The index of the Core of a group

Question

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!

Answer 1

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.