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I really looked all over the web and searched for an example I will understand. I don't understand how to complete a multiplication table! (all examples I found the Identity element was given)

$G=\{a,b,c\}$

and I know that $aa=b$

how do I start from here? (all I know that in each row and column every element must appear exactly once)

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3 Answers 3

up vote 6 down vote accepted

Certainly, $a$ is not the identity(Otherwise $aa=a$). We also know that $b$ is not the identity (Otherwise $|a|=2$, this contradicts Lagrange's theorem). We also note that every group of order 3 is cyclic. Thus, $a=a,b=a^2,c=a^3$ ($c$ is the identity). Now its easy to find the rest of the table.

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$G = \{a, b, c\}$, for group $G$, and we are assuming $a, b, c$ are distinct: i.e. $|G| = 3.$
And we know $aa = b$

  • Given $aa = b$, $a$ cannot be the identity. If it were, we'd have $aa = a$.
  • $b$ cannot the identity: if it were, then the order of $a$ would have to be $2$, which does not divide the order of $G = 3$ (Lagrange's theorem).
  • So $c$ must be the identity, since $G$ is a group, one of $a, b, c$ must be the identity element.

  • Every group order 3 is cyclic. So one element generates the other two (and of course, is its own generator!): we know $a = a, aa = b.$ Now, what must be the value of: $aaa = \;?\;$

  • The rest of the table should readily fall out. As you observe:

    Each element $a, b, c$ occurs once and only once in every column, and in every row.

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+1 for nicely doing the answer. :-) –  B. S. Aug 26 '13 at 7:44

This might be helpful to you: http://en.wikipedia.org/wiki/Cayley_table

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This might have been more suited as a comment. –  user50407 Dec 26 '12 at 22:09

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