Show that G is a group, if G is finite , has an identity, the operation is associative and cancellation law holds.

Question

Let $G$ be a non-empty finite set with an associative binary operation so that cancellation law holds, i.e. $ab=ac$ or $ba=ca$ implies $b=c$, for any choices of $a,b,c$ in $G$. Assume that there is an identity element $e$ in $G$. Show that $G$ is a group.

Proof: To show $G$ is a group, conditions must hold. Suppose that $ab=ac$, then $b=eb=(a^{-1}a)b =a^{-1}(ab)=a^{-1}(ac) =(a^{-1}a)c=ec=c$. So left cancellation holds in $G$.

2) Suppose $ba=ca$, then $b=be=b(aa^{-1}) =(ba)a^{-1}=(ca)a^{-1} =c(aa^{-1})=ce=c$. So right cancellation holds in $G$.

3) If $a$ exists in $G$ then $aa^{-1}=a^{-1}a=e$, where $a$ is an inverse of $a^{-1}$. Since inverses are unique, $(a^{-1})^{-1}=a$.

4) Let $x$ be the inverse of $ab$. Then $(ab)x=e$. By associativity we have $a(bx)=aa^{-1}$. Through left cancellation we have $bx=a^{-1}bx=ea^{-1}=b(b^{-1}a^{-1})$ and $x=b^{-1}a^{-1}$.

Thus $(ab)^{-1}=b^{-1}a^{-1}$. So all conditions hold, $G$ is a group.

Is this proof correct? I know to show $G$ is a group these conditions have to be met. This is all I have to show right?

Answer 1

Another proof makes use of the pigeonhole principle. Fix $a \in G$, and consider the map $$ f : G \to G, \qquad x \mapsto a x. $$ Since $G$ is cancellative, the map is injective, hence surjective because $G$ is finite, so that there is $b \in G$ such that $e = f(b) = a b$.

Similarly, there is $c \in G$ such that $c a = e$.

Thus $$c = c e = c (a b) = (c a) b = e b = b,$$ and $b = c$ is the required inverse of $a$.

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You need not assume that there is an identity, but can prove that one exists.

First show, with the arguments above, that any element $z$ of $G$ can be written in the form $z = x a$, for some $x \in G$. Then show that there is $s \in G$ such that $a s = a$. Now $$z s = x a s = x a = z, \quad\text{for all $z \in G$,}$$ so $s$ is a right identity. Similarly there is a left identity $t$ such that $$t z = z,\quad \text{for all $z \in G$,}$$ and finally $$t = t s = s$$ is the identity.

Answer 2

You know that cancellation holds in $G$, it's a hypothesis, so you have not to prove it. And, most important, you can't use inverses of elements without first proving they exist, which is exactly what you have to do.

Consider an element $a\in G$ and the map $f_a\colon G\to G$ defined by $$ f(x)=ax $$ This map is injective. Why?

This map is also surjective, because of an important hypothesis you have. Which one?

Now, what can you say about the element $x$ such that $f_{a}(x)=e$? It is…

Next, consider the map $g_{a}\colon G\to G$ defined by $g_{a}(x)=xa$. Repeat the reasoning to conclude.

Answer 3

No, you seem to be assuming things like the inverse would exist in your proof, which is the only thing you really need to prove (the other 3 conditions in the definition of a group are given to you).

So choose $a\in G$, and consider the set $$ \{a, a^2, a^3, \ldots \} \subset G $$ Since $G$ is finite, this set is finite, and so there exist $a^i, a^j$ such that $i>j$ and $$ a^i = a^j $$ Now consider $f = a^{i-j}$, then $$ fa^j = a^i = a^j = ea^j \Rightarrow f = e $$ Hence, if $b = a^{i-j-1}$, then $$ ab = e = ba $$ and hence $b = a^{-1} \in G$

Answer 4

Here is another way of seeing that a finite semigroup with cancellation is a group:

If $G$ is a semigroup, that is, a set with an associative binary operation, then there is a natural map $\lambda\colon S \to \mathrm{Map}(S):=\{f\colon S \to S\}$ given by $\lambda(s)(x) = sx$. The set of self-maps $\mathrm{Map}(S)$ of $S$ to itself is a semigroup under composition, and associativity shows that $\lambda$ is a homomorphism of semigroups.

Now suppose that left and right cancellation holds for $G$, that is, if for all $a,x,y \in S$ we have both $ax=ay$ implies $x=y$ and $xa=ya$ implies $x=y$. Then left cancellation implies that $\lambda(a)$ is injective for all $a \in S$, and right cancellation implies that $\lambda$ is injective. It follows that for a semigroup with left and right cancellation we obtain a map $\lambda\colon S \to \mathrm{Inj}(S)$ from $S$ to the subsemigroup of $\mathrm{Map}(S)$ consisting of injective self-maps of $S$ (a semigroup since the composition of injections is again clearly injective).

Now if $S$ is finite, then $\mathrm{Inj}(S) = \mathrm{Sym}(S)=G$ is the group of permutations of $S$, and hence $\lambda$ embeds $S$ as a sub-semigroup of the group of permutations of $S$. Thus to see $S$ is a group, it suffices to show that a subsemigroup $S$ of a finite group $G$ is a group. For this note that every element $g \in G$ has finite order, thus if $s \in S$ then the subsemigroup and subgroup it generates are equal. It follows that $S$ contains $s^{-1}$ and the identity of $G$, and hence it is a subgroup of $G$ as required.

Answer 5

The cancellation laws imply that left and right multiplication maps by a given $a\in G$ (say $l_a$ and $r_a$, respectively) are injective and hence (finiteness of $G$, see "pigeonhole principle") both are bijections on $G$. By the associativity, then, $l_al_b=l_{ab}$ and $r_ar_b=r_{ba}, \forall a,b\in G$, and hence (closure) $L:=\{l_a,a\in G\}\le\operatorname{Sym}(G)$ and $R:=\{r_a,a\in G\}\le\operatorname{Sym}(G)$. Therefore, $\exists e_l,e_r\in G$ such that $l_{e_l}=r_{e_r}=Id_G$ or, equivalently, $e_lg=ge_r=g, \forall g\in G$; as a particular case, $g=e_l$ and left cancellation imply $e_l=e_r=:e$ (same conclusion by assuming $g=e_r$ and right cancellation), so we finally get that $\exists e\in G$ such that $eg=ge=g, \forall g\in G$ (and hence the existence of the identity needs not to be assumed, but can be gotten from the finiteness and the cancellation laws).

About the inverses, note that by the surjectivity of $l_a$ and $r_a$ (for every $a\in G$), $\exists a_l,a_r\in G$ such that $e=l_a(a_r)=r_a(a_l)$. Now:

\begin{alignat}{1} e=aa_r &\Rightarrow a_r=a_re=a_raa_r \\ &\Rightarrow a_ra=e \\ &\Rightarrow a_ra=aa_r \\ &\Rightarrow a_ra=a_la \\ &\Rightarrow a_r=a_l=:a^{-1} \\ \end{alignat}

and hence left and right inverses exist and coincide for every $a\in G$. Therefore, $G$ is indeed a group.