MaBloWriMo 18: X is not a group

Yesterday we defined

X = \{ a + b \sqrt 3 \mid a,b \in \mathbb{Z}; 0 \leq a,b < q \}

along with a binary operation which works by multiplying and reducing coefficients \pmod q. So, is this a group? Well, let’s check:

  • It’s a bit tedious to prove formally, but the binary operation is in fact associative. Intuitively this follows from the fact that we could choose to do the reductions \pmod q immediately, or delay them until after completing several multiplications—and we know that normal multiplication is indeed associative.

  • 1 = 1 + 0\sqrt{3} is the identity for the operation.

  • Do all elements in X have inverses? Well… no! One simple counterexample is 0 \in X: there cannot possibly be any element x \in X such that 0x = 1. But 0 is not the only one. Depending on the choice of q, there can be many elements of X without inverses. For example, when q = 3, you can check that \sqrt 3 and 2 \sqrt{3} do not have inverses either (though the other elements do).

So X is a monoid (it has an associative binary operation with an identity) but it is not a group, because not every element has an inverse. Argh! But we really need a group, so that we have something to which we can apply all those nifty facts we proved! Well, fear not, it turns out that there is a simple way to turn any monoid into a group. Do you have an idea? Tomorrow I’ll explain how, and prove that it does in fact result in a valid group.


About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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One Response to MaBloWriMo 18: X is not a group

  1. Pingback: MaBloWriMo 20: the group X star | The Math Less Traveled

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