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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!

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1 Answer 1

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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 $x,y\in M$.

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    $\begingroup$ You need, also, to show that $M$ is nonempty. $\endgroup$
    – Shaun
    Oct 22, 2021 at 16:21
  • $\begingroup$ Can you please show me this shorter way details? $\endgroup$
    – Brain
    Oct 22, 2021 at 16:22
  • $\begingroup$ Can you please add the proof that it is nonempty? or this follows because it contains the identity? $\endgroup$
    – Brain
    Oct 22, 2021 at 16:22
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    $\begingroup$ @Brain generally the simplest way to show it is nonempty is to just show it has the identity element, if you can't simply show the identity is in your proposed subgroup, you have a problem! $\endgroup$
    – Alan
    Oct 22, 2021 at 16:26

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