# Showing that $AB$ is a subgroup.

I was trying to solve this question:

Let $$G$$ be a group and $$A,B$$ subgroups of $$G.$$ Define $$AB$$ as the set of all products $$ab,$$ where $$a \in A$$ and $$b \in B.$$ Prove that $$AB$$ is a subgroup of $$G$$ iff $$BA \subseteq AB.$$

Assume that $$G$$ is a group and $$A,B$$ subgroups of $$G.$$

$$\Rightarrow$$

Assume that $$AB$$ is a subgroup of $$G,$$ we want to show that $$BA \subseteq AB.$$ Let $$x \in BA,$$ then $$x = ba$$ where $$b \in B$$ and $$a \in A.$$ But then, $$x = ba = ((ba)^{-1})^{-1} = (a^{-1} b^{-1})^{-1}$$ which is in $$AB$$ as $$a^{-1} b^{-1} \in AB$$ as each of $$A$$ and $$B$$ are subgroups and because $$AB$$ is a subgroup by hypothesis then the inverse of each element in it is contained in it $$i.e., (a^{-1} b^{-1} )^{-1} \in AB.$$

$$\Leftarrow$$

Assume that $$BA \subseteq AB.$$ We want to show that $$AB$$ is a subgroup of $$G.$$ First assume that $$x,y \in AB,$$ we want to show that:

1- $$\forall x \in AB, x^{-1} \in AB.$$

2- $$e_G \in AB.$$

3- $$\forall x,y \in AB, xy \in AB.$$

I was able to prove the first one as follows:

Let $$x \in AB,$$ then $$x = ab$$ for some $$a \in A$$ and $$b \in B.$$ But then $$x^{-1} = (ab)^{-1} = b^{-1}a^{-1} \in BA$$ as each of $$A,B$$ is a subgroup of $$G.$$ But $$BA \subseteq AB$$ by assumption, then $$x^{-1} \in AB$$ as required.

But then I do not know how to show $$2$$ and $$3.$$ any help will be greatly appreciated!

2) $$e_G\in A$$ and $$e_G\in B$$. So $$e_G=e_Ge_G\in AB$$.
3) If $$x,y\in AB$$, then $$x=a_xb_x$$ and $$y=a_yb_y$$ for some $$a_x,a_y\in A$$ and $$b_x,b_y\in B$$; so $$xy=a_xb_xa_yb_y$$. But $$b_xa_y\in BA\subset AB$$, so $$b_xa_y=ab$$ for some $$a\in A$$, $$b\in B$$. Then $$xy=a_xabb_y\in AB$$.
Note: There is a shorter way, generally speaking, to show that a subset $$M\subset G$$ is a subgroup of a group $$G$$. It suffices to show that $$M\neq\emptyset$$ and $$xy^{-1}\in M$$ for every $$x,y\in M$$.
• You need, also, to show that $M$ is nonempty. Oct 22, 2021 at 16:21