0
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.