For finite groups $L \leq K$ we define $d(L,K)$ to be the least $n \in \mathbb{N}$ for which there exist $a_1, \dots, a_n \in K$ such that $\langle L, a_1 \dots, a_n \rangle = K$.

Is there some $m \in \mathbb{N}$ for which the following claim holds:

Let $G$ be a finite group and let $H \leq G$ be a subgroup with $d(H,G) = 1$. Suppose that $G_0$ is a subgroup of index $2$ in $G$ such that $H \leq G_0 \leq G$. Then $d(H,G_0) \leq m$.

  • $\begingroup$ "Suppose that $H\leq G_0\leq G$ is a subgroup of index $2$..." You mean "suppose that $G_0$, $H\leq G_0\leq G$, is a subgroup of order $2$", right? $\endgroup$ Apr 18, 2015 at 22:22
  • $\begingroup$ @ArturoMagidin I mean that $G_0$ is a subgroup of index $2$ in $G$, and that $G_0$ contains $H$. $\endgroup$
    – Pablo
    Apr 18, 2015 at 22:32
  • $\begingroup$ right: it's just that as you phrased it, the subject of the sentence is "$H\leq G_0\leq G$", for which "is a subgroup" makes no sense. $\endgroup$ Apr 18, 2015 at 22:32

1 Answer 1


I don't think so.

Let $K$ be any group, and let $e\neq t\in C_2$ act on $K\times K$ by exchanging coordinates. Let $G=(K\times K)\rtimes \langle t\rangle$, and let $H$ be generated by the elements $(k,1)$ with $k\in K$. Then $\langle H,t\rangle = G$, so $d(H,G) = 1$.

Now let $G_0 = K\times K$. Then $d(H,G_0)=d(1,K)$, and that can be arbitrarily large by a suitable choice of $K$.

  • 1
    $\begingroup$ Of course, this is just the wreath product $K\wr C_2$; you can see examples with arbitrary index for $G$ by taking $K\wr C_n$ instead. $\endgroup$ Apr 19, 2015 at 2:55

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