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Let’s for any bijection $f: A \to A$ define its support as $$supp(f) = \{a \in A| f(a) \neq a\}$$

Now, let’s define $S_\infty$ as the group of all bijections $\mathbb{N} \to \mathbb{N}$ with finite support. By Cayley Theorem any finite group is isomorphic to a subgroup of $S_\infty$. Therefore, for any finite group $G$ we can define its Cayley length as

$$len_c (G) = \min \{\sum_{\alpha \in A} |supp(\alpha)| | A \subset S_\infty \langle A \rangle \cong G \}$$

Now, we can define a following function:

$$CL(n) = \max \{len_c(G) | |G| \leq n \}$$

What is the asymptotic of $CL$?

I managed to derive the following two bounds:

$$CL(n) = O(n \log(n))$$

This is because any finite group $G$ has a generating set of size $O(\log(n))$ and the size of supports of permutations, corresponding to each of those generators under left multiplicative action is $n$.

$$CL(n) = \Omega(n)$$

Suppose $p$ is prime. Then $len_c(C_p) = p$. Indeed, all non-trivial elements of $C_p$ have order $p$, any permutation of order $p$ has size of support dividing $p$.

However, I do not know, whether any of those bounds is tight...

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    $\begingroup$ Why do you say $CL(n)\geq n$? For example isn't $CL(6)=5$? More generally, if $n=pq$, with $p\leq q$ primes, then I think $CL(n)=n-1$ if $p$ divides $q-1$, and $CL(n)=p+q$ otherwise. In particular, in the latter case, $CL(n)$ will be much smaller than $n$. You can generalise this to product of more than two distinct primes. If none of them is congruent to $1$ mod any of the other, then $CL(n)$ will be the sum of the primes, so much smaller than $n$. $\endgroup$ – verret Sep 15 '20 at 1:41
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    $\begingroup$ Conversely, if $n$ is a prime power, then $CL(n)\geq n$, as witness by the cyclic group of order $n$. I think generally, $CL(n)$ will be larger than this. For example, I think $CL(8)>8$, as witness by the quaternion group. So sometimes $CL(n)$ is much smaller than $n$, sometimes it is larger than $n$, depending very much on the factorisation of $n$. Given that the function is not very smooth, it's not clear what you mean by the asymptotics... $\endgroup$ – verret Sep 15 '20 at 1:42
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    $\begingroup$ @verret, $CL$ is taken as a maximum over all $|G| \leq n$. Therefore monotonously grows. I would like to know how fast does it grow. $\endgroup$ – Yanior Weg Sep 15 '20 at 5:26
  • $\begingroup$ Ah, I missed that, I thought it was just for groups of order $n$... $\endgroup$ – verret Sep 15 '20 at 5:51
  • $\begingroup$ What is your argument that shows $CL(n)$ is in $O(n)$? $\endgroup$ – verret Sep 15 '20 at 5:53
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Actually, $CL(n) = \Theta(n)$. Proof of $CL(n) = \Omega(n)$ can be found in the body of the question. To prove the bound $CL(n) = O(n)$ we will construct a "good" Cayley representation (a Cayley representation of $G$ is here a collection of permutations that generate a group isomorphic to $G$) for arbitrary group $G$ using the follwing recursive procedure:

Base: if $G \cong E$ then we do not need any permutations at all.

Step: Suppose, we have already done this for all groups of order less than $|G|$ and that for them all our Cayley representations satisfy the additional requirement of all generating permutations of any group $K$ being from $Sym(K)$. Now, suppose $H$ is some maximal normal subgroup of $G$ which is represented py permutations $p_1, ... , p_t$ from $Sym(H)$. Then $\frac{G}{H}$ is a simple group. Thus, as all simple groups are $2$-generated, we can take elements $g_1, g_2$ such that $\langle H \cup \{g_1, g_2\} \rangle = G$. Then $G$ can be represented by permutations $p_1, ... , p_t, (h \mapsto g_1 h), (h \mapsto g_2 h)$ from $Sym(G)$.

Now, let's demonstrate that the length (i.e. the sum of sizes of supports of all generating permutations) of Cayley representation constructed that way does not exceed $4|G|$:

Base: If $G \cong E$, then $0 \leq 4$

Step: If the inequality holds for every group of order less then $G$, then the length of the corresponding presentation for $H$ is $\leq 4|H| \leq 2|G|$. On the other hand the lengths of permutations $ (h \mapsto g_1 h)$ and $(h \mapsto g_2 h)$ are $|G|$ each. Thus the total length of this Cayley representation for $G$ $\leq 4|G|$.

Thus we can conclude, that $CL(n) \leq 4n$.

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