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

Question

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 $B$...

Answer 1

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