Show that Green's relations coincide on a group, in particular, that $\mathcal{L}=\mathcal{R}$.

Question

I am trying to prove that Green's relations on a group coincide, that is, $\mathcal{L}=\mathcal{R}=\mathcal{H}=\mathcal{D}=\mathcal{J}$. It can be easily deduced that $\mathcal{H},\mathcal{D},\mathcal{J}$ conicide with $\mathcal{L}$ and $\mathcal{R}$ once we establish that $\mathcal{L}=\mathcal{R}$. So, we first try to show that for any $a,b\in G$, $aG=bG \iff Ga=Gb$. It's this that I'm struggling with. (What makes a group different from a semigroup is that it has an identity and inverses. So, we should probably make use of this in the proof.) This is what I have so far:

Since $G$ has an identity, we have that $a\in aG$ and $b\in bG$. Thus, there exist $g,g'\in G$ such that $a=bg$ (1) and $b=ag'$ (2). Then $aG=bG\iff bgG=ag'G\iff GbgG=Gag'G$. Cancel $G$ on the left (we know this can be done since we have inverses): $Gbg=Gag'$. We are done if we can show $g=g'$. This is done by substituting equations 1 and 2 into each other: $a=bg=ag'g\iff 1=g'g$ and $b=bgg'\iff 1=gg'$ thus $gg'=g'g$ therefore $g=g'$.

Can anyone tell me if this is correct? Thanks.

Answer 1

Since $G\unlhd G$, we have $Ga=aG=bG=Gb$ as cosets.

Answer 2

Let $a \in G$. Then the map $x \to ax$ (respectively $x \to xa$) is bijective: just observe that, for each $a, b$, the equation $ax = b$ has a unique solution, namely $x = a^{-1}b$. It follows that $aG = Ga = G$ and hence $R(a) = L(a) = H(a) = D(a) = J(a)$. Thus $\cal R = \cal L = \cal H = \cal D = \cal J$ in $G$.