I have to prove the following:

Let $G$ be a group and $U$ be a subgroup of $G$. Then it holds: If $U$ has finite index, then $\text{Core}_G(U):=\bigcap\limits_{g\in G}gUg^{-1}$ has also finite index.

Would be nice if someone could give me some tips. Thanks!

  • $\begingroup$ The intersection of finitely many subgroups of finite index in $G$ also has finite index in $G$. $\endgroup$ – Derek Holt Jun 23 '14 at 11:56

Hint: $G$ acts on the right cosets of $U$ by right multiplication. This provides you a homomorphism from $G$ to $S_{index[G:U]}$, the permutation group on $index[G:U]$ elements. The kernel of the action is exactly $core_G(U)$, whence $|G/core_G(U)|$ divides $index[G:U]!$, in particular is finite.

  • $\begingroup$ Thanks for answer! I dont understand why the kernel of the provided map: G $\rightarrow S_{index[G:U]}$ is $core_G(U)$. Could u explain that for me? And how u get the last thing? $\endgroup$ – Marm Jun 23 '14 at 12:11
  • $\begingroup$ Yes I wrote a hint not the full solution, will do that later. For starter: let $\{U.1, Ug_2, \cdots, Ug_n\}$ be the right cosets of $U$, where $n=index[G:U]$. Then the action of $g \in G$, is defined by $(Ug_i)g=U(g_ig)$. $\endgroup$ – Nicky Hekster Jun 23 '14 at 12:25
  • $\begingroup$ Okay, i think i get it now. The kernel of the action is $Core_G(U)$ because: (I prefer to use the left cosets): $ker(\phi)=\{g \in G: gxU=xU \ \forall x \in G\}$=$\{g \in G: x^{-1}gxU=U \ \forall x \in G\}$=$\{g \in G: x^{-1}gx=U \ \forall x \in G\}$=$\{g \in G: g=xUx^{-1} \ \forall x \in G\}$=$Core_G(U)$ And now we can (with the homomorphism theorem) conclude that $\vert$G$/Core_G(U)$$\vert$=$\vert$G$/ker(\phi)$$\vert$=$\vert$im($\phi)$$\vert$<$\vert$G:U$\vert$!. The right side is finite, because $\vert$G:U$\vert$ is finite. The assertion follow. $\endgroup$ – Marm Jun 23 '14 at 13:18
  • $\begingroup$ Excellent, you got it, well done! And yes you can use left multiplication on left cosets too, that does not matter, it also generates a permutation on the cosets! $\endgroup$ – Nicky Hekster Jun 23 '14 at 13:56

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.