Let $G$ be a group with $B$ a subgroup of index $m$. There is a homomorphism that associates each element of $G$ to a permutation of the left cosets $xB\in G/B$, which is of cardinality $m$.

So these permutations live in $S_m$.

Why is the kernel of that homomorphism included in $B$?

$$\ker\phi=\{g\in G~:~gxB=xB \text{ for all } x\in G\}=\dotsb= \bigcap\limits_{g\in G}gBg^{-1}$$ I don't see why that is a subset of $B$...


Because take $g=e$ and you will get that $B$ is one of the groups in the intersection. Hence $B$ contains the intersection.


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.