}p^{-c_{1}(\lambda_{1} + 1) \cdots - c_{e}(\lambda_{e} + 1)},$$ which is convergent as it is bounded by a convergent geometric series. This proves the claim. $\Box$

Back to Step 1, we have $$\begin{align*}\sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G') &= \mathbb{E}( \#\mathrm{Hom}(X_{n}, G')) \\ &= \sum_{H \subset G'}\mathbb{E}(\#\mathrm{Sur}(X_{n}, H)),\end{align*}$$ where $H$ ranges over subgroups of (a representative of) fixed $G' \in \mathcal{A}_{m}.$ The second identity is evident, but the first identity needs an explanation. By definition, we have $$\begin{align*}\mathbb{E}(\#\mathrm{Hom}(X_{n}, G')) &= \int_{X \in \mathcal{S}} \#\mathrm{Hom}(X, G') \mu_{n}(X), \end{align*}$$ where $\mu_{n}$ means the given probability measure on the probability space $\mathcal{S}$ of finitely generated abelian groups, where $X_{n}$ takes place. Since $mG' = 0,$ we have $$\#\mathrm{Hom}(X, G') = \#\mathrm{Hom}(X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}, G')$$ for any $X \in \mathcal{S}$ so that $$\begin{align*}\mathbb{E}(\#\mathrm{Hom}(X_{n}, G')) &= \int_{X \in \mathcal{S}} \#\mathrm{Hom}(X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}, G') \mu_{n}(X) \\ &= \sum_{B \in \mathcal{A}_{m}}\int_{\substack{X \in \mathcal{S}: \\ (X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}) \simeq B}}\#\mathrm{Hom}(B, G')\mu_{n}(X) \\ &= \sum_{B \in \mathcal{A}_{m}}\int_{\substack{X \in \mathcal{S}: \\ (X \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z}) \simeq B}}\mu_{n}(X) \#\mathrm{Hom}(B, G') \\ &= \sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G'),\end{align*}$$ as desired.

By hypothesis, the limit $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, H))$$ exists for each subgroup $H$ of $G'$, so $$\lim_{n \rightarrow \infty}\sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G') = \sum_{H \subset G'}\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, H))$$ exists as well. In particular, there exists a bound $D_{G'} > 0$ independent of $n$ such that $$\sum_{B \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq B) \#\mathrm{Hom}(B, G') \leq D_{G'}.$$ Thus, for any fixed $H \in \mathcal{A}_{m},$ we have $$\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G') \leq D_{G'}.$$ This implies that $$\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \leq D_{G'}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')},$$ so $$\sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \leq D_{G'}\sum_{H \in \mathcal{A}_{m}}\frac{\#\mathrm{Hom}(H, G)}{\#\mathrm{Hom}(H, G')},$$ and by Claim the right-hand side converges as $n \rightarrow \infty.$ Thus, by the Lebesgue dominated convergence theorem, the left-hand side also converges and $$\begin{align*}&\lim_{n \rightarrow \infty}\sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \\ &= \sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G),\end{align*}$$ where we used our assumption that the limit $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists (which will be proven below). Since $$\#\mathrm{Hom}(H, G) = \sum_{B \subset G}\#\mathrm{Sur}(H, B)$$ where the sum is over subgroups $B$ of $G,$ subtracting the corresponding quantities for all subgroups of $G,$ we obtain $$\begin{align*}&\lim_{n \rightarrow \infty}\sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) \\ &= \sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G).\end{align*}$$ A similar computation as above gives $$\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) = \sum_{H \in \mathcal{A}_{m}}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G),$$ so $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G))$$ by the hypothesis. This finishes Step 1.

Step 2. We assume that $\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$ and $\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$ both converge as $n \rightarrow \infty$ for every $H \in \mathcal{A}_{m}.$  and $$\begin{align*}&\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) \\ &= \sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G)\end{align*}$$ for every $G \in \mathcal{A}_{m}.$ By Step 1, both sides are equal to $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) = \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(Y_{n}, G)).$$

Then we show that $$\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) = \lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$

Using a similar argument as in Step 1, we have $$\begin{align*}&\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \\ &= \sum_{B \subset G}\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Sur}(H, G) \\ &= \sum_{B \subset G} \lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)),\end{align*}$$ where $B$ ranges over subgroups of $G.$  We can also deduce a similar identity for $(Y_{n})_{n \geq 1}.$

We need two lemmas to finish Step 2, and this is where we need $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, G)) \leq |\wedge^{2}G|.$$

Lemma 1. Let $\lambda = (\lambda_{1} \geq \cdots \geq \lambda_{l})$ be a partition and $$G = \mathbb{Z}/p^{\lambda_{1}}\mathbb{Z} \times \cdots \times \mathbb{Z}/p^{\lambda_{l}}\mathbb{Z}.$$ Then $$\sum_{B \subset G}|\textstyle\bigwedge^{2}G| \leq F^{\lambda_{1}}p^{\sum_{i=1}^{\infty}\lambda'_{i}(\lambda'_{i}-1)/2},$$ where $B$ ranges over subgroups of $G$ and $$F = \frac{2}{1 - 2^{-1/8}} \prod_{i=1}^{\infty}\frac{1}{1 - 2^{-i}}.$$

Remark. For a later use, we note that the number of parts of $\lambda$ in Lemma 1 is at most $\lambda_{1}.$

For the next lemma, we need more notation. Fix any distinct primes $p_{1}, \dots, p_{s}$ and integers $e_{1}, \dots, e_{k} \geq 1.$ Let $M_{j}$ be the set of partitions with at most $e_{j}$ parts, and set $M_{(e_{1}, \dots, e_{k})} = M := M_{1} \times \cdots \times M_{k}.$ Given any $\mu \in M,$ we may write $$\mu = (\mu^{(1)}, \dots, \mu^{(k)}).$$ Since $\mu^{(j)} \in M_{j}$, we write $\mu^{(j)} = (\mu^{(j)}_{1} \geq \cdots \geq \mu^{(j)}_{e_{j}}),$ where $\mu^{(j)}_{i} \geq 0.$

Lemma 2. Assume the above notation. For each $\lambda \in M,$ suppose that we have $x_{\lambda}, y_{\lambda} \in \mathbb{R}_{\geq 0}$ such that $$\sum_{\mu \in M}x_{\mu}\prod_{j=1}^{k}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}\mu_{i}^{(j)}} = \sum_{\mu \in M}y_{\mu}\prod_{j=1}^{k}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}\mu_{i}^{(j)}} \leq \prod_{j=1}^{s}F^{e_{j}}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}(\lambda_{i}^{(j)}-1)/2},$$ for some constant $F > 0$ that does not depend on $\lambda.$ Then we have $x_{\lambda} = y_{\lambda}$ for all $\lambda \in M.$

Going back to Step 2, prime factorize $m = p_{1}^{e_{1}} \cdots p_{k}^{e_{k}}$ with $e_{j} \geq 1.$ Then there is a bijection between $M_{(e_{1}, \dots, e_{k})} = M$ and $\mathcal{A}_{m}$ by mapping each $$\lambda = (\lambda^{(1)}, \dots, \lambda^{(k)}) \in M$$ to a finite abelian group, whose $p_{j}$-part corresponds to the conjugate of the partition $\lambda^{(j)}$ for $1 \leq j \leq k.$ (This is where we need abelian groups in $\mathcal{A}_{m}$ to be finite.) What's important here is that if $G, H \in \mathcal{A}_{m}$ correspond to $\lambda, \mu \in M$ respectively under this bijection, then $$\prod_{j=1}^{k}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}\mu_{i}^{(j)}} = \#\mathrm{Hom}(H, G).$$ Now, if we let $$x_{\lambda} := \lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ and $$y_{\lambda} := \lim_{n \rightarrow \infty}\mathrm{Prob}(Y_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H),$$ then to finish Step 2 with the help of Lemma 2, we only need to show that $$\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) \leq \prod_{j=1}^{s}F^{e_{j}}p_{j}^{\sum_{i=1}^{\infty}\lambda_{i}^{(j)}(\lambda_{i}^{(j)}-1)/2},$$ for some constant $F > 0$ that does not depend on $G \in \mathcal{A}_{m}.$ The left-hand side is $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Hom}(X_{n}, G)) = \sum_{B \subset G}\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, B)),$$ where $B$ varies over subgroups of $G,$ and we now use the assumption that $$\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, B)) \leq |\wedge^{2} B|,$$ which gives us $$\begin{align*}\sum_{H \in \mathcal{A}_{m}}\lim_{n \rightarrow \infty}\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \#\mathrm{Hom}(H, G) &= \sum_{B \subset G}\lim_{n \rightarrow \infty}\mathbb{E}(\#\mathrm{Sur}(X_{n}, B)) \\ &\leq \sum_{B \subset G} |\wedge^{2}B| \\ &\leq \prod_{j=1}^{k} F^{e_{j}}p^{\sum_{i=1}^{\infty}\lambda'_{i}(\lambda'_{i}-1)/2},\end{align*}$$ where $$F = \frac{2}{1 - 2^{-1/8}} \prod_{i=1}^{\infty}\frac{1}{1 - 2^{-i}}.$$ We used Lemma 1 along with how direct sums work with wedge products. This finishes Step 2.

Remark. Step 2 is the most nontrivial step, and Lemma 1 and Lemma 2 are where interesting bounds are analyzed. Lemma 2 seems to be where the nontriviality lies. We will cover these two lemmas in a different post, and think about whether the bound given here can be improved.

 Step 3. We show that $\mathrm{Prob}(X_{n} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$ converges as $n \rightarrow \infty.$

Since the sequence in question is bounded in $[0, 1],$ by the Bolzano--Weierstrass theorem, there must be a subsequence $X_{n_{1}}, X_{n_{2}}, X_{n_{3}}, \dots$ such that the limit $$L = \lim_{i \rightarrow \infty}\mathrm{Prob}(X_{n_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists for any $H \in \mathcal{A}_{m}.$

For contradiction, suppose that the sequence in question does not converge for some $H \in \mathcal{A}_{m}.$ This means that there exists $\epsilon > 0$ and a subsequence $X_{m_{1}}, X_{m_{2}}, X_{m_{3}}, \dots$ such that $$|\mathrm{Prob}(X_{m_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) - L| \geq \epsilon.$$ We may refine this subsequence by the Bolzano--Weirstrass theorem to assume that the limit $$\lim_{i \rightarrow \infty}\mathrm{Prob}(X_{m_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H)$$ exists. The contradiction comes from the fact that $$\lim_{i \rightarrow \infty}\mathrm{Prob}(X_{m_{i}} \otimes_{\mathbb{Z}} \mathbb{Z}/m\mathbb{Z} \simeq H) \neq L,$$ because if we go through Step 1 and Step 2 with these two subsequences, Step 2 tells us that the two limits must be equal. This finishes the proof of Step 3.