Let $*$ be a binary operation defined on a non empty, finite set $G$ such that it follows associative, commutative and cancellation law.Show that $G$ under the operation $*$ is abelian.
Now for $G$ to be abelian, it is missing identity and inverse existence. But above statement doesn't seem to contribute anything for proving above. Also there is the bit about the group $G$ being finite, the importance of which I fail to recognize.
$\begingroup$
$\endgroup$
2
asked Sep 10, 2012 at 16:09
-
$\begingroup$ Do you mean you want to prove $G$ is an abelian group? Proving it is abelian doesn't make much sense on its own. $\endgroup$– Chris EagleSep 10, 2012 at 16:33
-
$\begingroup$ Yup, an abelian group. $\endgroup$– Abhishek AnandSep 10, 2012 at 16:33
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
6
A start: Let $a$ be any element of our set.
Consider the sequence that goes $a$, $a\ast a$, $a\ast(a\ast a)$, and so on like that forever. Let's switch to more standard juxtaposition notation for the product, and exponential notation. So we are looking at the sequence $a,a^2,a^3,\cdots$.
Because our set $G$ is finite, there are positive integers $i$ and $j$, with $i\lt j$, such that $a^i=a^j$. In fact, if $G$ has $n$ elements, there will be such $i$, $j$ with $j\le n+1$.
Thus for any $x$, $a^i x=a^j x$. By cancellation we have $a^{j-i}x=x$. We have established the existence of an identity element. It remains to show the existence of inverses.
answered Sep 10, 2012 at 16:18
-
$\begingroup$ Are you assuming $*$ to be multiplicative function? $\endgroup$ Sep 10, 2012 at 16:33
-
$\begingroup$ @AbhishekAnand: I changed things a bit, starting with you $\ast$ and then switching to more common notation. If you want to be absolutely precise, define $a^n$ by induction, $a^1=a$, $a^{n+1}=a\ast a^n$. $\endgroup$ Sep 10, 2012 at 16:40
-
-
$\begingroup$ I am a little confused: this approach certainly proves that $G$ is a group, but what about the abelian part of the question? Surely there are finite non-abelian groups? $\endgroup$– Old JohnSep 10, 2012 at 17:02
-
$\begingroup$ @JohnSenior: The question specified that the operation $\ast$ is commutative. $\endgroup$ Sep 10, 2012 at 17:09
$\begingroup$
$\endgroup$
inverse existence
is $a \in G$ we know there is $0 <k <n +1$ such that $a ^ k$ is neutral, then the inverse of $a$ is $a^{k-1}$ because $aa^{k-1} = a^{k-1}a = a^k$.
