## MaBloWriMo 20: the group X star

So, where are we? Recall that we are assuming (in order to get a contradiction) that $M_n$ is not prime, and we picked a smallish divisor $q$ (“smallish” meaning $q^2 \leq M_n$). We then defined the set $X$ as $X = \{ a + b\sqrt 3 \mid a, b \in \mathbb{Z}; 0 \leq a, b < q \}$

that is, combinations of $1$ and $\sqrt 3$ where the coefficients are between $0$ and $q-1$. We defined a binary operation on $X$ which works by multiplying and then reducing the coefficients $\pmod q$. This is not a group, since for example $0$ doesn’t have an inverse. But in the last post we saw that we can make a group $X^*$ simply by including only the elements from $X$ that do have inverses.

Recall that $\omega = 2 + \sqrt 3$ is in $X$ (as long as $q$ is not $2$—which it can’t be, since $M_n = 2^n - 1$ is odd), and we know that $\omega \overline{\omega} = 1$, so $\omega$ has an inverse and we conclude $\omega \in X^*$.

You might enjoy figuring out what $X^*$ looks like in the case $q = 3$. Tomorrow, we’ll start thinking about the implications of our assumption that $s_{n-2}$ is divisible by $M_n$, and in particular what it means about the order of $\omega$ in $X^*$. 