Let $G$ be a group and $H$ a subgroup. Consider the ascending chain of iterated normalizers: $$ H \trianglelefteq N_G(H) \trianglelefteq N_G(N_G H) \trianglelefteq \cdots \trianglelefteq N^{(k)}_G(H) \trianglelefteq \cdots. $$ Is there an example where all the terms are finite, but the chain fails to stabilize?
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2$\begingroup$ Of course, Wielandt's Theorem implies that the chain stabilizes if $C_{G}(H) = 1$. $\endgroup$– Simon ThomasCommented Sep 1, 2011 at 13:15
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1 Answer
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Can't we just take $G$ to be the union of the chain $H_1<H_2<\cdots$, where $H_1$ is a Klein 4-group and $H_i$ is the dihedral group of order $2^{i+1}$? Then $N_{G}(H_i) = H_{i+1}$.
So
$G = \langle x_i (i \ge 0), y \mid x_1^2=1, x_{i+1}^2=x_i (i\ge 1), y^2=1, (yx_i)^2 = 1 (i \ge 1) \rangle$,
where $H_i$ is the subgroup $\langle x_i, y \rangle$.
answered Sep 2, 2011 at 9:11
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$\begingroup$ For your idea, it seems one can consider a slightly bigger group, $\hat{G} = O(2) = U(1) \rtimes_{c} C_2$, where $c$ is complex-conjugation, and the subgroup $H = D_2 = O(1) \rtimes_{c} C_2$. However $H \triangleleft \hat{G}$; hence $H \triangleleft G$. $\endgroup$ Commented Sep 5, 2011 at 1:49
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$\begingroup$ But do you have a problem with the example I gave? $\endgroup$ Commented Sep 5, 2011 at 8:54
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$\begingroup$ Maybe I'm misunderstanding, but I got $H_3 \subseteq N_G(H_1)$ ... $\endgroup$ Commented Sep 6, 2011 at 0:43
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$\begingroup$ The normalizer of a Klein 4-group in a dihedral group of order $16$ has order 8, so $H_1$ is not normal in $H_3$. I have added a more precise definition of the groups. $\endgroup$ Commented Sep 6, 2011 at 10:07
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$\begingroup$ I believe that "$i \geq 0$" should be "$i \geq 1$" in the edit. Note: [ x_3^{-1} x_1 x_3 = x_3^{-1} x_3^4 x_3 = x_3^4 = x_1. ] $\endgroup$ Commented Sep 10, 2011 at 20:53