Let $G$ be a finite group, and $1 \lhd N \lhd G$. With $G = \langle A \rangle$ and $N = \langle B \rangle$ be minimal. Is it possible for $|B|>|A| $?
Main motivation behind this question was comparison between $A_n$ and $S_n$. Both groups can be generated via 2 elements, but in some sense, it is much harder to come up with 2 element generator for $A_n$ than $S_n$, and I am wondering if the inequality is actually possible to attain?
And of course we need the assumption that $G$ is finite because all sorts of funny business happens with infinite groups.
Edit: I have discovered that $D_8 \times C_2 < S_6$, if i throw normality condition away, it is possible
Let me generalize @ahulpke's example: Let $V$ be the additive group and $H$ be the multiplicative group of a finite field $\mathbb{F}_{p^n}$. Then $H$ acts on $V$ by multiplication and we can define $G=V\rtimes H$. It is well-known that $H$ is cyclic and acts transitively on $V\setminus\{0\}$. Hence, $G$ can be generated by two elements. On the other hand, the elementary abelian group $V$ cannot be generated by less than $n$ elements.