Theorem 18 in *Abstract Algebra* by Dummit and Foote

Question

In the proof of this theorem on page 97, it is said:

i.e., $B$ is a normal subgroup of the subgroup $AB$.

But I don't understand how this follows from the fact that $AB \leq N_G(B)$. Can someone please clarify this?

Answer 1

As $B=eB\subset AB$ we have $B\leq AB$.

If $g\in AB\leq N_G(B)$, then $g\in N_G(B)$, so $gBg^{-1}=B$.

Thus $B$ is normal in $AB$.

In general, if $B\leq C\leq N_G(B)$, then $B$ is normal in $C$.

Answer 2

It is better if we try to express everything in words. I will read "$AB\leq N_G(B)$" as "$AB$ normalizes $B$" which exactly means "$B$ is normal in $AB$" (since $B$ is subgroup of $AB$).

On the other hand, in Klein-4 group $\{e,a,b,c\}$, the subgroup $\{e,a\}$ normalizes subgroup $\{e,b\}$ but we can not say $\{e,a\}$ is normal in $\{e,b\}$ because of no subgroup relation between them.