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The pronormaliser of a subgroup $H$ in a group $G$, denoted $P_G(H)$, is defined to be the set of elements of $G$ that pronormalise $H$. That is, $$P_G(H) = \{g \in G \; | \; \exists x\in \langle H, H^g \rangle\;\,\text{such that} \,\;H^x = H^g \}$$ However, the pronormaliser is not a subgroup of $G$ in general.

Lemma: Let $G$ be group with $H\leq G$. If $x \in P_G(H)$, then the cosets $N_G(H)x$ and $xN_G(H)$ are contained in $P_G(H)$.

Theorem: Let $G$ be a group and $K \leq G$. If there exists an ascendant subgroup $H$ of $G$ such that $K \leq H \subseteq P_G(K)$, then $P_G(K) = N_G(K)H = HN_G(K)$. In particular, $P_G(K)$ is a subgroup of $G$

Proof: By the above Lemma it is evident that $HN_G(K) \subseteq P_G(K)$ and $N_G(K)H \subseteq P_G(K)$.

Let $x \in P_G(K)$. Since $H$ asc $G$, there exists an ascending series given by $$ H =H_0 \unlhd H_1 \unlhd \ldots \unlhd H_\alpha \unlhd H_{\alpha +1} \unlhd \ldots \unlhd H_\beta = G.$$ Let $\alpha \leq \beta$ be the least such ordinal that $x \in H_\alpha N_G(K)$. Assume by contradiction that $\alpha > 0$. Now $\alpha$ is not a limit ordinal. To see this, if $\alpha$ was a limit ordinal, then by definition of the ascending series, $H_\alpha = \bigcup \limits_{ \tau < \alpha} H_\tau$, and thus $H_\alpha N_G(K) = \bigcup \limits_{ \tau < \alpha} H_\tau N_G(K)$. So $g\in H_\alpha N_G(K)$ implies that there is some $\tau < \alpha$ such that $x\in H_\tau N_G(K)$, contradicting the minimality of the choice of $\alpha$.

Therefore, $\alpha$ is a non-limit ordinal, and so there exist some unique $\lambda < \alpha$ such that $\alpha = \lambda + 1$. In other words, the ordinal $\alpha-1$ exists and $x \notin H_{\alpha - 1}N_G(K)$. We may express $x \in H_\alpha N_G(K)$ as $x=ha$, where $h \in H_\alpha$ and $a\in N_G(K)$.

We then obtain that $K^x = (ha)K(a^{-}h^{-1}) = hKh^{-1} = K^h$. Hence, $\langle K, K^x \rangle = \langle K, K^h \rangle \leq H \leq H_{\alpha -1}$, because $H_{\alpha -1} \unlhd H_\alpha$ . Now since $x \in P_G(K)$, there exists $y \in H_{\alpha -1}$ such that $K^y = K^x$. Then $y^{-1}x \in N_G(K)$, and so $x \in yN_G(K) \subseteq H_{\alpha -1}N_G(K)$, which is a contradiction. Hence, $a =0$, and therefore $x \in HN_G(K)$. From this, we have obtained the inclusion $P_G(K) \subseteq HN_G(K)$.

Question: How can I deduce that $P_G(K) \subseteq N_G(K)H$, because the author does not cover this case.

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    $\begingroup$ That's because $P_G(K)$ is closed under inversion, i.e. $x \in P_G(K) \Leftrightarrow x^{-1} \in P_G(K)$. $\endgroup$ – Derek Holt Mar 21 '17 at 11:46
  • $\begingroup$ Thanks. I have managed to prove your claim. I am having a problem using it in the argument. Let $x \in P_G(K) = HN_G(K)$, then $x= ha$ for some $h \in H$ and $a\in N_G(K)$. Now $x^{-1} = a^{-1}h^{-1} \in P_G(K)$. I know that $a^{-1}h^{-1} \in N_G(K)H$. How does $P_G(K) \subseteq N_G(K)H$ follow from this? $\endgroup$ – R Maharaj Mar 21 '17 at 14:16
  • $\begingroup$ I think you should be able to figure that out for yourself! $\endgroup$ – Derek Holt Mar 21 '17 at 14:22
  • $\begingroup$ Would this argument be valid? Since $P_G(K)$ is closed under inversion, $(P_G(K))^{-1} = P_G(K)$. Then using that $P_G(K) \subseteq HN_G(K)$, we have that $[P_G(K)]^{-1} \subseteq [HN_G(K)]^{-1}$ $\implies$ $P_G(K) \subseteq (N_G(K))^{-1}H^{-1}$. However, since $N_G(K)$ and $H$ are subgroups of $G$, $(N_G(K))^{-1}= N_G(K)$ and $H^{-1}=H$. Thus, $P_G(K) \subseteq N_G(K)H$ $\endgroup$ – R Maharaj Mar 21 '17 at 14:34

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