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. A paper I have been looking at mentions the subgroup $H = \langle (12)(34) \rangle \cong \mathbb{Z}_2$ of the Alternating group $A_5$ as a counterexample. I want to verify this claim explicitly. I have shown that $(12)(345) \in P_{A_5}(H)$ as well as $(45)\in P_{A_5}(H)$. I want to show that $P_{A_5}(H)$ fails to be a subgroup of $A_5$ but I'm unable to get the desired result as counterexample.
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2$\begingroup$ Neither $(1\,2)(3\,4\,5)$ nor $(4\,5)$ lies in $A_5$. $\endgroup$– LSpiceMar 18, 2017 at 14:31
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1$\begingroup$ Let $S$ be the $8$ elements of order $3$ that permute the set $\{1,2,3,4\}$. Then, in your example, $P_G(A_5) = A_5 \setminus S$, so it has $52$ elements and cannot be a subgroup. That's because, for all $g \in A_5 \setminus S$, $H$ and $H^g$ are Sylow $2$-subgroups of $\langle H,H^g \rangle$. $\endgroup$– Derek HoltMar 18, 2017 at 16:16
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$\begingroup$ @DerekHolt : I think there is a small typo in your comment. $\endgroup$– Geoff RobinsonMar 19, 2017 at 17:09
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$\begingroup$ @GeoffRobinson Yes, replace $P_G(A_5)$ by $P_G(H)$. Thanks! $\endgroup$– Derek HoltMar 19, 2017 at 17:20
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