# Cancellation Laws-groups

• Aug 7th 2010, 11:59 PM
prashantgolu
Cancellation Laws-groups
if a set is closed and associative with respect to an operation *...and both cancellation laws hold...proove that the set is a group with respect to *.thnx in advance
• Aug 8th 2010, 01:18 AM
Vlasev
So for a set S to be a group it must:

- be closed with respect to $*$ (check)
- associative $(ab)c = a(bc)$ (check)
- identity $ae=ea =a$
- inverses

So you are given that both cancellation laws hold. I assume that you mean that left and right cancellations hold. First we need to show that there exists an identity because we need it for the inverses.

If the set is finite, we must have that for some $k \in \mathbb{N}$

$a^k = a$

From associativity you have

$a\,a^{k-1}= a$ and $a^{k-1}a =a$. So now $b = a^{k-1}$ is our identity.

So now that $e = a^{k-1}$, we must have that

$a^{k-2}a = e$ and $a\,a^{k-2} = e$, so we get that $a^{-1}$ exists and is equal to $a^{k-2}$. Thus the set is a group.

Now, we cannot use the same argument if the set is infinite. We need something else there. In the above argument we have only used the definitions of the identity and the inverses and we haven't used the cancellation laws yet.
• Aug 8th 2010, 01:42 AM
prashantgolu
thank you very much..
but then the next ques in my textbook says that if only one of the cancellatinon law holds then this will not be the case...so that means cancellations laws are an imp part of this ques...
a^k=a
a^(k-1)=e
we can call this cancellation
so i assume cacncellation is being used..correct me if u think i am wrong...thnx...
• Aug 8th 2010, 01:55 AM
Vlasev
Oh, sorry, I missed a step. The definition of the identity is that there exists and element e in the set S such that $ae = ea$ FOR ALL $a \in S$. My proof above shows this is true only for a CERTAIN element $a \in S$.

Now, since $a^k = a$, let b be an arbitrary element in $S$.

We have that $a^kb = ab$ but also, $a^kb = a a^{k-1} b$, so

$ab = a a^{k-1}b$. Then cancel a on the left to get $b = a^{k-1}b$. To show it on the right hand side, just use $b\,a^k$.

Thus, $a^{k-1}$ works for all elements in S. Only now we have the conclusion that the identity exists. So yes, the cancellation is necessary.
• Aug 8th 2010, 01:17 PM
alunw
Only true for finite sets.
Quote:

Originally Posted by prashantgolu
if a set is closed and associative with respect to an operation *...and both cancellation laws hold...proove that the set is a group with respect to *.thnx in advance

This is not necessarily true for infinite sets. The set of positive integers is closed under multiplication and addition, both cancellation laws hold in each case, but it is not a group under either operation
• Aug 8th 2010, 01:34 PM
Vlasev
Ohh, then that makes is simple then, because we already proved the finite case!