What's the determinant of this lattice?

**amanda19** · January 5th 2011, 04:31 AM

Say $a$ and $b$ are both coprime to $c$ . Let $L$ be the lattice

$L = \{ (i,j) \in \mathbb{Z}^2: a^i b^j = 1 \text{ in } (\frac{\mathbb{Z}}{c \mathbb{Z}})^{\times} \}$ .

What is the determinant of $L$ ? I think the answer may be that it's the cardinality of the subset of $(\frac{\mathbb{Z}}{c \mathbb{Z}})^{\times}$ generated by $a$ and $b$ but I can't prove it. Is this a standard result?

**Drexel28** · January 5th 2011, 11:40 AM

Originally Posted by amanda19

Say $a$ and $b$ are both coprime to $c$ . Let $L$ be the lattice

$L = \{ (i,j) \in \mathbb{Z}^2: a^i b^j = 1 \text{ in } (\frac{\mathbb{Z}}{c \mathbb{Z}})^{\times} \}$ .

What is the determinant of $L$ ? I think the answer may be that it's the cardinality of the subset of $(\frac{\mathbb{Z}}{c \mathbb{Z}})^{\times}$ generated by $a$ and $b$ but I can't prove it. Is this a standard result?

What does any of this mean? It seems like this is bound to be an infite set, so how does one define the determinant here? What is $\displaystyle \frac{\mathbb{Z}}{c\mathbb{Z}}$ , a weird way of writing $\mathbb{Z}/c\mathbb{Z}$ ?

**amanda19** · January 5th 2011, 12:08 PM

Yes, $\frac{\mathbb{Z}}{c\mathbb{Z}}$ means $\mathbb{Z}/c\mathbb{Z}$ , but I'll write $\mathbb{Z}/c\mathbb{Z}$ if it's confusing. And yes, the lattice is going to be an infinite set of points (and then the determinant is defined in the standard way - see eg. Lattice (group) - Wikipedia, the free encyclopedia )

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