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Suppose $B$ is a group, $A\leq B$ with $[B : A] = b < \infty$ (here $[B : A]$ is the cardinality of $B/A := \{bA : b \in B\}$). Show that there is a normal subgroup $C\unlhd B$ with $[B : C] \leq b!$ and $C\leq A.$

I know various theorems that might help with this. For example, I know Cayley's theorem, which states that if $B$ is a finite group, then $B$ is isomorphic to a subgroup of $S_B,$ the set of invertible functions from $B$ to $B$. Also, I know that if $B$ is a finite group, and $p$ is a prime dividing $|B|,$ then $B$ has an element of order $p$. I also know that if $B$ is a finite group and $A\leq B$ has index $[B : A] = p,$ where $p$ is the smallest prime dividing $|B|,$ then $A\unlhd B.$ However, I'm not sure which facts to use to find $C$. With $b!$, it seems that some permutations may be involved. It also doesn't seem very useful to come up with a contradiction.

Thomas11441
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    This question is asked frequently. The normal subgroup $C$ is the kernel of the action of $B$ by left multiplication on the left cosets of $A$ in $B$. – Derek Holt Jun 11 '21 at 22:07
  • Thanks. So could I define $\phi : B\to S_{B/A}, \phi(b)$ is the function so that $\phi(b)(x) = b\cdot x$ for $x\in B/A$ and then let $C= \ker \phi$? Then $B/C$ is isomorphic to a subgroup of $S_{B/A}, \Rightarrow |B/C| \leq |S_{B/A}| = b!$ and the rest of the results seem fairly simple to show. – Thomas11441 Jun 11 '21 at 23:06
  • Yes that's right. – Derek Holt Jun 12 '21 at 08:10

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