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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

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    $\begingroup$ Yes, this is possible, and for arbitrary large difference. A (random) example would be the group $G=\langle (2,16,9,5,3)(4,15,8,13,7)(6,14,10,12,11), (1,6)(2,5)(3,8)(4,7)(9,14)(10,13)(11,16)(12,15)\rangle\cong C_2^4\rtimes C_5$ which has.a normal subgroup $N\cong C_2^4$ that requires $4$ generators. $\endgroup$
    – ahulpke
    Nov 12, 2023 at 4:39
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    $\begingroup$ You need to make it clearer what you are asking. The questions in the title and in the first paragraph seem to be different. Although you have been given sensible answers, the answer to the question in the first paragraph is trivially yes, even in a cyclic group, because we could take $B=N$. Also "minimal generating" is notoriously ambiguous. $\endgroup$
    – Derek Holt
    Nov 12, 2023 at 6:15
  • $\begingroup$ @ahulpke Even short answers should be posted as answers: comments can be deleted at any time, and it'd be a shame to lose such a clear and concise answer. $\endgroup$
    – wizzwizz4
    Nov 13, 2023 at 1:38

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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.

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  • $\begingroup$ Didn't think I would see an affine group here!! It is not entirely the same but the geometric intuition I have gotten is that $\mathbb{Z}^2$ is two dimensional over $\mathbb{Z}$ but the special affine orthogonal group over $\mathbb{Z}^2$ only needs 1 translation since the rotation along with translation lets you translate into any direction. $\endgroup$
    – Leon Kim
    Nov 12, 2023 at 10:08

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