Prove that the following language L over alphabet $\{1\}$ is not regular. $L = \{w \mid |w| = k, \text{ where } k \text{ is a prime number}\}$
Suppose the language is regular for contradiction. Since $L$ is regular, we apply the pumping lemma. Let $p$ be given by PL such that the decomposition of the string $w = xyz$:
for each $i \geq 0$, $xy^iz \in L$;
$|y| > 0$;
$|xy| \leq p$.
There are infinitely prime numbers, then there exists a string $w \in L$, such that $|w| = k$ is the first prime number $> p$. Hence $w \in L$ and $|w| > p$ and we can decompose $w = xyz$ in order to satisfy the above conditions. Consider $v = xy^{k+1}z$, by 1) above $|v|$ must be a prime. But $|v| = k(1+|y|)$. But $k$ divides $k(1 + |y|)$ and $k > 1$. Hence $|v|$ is not a prime. We reach a contradiction. $L$ is not regular.
How can I use 2) and 3) conditions?