Consider the following two languages over the alphabet $\{a, b\}$: $$L_1 = \{ \alpha\beta\alpha \mid \alpha \in \{a, b\}^+ \text{ AND } \beta \in \{a, b\}^+ \}$$ $$L_2 = \{ \alpha\beta\alpha \mid \alpha \in \{a\}^+ \text{ AND } \beta \in \{a, b\}^+ \}$$ Which ONE of the following statements is CORRECT?

To determine the regularity of both languages:

1. Analysis of $L_2$:

   • In $L_2$, $\alpha \in \{a\}^+$, which means $\alpha = a^k$ for some $k \ge 1$.

   • Any string in $L_2$ has the form $w = a^k \beta a^k$ with $k \ge 1$ and $\beta \in \{a, b\}^+$.

   • Notice that if $k > 1$, we can rewrite $w = a(a^{k-1}\beta a^{k-1})a = a\beta^{\prime} a$, where $\beta^{\prime} = a^{k-1}\beta a^{k-1} \in \{a, b\}^+$.

   • Conversely, any string of the form $a \beta^{\prime} a$ with $\beta^{\prime} \in \{a, b\}^+$ can be generated by choosing $\alpha = a$ ($k = 1$) and $\beta = \beta^{\prime}$.

   • Therefore, $L_2 = a(a + b)^+a$, which is generated by a regular expression and recognized by a Finite Automaton. Thus, $L_2$ is a regular language.

2. Analysis of $L_1$:

   • In $L_1$, $\alpha \in \{a, b\}^+$ and $\beta \in \{a, b\}^+$.

   • Intersect $L_1$ with the regular language $R = a b^+ a b^+$:

     $$L_1 \cap R = \{ a b^m a b^n \mid m > n \ge 1 \}$$

     *(since any prefix $\alpha$ must match a suffix $\alpha$, the only suffix starting with $a$ is $a b^n$, requiring the prefix to be $a b^n$ and thus $m = n + |\beta| > n$)*.

   • By the Pumping Lemma, the language $\{ a b^m a b^n \mid m > n \ge 1 \}$ is non-regular.

   • Since regular languages are closed under intersection with regular languages, $L_1$ is not a regular language.

Hence, $L_1$ is not regular and $L_2$ is regular, which corresponds to option (C).