Raising the index of accessibility

Under GCH, if $\lambda < \mu$ are regular cardinals, then $\lambda \lhd \mu$ implies $\lambda \ll\mu$. The proof uses the following standard fact:

Lemma. Suppose $\lambda \leq \gamma$ are infinite cardinals. Then $\gamma^{<\lambda} = 2^{<\lambda} \cdot \text{cf}(P_\lambda(\gamma))$.

Proof. Recall that since $\lambda \leq \gamma$, $ |P_\lambda(\gamma)| = \gamma^{<\lambda}$. The inequality $\gamma^{<\lambda} \geq 2^{<\lambda} \cdot \text{cf}(P_\lambda(\gamma))$ comes from the fact that $\gamma^{<\lambda} = |P_\lambda(\gamma)|\geq \text{cf}(P_\lambda(\gamma))$. Let us prove the reverse inequality. Let $C\subseteq P_\lambda(\gamma)$ be a cofinal set of cardinality $\text{cf}(P_\lambda(\gamma))$. Note that $P_\lambda(\gamma) = \bigcup_{\sigma\in C} P(\sigma)$, so $\gamma^{<\lambda} = |P_\lambda(\gamma)| \leq |C|\cdot \sup_{\sigma\in C}|P(\sigma)| \leq \text{cf}(P_\lambda(\gamma)) \cdot 2^{<\lambda}$, as desired.

Under GCH, this lemma has the following corollary:

Corollary (GCH). Suppose $\lambda \leq \gamma$ are infinite cardinals and $\lambda$ is regular. Then $\gamma^{<\lambda} = \text{cf}(P_\lambda(\gamma))$.

Proof. By the lemma, it suffices to show that $\text{cf}(P_\lambda(\gamma))\geq 2^{<\lambda}$. By GCH, $2^{<\lambda} =\lambda$. But $\text{cf}(P_\lambda(\gamma)) \geq \text{cf}(P_\lambda(\lambda)) \geq \lambda$ since $\lambda$ is regular.

Proposition (GCH). If $\lambda < \mu$ are regular cardinals, then $\lambda \lhd \mu$ implies $\lambda \ll\mu$

Proof. Suppose $\lambda < \mu$ are regular cardinals such that $\lambda \lhd \mu$. In other words $\text{cf}(P_\lambda(\mu')) < \mu$ for all $\mu' < \lambda$. By the corollary, $(\mu')^{<\lambda} < \mu$ for all $\mu' < \mu$. In particular, $(\mu')^{\lambda'} < \mu$ for all $\mu' < \mu$ and $\lambda' < \lambda$. Thus $\lambda \ll\mu$.

More generally, the following is Fact 2.5 of Lieberman, Rosický, and Vasey - Internal sizes in $\mu$-abstract elementary classes:

Theorem. Assume $\lambda$ and $\mu$ are regular cardinals and $2^{<\lambda} < \mu$. Then $\lambda \triangleleft \mu$ if and only if $\lambda \ll \mu$.

Assuming GCH, $2^{<\lambda} = \lambda$, so we recover Gabe's answer (the proof is the same).