As this question was closed as a duplicate of Existence of simultaneously normal finite index subgroups, I am opening a modified version here.

$G$ is a group, $A$ and $B$ are two subgroups of $G$. Suppose that $A∩B$ has finite index in both $A$ and $B$. It has been shown here that $A\cap B$ need not have a subgroup of finite index which is normal in both $A$ and $B$.

Question : If the set $A\cup B$ is normalised by $G$, does $A\cap B$ has a subgroup of finite index which is normal in both $A$ and $B$.


The fact is true. Here is the proof that if $A\cup B$ is normal in $G$, then $A\cap B$ is normal in $G$ or $A^g$ is inside $B$ for some $g$. Suppose that $A^g$ is not inside $B$ for any $g$. If $A\cup B$ is normal in $G$, then $A$ and $B$ are normalized by an index at most 2 subgroup $G_1$ of $G$. Indeed, for every $g\in G$, we have $A^g$ is a union of two subgroups $A^g\cap A$ and $A^g\cap B$. That can happen only if $A^g=A$ or $A^g=B$, same for $B$. Now if $G_1=G$, there is nothing to prove. If $G_1$ is of index 2, then there exists an element $g$ with $A^g=B$, $B^g=A$ and $G=\langle G_1, g\rangle$. Therefore $A\cap B$ is normal in $G$. In particular, it is normal in $A$ and $B$. Now suppose that $A^g$ is inside $B$ for some $g$. Notice that for every $h\in G$ either $A^h=A$ or $A^h\subseteq B$ and similarly either $B^h=B$ or $B^h\subseteq A$. If $A^h\subseteq B$ and $B^h \subseteq A$ then $(A\cap B)^h=A\cap B$. Therefore the normalizer $C$ of $A\cap B$ together with the normalizer $D$ of $B$ gives the whole $G$. This again implies that either $C=G$ and $A\cap B$ is normal in $G$ or $D=G$, and $B$ is normal in $G$. Thus the only case left is when $A^g\subseteq B$ for some $g$ and $B$ is normal in $G$. In that case $A\subseteq B^{g^{-1}}=B$, so $A\cap B=A$ and since it must be of finite index in $B$ by assumption, we conclude that $A=A\cap B$ has a normal subgroup of finite index both in $A$ and in $B$.

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  • $\begingroup$ @Mark Sapir : Thanks ! You are proving even more : if some group K normalises $A\cup B$, then it normalises $A\cap B$, and your key point is that a union of two distinct subgroups is not a group. $\endgroup$ – Drike Nov 21 '11 at 12:44
  • $\begingroup$ @Drike: Reading Mark's answer carefully also shows that the correct spelling is "normalize" (not "normalise"). $\endgroup$ – Todd Leason Nov 21 '11 at 13:25
  • $\begingroup$ @Todd: perhaps @Drike is a native English speaker (as opposed to a native, or otherwise, American speaker)? Google "normalize" and you will be enlightened. $\endgroup$ – Igor Rivin Nov 21 '11 at 15:44
  • $\begingroup$ I am not a native English speaker, so my spelling cannot be trusted. Google gives 10 mil links for "normalize" and 3 mil links for "normalise". I guess this makes both versions possible. $\endgroup$ – user6976 Nov 21 '11 at 16:03
  • $\begingroup$ I think Drike (deleted comment) and the others are right: "normalize" is American English and "normalise" British English. More examples can be found on spelling.org/free/instructional/…. An example in mathematical context is the announcement talks.cam.ac.uk/talk/index/34171 where "generalise" appears. Sorry, if my comment led to confusion. $\endgroup$ – Todd Leason Nov 21 '11 at 17:08

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