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

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.

• I'm not sure about the step of cancelling $G$. Wouldn't that imply $a=b$?
– Shaun
Commented Oct 28, 2021 at 14:24
• Oh - if we could cancel $G$, then we could cancel $G$ from the start to just have $a=b$. Right. Commented Oct 28, 2021 at 14:30

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

• Okay, I agree. Since G is a normal subgroup of itself, its left cosets are equal to its right cosets. Thanks! Commented Oct 28, 2021 at 14:42
• You're welcome, @tashakinns.
– Shaun
Commented Oct 28, 2021 at 14:43
• I now need to show that the relation $\mathcal{L}$ is equivalent to the relation $G\times G$. I am trying to prove by contradiction (i.e., assume there is one element of $G\times G$ such that $a$ is not $mathcal{L}$-related to $b$. But I get nowhere with this. Any thoughts? Commented Oct 28, 2021 at 18:56
• I suggest you ask that as a separate question, @tashakinns. However, you could use the fact that a semigroup $S$ is a group if and only if $aS=S=Sa$ for all $a\in S$.
– Shaun
Commented Oct 28, 2021 at 19:09
• I essentially just worked that out a few moments ago, although through visualising a group's Cayley table as opposed to recognising the equality itself. Thanks for your help! :) Commented Oct 28, 2021 at 19:27

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 $$G$$.