Does the critical sequence for subalgebras of elementary embeddings with finitely many generators have order type $\omega$?

Question

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.

Answer 1

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 $\omega$.