Is this relation about elementary embedding transitive? For ordinals $\alpha<\beta$, we say $\alpha<_{el}\beta$, if there is an elementary embedding with domain $L_\beta$ and critical point $\alpha$.
Is $<_{el}$ transitive?
 A: EDIT: If the codomain $N$ is allowed to be illfounded, then the answer is yes.
(Here if $\kappa=\mathrm{crit}(j)$ then this will mean that $\kappa\subseteq N$, but it might be that $N$ is illfounded and $\kappa$ is exactly the wellfounded part of the codomain, in which case $\kappa\notin N$.)
For let $\kappa_0<\beta_0=\kappa_1<\beta_1$ be ordinals and $j_i:L_{\beta_i}\to N_i$ be elementary (for $i=0,1$) with $\mathrm{crit}(j_i)=\kappa_i$. Let $U_0$ be some non-principal ultrafilter derived from $j_0$ (let $x\in N_0$ such that $N_0\models$"$\alpha\in x\in j_0(\kappa_0)$" for each $\alpha<\kappa_0$, and let $U_0$ be the filter over $\kappa_0$ derived from $x$). Then $U_0$ is an $L_{\beta_0}$-ultrafilter over $\kappa_0$ which is $L_{\beta_0}$-$\kappa_0$-complete (i.e. closed under length ${<\kappa_0}$-intersections of sequences $\left<X_\alpha\right>_{\alpha<\gamma}\in L_{\beta_0}$). And because $\kappa_1=\mathrm{crit}(j_1)$, $\kappa_1$ is a (regular) cardinal in $L_{\beta_1}$. Therefore $\mathcal{P}(\kappa_0)\cap L_{\beta_1}\subseteq L_{\beta_0}=L_{\kappa_1}$. But then $U_0$ is also an $L_{\beta_1}$-ultrafilter over $\kappa_0$ which is $L_{\beta_1}$-$\kappa_0$-complete. So letting $N=\mathrm{Ult}(L_{\beta_1},U_0)$ and $j:L_{\beta_1}\to N$ the ultrapower map, then $j$ is elementary and $\mathrm{crit}(j)=\kappa_0$.

EDIT 2: On the other hand, if the codomain is required to be wellfounded, and $0^\sharp$ exists, the answer is no. For let $\kappa_0$ be the least $L$-indiscernible and let $\pi:L_{\omega_1}\to L_{\omega_1}$ be elementary with $\mathrm{crit}(\pi)=\kappa_0$. (Here $\omega_1$ is as computed in $V$.)
There is an elementary $\sigma:L_{\kappa_0}\to L_{\kappa_0}$. For there is an elementary $\sigma':L_{\kappa_\omega}\to L_{\kappa_\omega}$ where $\kappa_\omega$ is the $\omega$th $L$-indiscernible, for example the one with $\sigma'(\kappa_n)=\kappa_{n+1}$. But the existence of some such $\sigma'$ is forced over $L$ by $\mathrm{Coll}(\omega,\kappa_\omega)$, and so by indiscernibility, the existence of a $\sigma$ as stated is forced over $L$ by $\mathrm{Coll}(\omega,\kappa_0)$, and therefore there is actually some such embedding $\sigma$ in $V$.
So letting $\kappa=\mathrm{crit}(\sigma)$, then $\kappa<_{\mathrm{el}}\kappa_0<_{\mathrm{el}}\omega_1$. But I claim $\kappa\not<_{\mathrm{el}}\omega_1$.
For since $\kappa<\kappa_0$, $\kappa$ is not an $L$-indiscernible. But if $j:L_{\omega_1}\to L_{\lambda}$ is elementary (where $\lambda$ is some ordinal) then $\mathrm{crit}(j)$ is an $L$-indiscernible. (This is a standard fact. The key is that if $U$ is the normal ultrafilter derived from $j$, then $\mathrm{Ult}(L,U)$ is wellfounded.)

These observations leave open the version where the codomain is required to be wellfounded, but $0^\sharp$ does not exist.
