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If $\{a,b\}$ with operation $*$ is to be a group, with $a$ the identity element, then what must the Cayley table be?

I thought the following:

\begin{array}{c|cc} * & a & b\\ \hline a & a & b\\ b& b& ? \end{array} Can the question mark be either $a$ or $b$? Or does it have to be $a$? The information given to me does not seem to make it clear that it has to be one or the other.

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  • $\begingroup$ Hint: if $a \ast b = b \ast b$, what can we say about $a$ and $b$? $\endgroup$ – user64687 Apr 22 '15 at 14:36
  • $\begingroup$ @AsalBeagDubh That they're equivalent? So are you saying that assumption would be erroneous thus $b*b$ should be $b$? $\endgroup$ – fancynancy Apr 22 '15 at 14:38
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    $\begingroup$ @Travis You are clearly pushing me to conclude that $b*b=a$, but I haven't gotten to a place in my text where I can use left- or right-cancellation to make the deductions we are talking about. This is the first section on an introduction to groups. So I imagine I'm supposed to use associativity, identity, or inverse of elements to conclude that $b*b=a$. But I'm not seeing how to use one of those properties. $\endgroup$ – fancynancy Apr 22 '15 at 14:45
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    $\begingroup$ @fancynancy Left- and right-cancellation themselves just use those properties. Given any group $G$ (say with identity $e$), suppose we have elements $x, y, z \in G$ such that $x z = y z$. Any element, in particular $z$, has an inverse, and multiplying by $z^{-1}$ (on the right) gives $(xz)z^{-1} = (yz)z^{-1}$, and by associativity we have $(xz)z^{-1} = x(zz^{-1}) = xe = x$ and likewise for the the r.h.s., and hence $x = y$. Left-cancellation works analogously. $\endgroup$ – Travis Apr 22 '15 at 15:10
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    $\begingroup$ Using the kind of reasoning Asal's hint suggests, one can show that for any group, each group element appears precisely once in each column and each row of the Cayley table, that is, the Cayley table must be a Latin square. See en.wikipedia.org/wiki/Latin_square for much more. $\endgroup$ – Travis Apr 22 '15 at 15:13
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Every element in a group has an inverse. In your case, there must be some element $x \in \{a,b\}$ for which $b * x = a$.

To see this another way, note that no two elements in a row are equal. This is because if $b * x = b * y$, then by multiplying on the left by $b^{-1}$, we have $x = y$.

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    $\begingroup$ This makes perfect sense. Thanks! $\endgroup$ – fancynancy Apr 22 '15 at 14:50

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