Question

If A, B are subgroups of G such that b^(-1) Ab⊇A for any b ϵ B, show that AB≤G.

Answer 1

Question 1. If $A, B$ are subgroups of $G$ such that $b^{-1}Ab \subset A$ for any $b \in B$, show that $AB \leq G$.

Solution. Let $a_1b_1, a_2b_2 \in AB$, where $a_1, a_2 \in A$ and $b_1, b_2 \in B$. Then

$$a_1 b_1 \cdot a_2 b_2 = a_1 (b_1 a_2 b_1^{-1}) (b_1 b_2) = a_1 ((b_1^{-1})^{-1} a_2 b_1^{-1}) (b_1 b_2) \in AB,$$

because $(b_1^{-1})^{-1} a_2 b_1^{-1} \in A$ and $b_1 b_2 \in B$. Furthermore, for any $ab \in AB$

$$(ab)^{-1} = b^{-1} a^{-1} = (b^{-1} a^{-1} b) b^{-1} \in AB.$$

And finally, since the identity $e$ of $G$ belongs both to $A$ and to $B$, we conclude that $e = e \cdot e \in AB$. Thus, $AB$ is a subgroup of $G$.