Let $G$ be a group with $H, D \leq G$ such that $D$ is an $H$-invariant subgroup and $D$ satisfies the normalizer condition ie. every subgroup of $D$ is properly contained in its normalizer. Suppose that every intermediate subgroup for $H$ coincides with its normalizer.

Let $d\in D$. Denote $K = \langle H,H^{d}\rangle$ and $B = D \cap K$. If $d\notin B$. Since $D$ satisifies the normalizer condition, $B_1 = N_D(B)\not= B $. It is also easy to see that $B_1$ is $H$-invariant

According to the paper, they define an ascending series: $$ B=B_0 \leq B_1 \leq \ldots \leq B_\alpha \leq B_{\alpha +1} \leq \ldots B_\gamma $$ by the following rule $B_{\alpha+1} = N_D(B_\alpha)$ for every $\alpha < \gamma$ and $B_\lambda = \bigcup\limits_{\lambda < \mu} B_\mu$ for a limit ordinal $\lambda$.

My question: What are limit ordinals and why have they defined the series in this way?

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    $\begingroup$ I am not sure why you are asking what are limit ordinals in a question tagged as group theory. They are more of a topic in set theory. You could start with the wikipedia page. As for why they have defined it that way, it is hard to know without knowing more context, but it s the natural definition to make if you want to consider an infinite ascending chain os subgroups. You would start with $B_0 < B_1 < B_2$ and then take the union $B_\omega$ of the $B_i$ over all $i \in {\mathbb N}$, but then you would naturally want to restart with $B_\omega$. $\endgroup$ – Derek Holt Mar 1 '17 at 8:10
  • $\begingroup$ Okay if $\lambda$ is a limit ordinal then for any $\beta < \lambda$ its successor $\beta +1$ is also less than $\lambda$. Using the defined sequence above, if for instance $\lambda = 3$, then $B_3 = B_1 \cup B_2$? $\endgroup$ – R Maharaj Mar 1 '17 at 10:59
  • $\begingroup$ No, $3$ is not a limit ordinal. $\endgroup$ – Derek Holt Mar 1 '17 at 11:50
  • $\begingroup$ I'm not sure how to use the definition of the limit ordinal. In the context of the above, $d$ belongs to some term of the series since $D = B_\gamma$. Is it because $B_\gamma = \bigcup \limits_{\mu < \gamma} B_\mu$? $\endgroup$ – R Maharaj Mar 1 '17 at 12:11
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    $\begingroup$ That depends on whether $\lambda$ is a limit ordinal. If so then yes. If not, then $\lambda = \mu+1$ for some ordinal $\mu$ and $B_\mu \lhd D$. $\endgroup$ – Derek Holt Mar 1 '17 at 22:03

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