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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!

user35603
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Marm
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1 Answers1

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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.

Nicky Hekster
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  • 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? – Marm Jun 23 '14 at 12:11
  • 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 $(Ug_i)g=U(g_ig)$. – Nicky Hekster Jun 23 '14 at 12:25
  • 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.

     – Marm Jun 23 '14 at 13:18
  • 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! – Nicky Hekster Jun 23 '14 at 13:56