# Thread: equivalence relation g~h for all g,h => group trivial

1. ## equivalence relation g~h for all g,h => group trivial

Good morning! I have just read the following: Let $G$ be a group and ~ be an equivalence relation in $G$ so that $g$~ $h \, \forall g,h \in G$. Then $G$ is the trivial group.

I have tried to proof that but failed. Can somebody please give me a hint?

2. Originally Posted by Banach
Good morning! I have just read the following: Let $G$ be a group and ~ be an equivalence relation in $G$ so that $g$~ $h \, \forall g,h \in G$. Then $G$ is the trivial group.

I have tried to proof that but failed. Can somebody please give me a hint?
that's not correct! on every set the relation, as you defined, is an equivalence relation. so i'm sure the relation that you're talking about is already defined somewhere in your book

but you missed it. you need to find out how exactly the relation is defined, i.e. g ~ h if and only if what? for example you may define g ~ h iff gk = kh, for some k in G. then, with this

definition, ~ becomes an equivalence relation in G and if g ~ h for all g and h in G, then G has to be trivial because g ~ 1 if and only if gk = k, for some k in G. since G is a group, we'll

get that g = 1. so G has only one element, i.e. the identity element 1.

3. Originally Posted by NonCommAlg
that's not correct! on every set the relation, as you defined, is an equivalence relation. so i'm sure the relation that you're talking about is already defined somewhere in your book

but you missed it. you need to find out how exactly the relation is defined, i.e. g ~ h if and only if what? for example you may define g ~ h iff gk = kh, for some k in G. then, with this

definition, ~ becomes an equivalence relation in G and if g ~ h for all g and h in G, then G has to be trivial because g ~ 1 if and only if gk = k, for some k in G. since G is a group, we'll

get that g = 1. so G has only one element, i.e. the identity element 1.
Hi! Well, you are completely right. In any set A a~b iff a and b are in A defines an equivalence relation.

It is actually strange because in the book the argument is expressed for a general equivalence relation in any group and then used for a special one.

This is a little confusing.