# Prove that if the cancellation law holds from both sides in a set, it forms a group.

## Homework Statement

Let G be a finite nonempty set with an operation * such that:
1. G is closed under *.
2. * is associative.
3. Given a,b,c in G with a*b=a*c, then b=c.
4. Given a,b,c in G with b*a=c*a, then b=c.

Prove that G must be a group under *.

## The Attempt at a Solution

It's obvious that identity element satisfies the conditions 3 and 4, but I don't know whether that proves that the identity element is contained in G or not? moreover, How can I show that the inverse of any element in G is contained in G?

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Dick
Homework Helper

You have to prove there IS an identity first. Start by picking any element g and defining f(x)=gx. Show using cancellation that f is an injective map from G->G. Since G is finite that makes it a bijection. So there exists an element e such that ge=g. Can you show eg=g? Can you show eh=he=h for ANY element of G, not just g?

Start by picking any element g and defining f(x)=gx. Show using cancellation that f is an injective map from G->G.
let's take $f(x_1)=f(x_2)$, hence, $g*x_1=g*x_2$ and the third axiom implies $x_1=x_2$. that proves f is injective.
Since G is finite that makes it a bijection.
if I've understood you correctly then I disagree with this part, f can be injective and finite, and yet it fails to be surjective. It's obvious from the definition of f that it's bijective though, so let's continue.

So there exists an element e such that ge=g.
sounds fine.

Can you show eg=g?
sure, I just need to define $f':G \to G , f(x)=xg$ and prove that f' is bijective. true?

Can you show eh=he=h for ANY element of G, not just g?
well, g was no sacred element and we had stated 'pick any element g in G'. Doesn't that already suffice to conclude that? all we need to do is that we should define a new f. so if we denote the dependence of f on g by $f_g: G \to G$ that would be fine. right?

OK. so now it's also easy to verify that the inverse of any element exists, It deploys the same logic about bijectivity but I'm failing at formalizing it. Could you help please?

Dick
Homework Helper

let's take $f(x_1)=f(x_2)$, hence, $g*x_1=g*x_2$ and the third axiom implies $x_1=x_2$. that proves f is injective.

if I've understood you correctly then I disagree with this part, f can be injective and finite, and yet it fails to be surjective. It's obvious from the definition of f that it's bijective though, so let's continue.

sounds fine.

sure, I just need to define $f':G \to G , f(x)=xg$ and prove that f' is bijective. true?

well, g was no sacred element and we had stated 'pick any element g in G'. Doesn't that already suffice to conclude that? all we need to do is that we should define a new f. so if we denote the dependence of f on g by $f_g: G \to G$ that would be fine. right?

OK. so now it's also easy to verify that the inverse of any element exists, It deploys the same logic about bijectivity but I'm failing at formalizing it. Could you help please?
f is bijective because G is finite. Since f is injective the image f(G) contains the same number of elements as G and is contained in G. And sure, for any two elements of the group g and h you can find elements such that g*e_g=g and h*e_h=h. But now you have to show e_g=e_h.

f is bijective because G is finite. Since f is injective the image f(G) contains the same number of elements as G and is contained in G.
Alright. I forgot that f was from G to G. that's right.
And sure, for any two elements of the group g and h you can find elements such that g*e_g=g and h*e_h=h. But now you have to show e_g=e_h.
Now I see what you mean. makes sense. I'll think about it.
How about proving the existence of inverses?

Dick