Suppose that $\lambda$ is a cardinal. Let $\mathcal{E}\_{\lambda}$ be the set of all elementary embeddings from $V\_{\lambda}$ to $V\_{\lambda}$. If $j,k\in\mathcal{E}\_{\lambda}$, then define
$j[k]=\bigcup\_{\alpha<\lambda}j(k|\_{V\_{\alpha}})$. Suppose that $j\_{1},...,j\_{n}\in\mathcal{E}\_{\lambda}$. Let $\langle j\_{1},...,j\_{n}\rangle$ denote the smallest subset of $\mathcal{E}\_{\lambda}$ that contains $\{j\_{1},...,j\_{n}\}$ and is closed under the operation $(j,k)\mapsto j[k]$. Let $K=\{\mathrm{crit}(j)|j\in\langle j\_{1},...,j\_{n}\rangle\}$. 


1. Then does $K$ necessarily have order type $\omega$?


If $\gamma<\lambda$ is a limit ordinal, then let $\equiv^{\gamma}$ to be the equivalence relation on $V\_{\lambda}$ where $j\equiv^{\gamma}k$ if and only if $j(x)\cap V\_{\gamma}=k(x)\cap V\_{\gamma}$ whenever $x\in V\_{\gamma}$. Then $\equiv^{\gamma}$ is a congruence on $\mathcal{E}\_{\lambda}$.


2. Is the quotient algebra $\langle j\_{1},...,j\_{n}\rangle/\equiv^{\gamma}$ finite whenever $\gamma$ is a limit ordinal with $\gamma<\lambda$?


An affirmative answer to question 2 will give an affirmative answer to question 1 as well.



Does the critical sequence for subalgebras of elementary embeddings with finitely many generators have order type $\omega$?
I claim that the answer to both questions is **yes**
.
Suppose that $A$ is a finite set, $j\_{a}\in\mathcal{E}\_{\lambda}$ for each $a\in A$ and $\gamma<\lambda$ is a limit ordinal. Then I claim that $\langle\{j\_{a}|a\in A\}\rangle/\equiv^{\gamma}$ is finite and the sequence $\{\textrm{crit}(j)|j\in\langle\{j\_{a}|a\in A\}\rangle\}$ has order type $\omega$.


Let $M$ be the set of all strings $a\_{1}...a\_{n}$ in $A^{+}$ so that either $n=1$ or if $m<n$, then $\textrm{crit}(j\_{a\_{1}}[j\_{a\_{2}}]...[j\_{a\_{m}}])<\gamma$. Then $M$ is finite; if $M$ were infinite, then the tree $M$ would have an infinite branch consisting of all substrings of $(a\_{n})\_{n=1}^{\infty}$, but for such an infinite branch, the sequence of elementary embeddings $(j\_{a\_{1}}[j\_{a\_{2}}]...[j\_{a\_{n}}])\_{n=1}^{\infty}$ contradicts the Laver-Steel theorem.


Let $F$ be the set of all maximal strings in $M$ (in other words, $x\in F$ if and only if $xa\not\in F$ whenever $a\in A$) and let $L=M\setminus F$. Then define a binary operation $\*$ on $M$ according to the following rules:


i. $x\*y=y$ whenever $x\in F$


ii. $x\*a=xa$ whenever $x\in L,a\in A$


iii. $x\*y=(x\*z)\*xa$ whenever $y=za$ and $x\in L$.


One can show that $\*$ produces a well-defined total operation on $M$ the same way that one can show without resorting to large cardinals that the Laver table operation operation on $\{1,...,2^{n}\}$ is well-defined. Indeed, the structure $(M,\*)$ is a generalization of the notion of a Laver table (the structure $(M,\*)$ is self-distributive although we shall not use the self-distributivity of $(M,\*)$ in this answer).


Define a mapping $\Phi:(M,\*)\rightarrow(\mathcal{E}\_{\lambda}/\equiv^{\gamma})$ by letting
$\Phi(a\_{1}...a\_{n})=j\_{a\_{1}}[j\_{a\_{2}}]...[j\_{a\_{n}}]$ whenever $a\_{1}...a\_{n}\in M$. Then one can show that $\Phi(x)\*\Phi(y)=\Phi(x\*y)$ by a double induction which is descending on the length of $x$ and for each individual $x$ one uses an induction which is ascending on the length of $y$. In other words, $\Phi$ is a homomorphism. However, we have $\Phi[M]=\langle\{j\_{a}|a\in A\}\rangle/\equiv^{\gamma}$ since the algebra $M$ is generated by $A$. Therefore the algebra $\langle\{j\_{a}|a\in A\}\rangle/\equiv^{\gamma}$ is a finite algebra.


Since each algebra $\langle\{j\_{a}|a\in A\}\rangle/\equiv^{\gamma}$ is finite, the set $\{\textrm{crit}(j)|j\in\langle\{j\_{a}|a\in A\}\rangle\}\cap\gamma$ is finite. Therefore, the set
$\{\textrm{crit}(j)|j\in\langle\{j\_{a}|a\in A\}\rangle\}$ has order type $\omega$.


