Let $G$ be an arbitrary group, and suppose that $H=C_G(g_1,\ldots,g_n)$ is also the intersection of all centralizers of finite index in $G$, and furthermore $[G:H]<\infty$. Is it true that $H$ is a normal subgroup of $G$?

Probably related: is it true that if a group $G$ is virtually-abelian then it is abelian-by-finite?

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    $\begingroup$ The answer to your second question is "yes", but the abelian subgroups are not necessarily the same. This is because if $H\leq_fG$ then there exists a subgroup $K\leq_fH$ such that $K\unlhd G$ ($K$ is the intersection of the conjugacy classes of $H$ in $G$). $\endgroup$ – user1729 Apr 7 '16 at 18:26
  • $\begingroup$ @user1729 Why $K$ has finite index in $H$? $\endgroup$ – Darío G Apr 7 '16 at 18:51
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    $\begingroup$ What are $g_1,\ldots,g_n$, and what does $C_G(g_1,\ldots,g_n)$ mean? $\endgroup$ – Derek Holt Apr 7 '16 at 19:06
  • $\begingroup$ @DerekHolt $C_G(g_1,\ldots,g_n)$ is the set of elements in $G$ that commute with $g_1,\ldots,g_n$ (some fixed elements in $G$). On the other hand, could you tell me where can I find the reference of the statement "virtually-P $\Leftrightarrow$ P-by-finite"? $\endgroup$ – Darío G Apr 7 '16 at 19:27
  • $\begingroup$ The intersection of the centralizers of all elements in $G$ is certainly normal in $G$ - in fact it's characteristic. $\endgroup$ – Derek Holt Apr 7 '16 at 20:26

Sorry, I need to correct my comment about virtually $P$ $\Leftrightarrow$ $P$-by-finite. Let $P$ be a property of groups such that, if $G$ has property $P$ and $H$ is a subgroup of $G$ of finite index, then $H$ has $P$. So, all subgroup-closed properties, like free, abelian, nilpotent, solvable, satisfy this condition.

Then $G$ is virtually-$P$ if and only if $G$ is $P$-by-finite. Clearly $P$-by-finite implies $G$ virtually-$P$.

Conversely, if $G$ is virtually-$P$, then it has a subgroup $H$ of finite index such that $H$ has $P$. Let $K = {\rm Core}_G(H) = \cap_{g \in G} g^{-1}Hg$. Then $K \le H \le G$ with $K \unlhd G$ and $|G:K|$ finite. So, by the assumption about $P$, $K$ has $P$ and hence $G$ is $P$-by-finite.


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