Given a subgroup $B$ of a group $A$, show that $a\in B$ iff $Ba = B$

Question

Let's suppose that $A$ is a group, and that $B$ is a subgroup of $A$; examining Lagrange's Theorem and the idea of cosets let's also assume that $a \in A$ and then prove that $a \in B$ iff $Ba = B$

My idea:

• In order for $a$ $\in$ $B$, $a$ would have to be the 'generator' of that coset $Ba$ such that it's an element of $A$ that is used to scale $B$ into another subgroup that's still contained by $A$; and if $Ba = B$ is to hold true, then $a$ must be the identity element and thus still is one of the cosets that are listed in the enumeration of $A_{/B}$ while still satisfying the original statement

I'm still new with this kind of problem solving and I'm having trouble seeing the connection between the scaling element of the original group and the coset equaling itself after being scaled.

So any and all help is appreciated !!

Answer 1

If $a \in B$, then since $B$ is a subgroup, $\forall x \in B~(xa \in B)$ so $Ba \subseteq B$. Also, $\forall x \in B~(x=xa^{-1}a)$ and since $a \in B, a^{-1} \in B$ so $x = xa^{-1}a \in Ba$ and $B \subseteq Ba$.

Conversely, if $Ba=B$, then $1 \in B$ (since $B$ is a subgroup) so $a=1a \in Ba=B \Rightarrow a \in B$.

Answer 2

Hints.

$\Rightarrow$: use the fact that $a$ has an inverse to show $Ba=B$ (show both inclusions).

$\Leftarrow$: $B$ is a subgroup, so it contains the neutral element of $G$. But $Ba\subset B$, so ...