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

Question

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).

Please help me, I can't think of anything.

Answer 1

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.

Answer 2

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.