# counter example that shows if set $G$ is infinite is not a group. [closed]

let $$G$$ be a finite set and let $$*$$ be an associative binary operation on $$G$$ satisfying $$\forall a,b,c \in G$$:

1. if $$a*b = a*c$$ then $$b=c$$
2. if $$b*a = c*a$$ then $$b=c$$

then G must be group under * (Also provide a counter example that shows that this is false if G is infinite).

• Are you sure the first line isn’t $a * b = a * c$? That would be the typical definition for a cancellative semigroup. Commented Sep 21, 2020 at 20:37

Hint: Select any $$a \in G$$. If we define $$a^n = \overbrace{a*\cdots*a}^n$$, consider the fact that there must exist positive integers $$m \neq n$$ for which $$a^m = a^n$$.

A counterexample in the case that $$G$$ is infinite: consider $$G = \{0,1,2,3,\dots,\}$$ with the usual addition operation.

Sketch/hints: you need to find the identity and prove existence of inverses. The idea is that conditions 1 and 2 say that $$L_a:G\to G$$ and $$R_b:G\to G$$ given by $$L_a(x)=ax$$ and $$R_b(x) = xb$$ are injective. Since $$G$$ is finite, $$L_a$$ and $$R_b$$ are also bijective. So fix $$a\in G$$. There is $$e_a \in G$$ such that $$L_a(e_a)=a$$. This $$e_a$$ will be the identity. So you have to show that:

i) $$R_a(e_a)=a$$ as well (here associativity comes in)

ii) Given $$a,b\in G$$, we have $$e_a = e_b$$. So pick any of those $$e_a$$ and call it $$e$$: this will be the identity.

Now, for every $$a \in G$$ there is $$a'\in G$$ such that $$L_a(a') = e$$. Then show:

(iii) $$R_a(a')=e$$. This allows you to denote $$a'=a^{-1}$$.

I'll let you, based on the above, think of a counter-example for infinite $$G$$. I can try to provide more details later if I find the time.