Suppose language $L = \{\,a^{i} b^{k} : k \text{ divides } i\,\}$.
Some strings in $L$ include …
- $\,a^{0} b^{1} = b \in L\,$ since $1 \text{ divides } 0$
- $\,a^{1} b^{1} = ab \in L\,$ since $1 \text{ divides } 1$
- $\,a^{2} b^{1} = aab \in L\,$ since $1 \text{ divides } 2$
- $\,a^{0} b^{2} = bb \in L\,$ since $2 \text{ divides } 0$
- $\,a^{2} b^{2} = aabb \in L\,$ since $2 \text{ divides } 2$
- $\,a^{4} b^{2} = aaaabb \in L\,$ since $2 \text{ divides } 4$
- $\,a^{0} b^{3} = bbb \in L\,$ since $3 \text{ divides } 0$
- $\,a^{3} b^{3} = aaabbb \in L\,$ since $3 \text{ divides } 3$
- $\,a^{6} b^{3} = aaaaaabbb \in L\,$ since $3 \text{ divides } 6$
- $\,\,\,\,\ldots\,\text{plus infinitely many additional strings that are members of }L\,\ldots\,\,\,$
Is $L$ a regular language?
Let $K = \{b^k : k \gt 0\} = \{b,bb,bbb,bbbb,\ldots\}$. One may establish that $K$ is regular by (for example) constructing a deterministic finite accepter $M$ such that $K = L(M)$, the language accepted by $M$.
If $L$ were regular, then (by the closure of the family of regular languages under complementation and intersection) $L - K = L \cap \overline{K}$ would also be regular.
Next, apply the Pumping Lemma for Regular Languages to the infinite regular language $L - K$ and derive a contradiction. To derive a contradiction, one may consider $a^{\,p} \, b^{\,p}$ where $p$ is as described on Wikipedia.
Once a contradiction is reached, one may conclude that $L$ cannot be regular.