# Why is the kernel of $\phi : G\mapsto S_m$ included in $B\le G$ where $\phi(g)=g:xB\mapsto gxB$?

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 $$B$$...

## 1 Answer

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