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I am having a hard time proving the following with pumping lemma: Is $L_2 = \{a^n \mid \text{$n$ is a product of one or more primes}\}$ regular?

Here's what I have so far:

  • Suppose $L$ is regular, and let $m$ be the constant pumping lemma.

  • Let then $w = a^m$ where $m$ is a product of at least one prime number, therefore $w$ is in the language $L_2$ and $|w| \geq m$.

  • Then $w = xyz$, where $|xy| \leq m$ and $|y| \geq 1$.

  • Then $y = a^j$ where $j \geq 1$.

This is where I'm not sure what to do: Pumping down we get $w_2 = a^{m-j}$

I don't know where to keep going from here since I don't even know if pumping down is the right thing to do.

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    $\begingroup$ Can you describe some words which are not in L2? $\endgroup$ – Yuval Filmus Oct 14 '19 at 13:17
  • $\begingroup$ I suppose $a$ (i.e., $n=1$) would be one such, since nowadays most people would call 1 a unit, not a prime. $\endgroup$ – Rick Decker Oct 14 '19 at 23:41
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According to the Fundamental Theorem of Arithmetic, any integer $>1$ can be written as a product of one or more primes (in a unique way). So, it seems that your language can be simplified as $\{a^n\mid n\geq 2\}$.

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As all words of length $>1$ and only consisting of a's should be contained in L2, there is a simple finite automaton that recognizes it. So your attempt at using the pumping lemma is futile, as the pumping lemma only helps you prove that a language is irregular if it is, and doesn't tell you anything about languages that are regular.

Maybe I'm also understanding your description of L2 wrong, in that case please add a comment or edit your question.

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