Number of ways of factoring a group

Question

I am looking for an example of a finite group $G$ for which number of ways to factor it is very high. For example I can factor $C_{15} = C_3 \times C_5$. I am looking for an example of group for which number of ways of factoring that group is high. I want an example of group of size $n$ for which number of ways to decompose that are superpolynomial in $n$.

Answer 1

Consider $G = C_2^k$, which has order $n=2^k$. It has at least $2^{k^2/4}$ subgroups of order $2^{k/2}$ (let's suppose $k$ is even for simplicity), and each of those has a comparable number of complements, so we have $\Omega(2^{k^2/2}) = \Omega(n^{t\log n})$ factorizations for some $t>0$, which is superpolynomial in $n$.

Answer 2

Take advantage of the fact$^\dagger$ that $\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn} \iff \gcd(m, n) = 1$. For intuition-building, let's suppose you have $\displaystyle n = \prod_k p_k$ for distinct primes $p_k$. Count the number of ways you can distribute this collection of $p_k$ into "boxes" $\{B_j\}$ where each box gets at least one prime, and let $l_j$ denote the product of the primes in the box $B_j$. Notice that the set of integers $\{l_j\}$ are mutually coprime, so given such a distribution, we can write $\mathbb{Z}_n \cong \displaystyle \prod_{j} \mathbb{Z}_{l_j}$. In other words, there's a bijection between the number of distinct ways you can distribute those primes into boxes and the number of ways you can factor $\mathbb{Z}_n$. Also, think about what you want "distinct factorings" to mean; this is to say, is $A \times B$ considered the same as $B \times A$? Whether the answer is 'yes' or 'no' won't drastically change the above reasoning, but you'll want to take it into account.

Now generalize the idea for arbitrary $n \in \mathbb{N}$.

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$^\dagger$From this fact quickly follows the more general statement $\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \cdots \times \mathbb{Z}_{n_k} \cong \mathbb{Z}_{n_1n_2...n_k} \iff \operatorname{gcd}(n_k, n_j) = 1$ whenever $k \neq j$. I provide a proof here. This helps motivate the fundamental theorem of finitely generated abelian groups, which gives two "canonical" ways to express finite abelian groups in particular as a direct product of finite cyclic groups.