How to show that for the Schläfli symbol$\{m,k\}$ the polygon is non-degenerate if $m$ and $k$ are coprime?

Question

I know by definition that when the elements of the Schläfli symbol $\{m,k\}$ are coprime then the polygon is non-degenerate, i.e. can be traced without lifting a pencil off of the paper. Is it possible to prove or demonstrate this mathematically, perhaps by using modular arithmetic? (I only have a basic understanding of modular arithmetic, so explanation of the notation and logic would be really useful.)

Answer 1

Short answer: The question boils down to whether $k$ is a zero divisor, modulo $m$. This happens precisely when $k$ and $m$ are not coprime, so that $\{m/k\}$ is indeed connected (it takes $m$ full steps of size $k$ to get back to where we started) whenever $m$ and $k$ are coprime. But if you're not sure about some of this, keep reading.

Let's look at star polygons with symbol $\{12/k\}$, rather than an arbitrary $m$. We'll see later that the argument generalizes to all $m$ with no additional effort. Now you can just imagine a standard clock, but a good mathematician's clock, numbered $0$ through $11$ (so $0$ at the top rather than $12$). Now we're doing modular arithmetic, so we'll use $\equiv$ rather than $=$ occasionally.

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Modular Arithmetic Interlude

Two numbers, $x$ and $y$, are congruent modulo $12$, written $x \equiv y \pmod {12}$, if they differ by a multiple of $12$; in symbols, if $12 \mid (x - y)$. That's why we can replace $12$ with $0$, as $12 \mid (12 - 0)$, or $0 \equiv 12 \pmod {12}$. Everything that follows (or has preceded) works using an arbitrary modulus $m$ rather than $12$.

A lot of arithmetic works the same, working modulo $12$: For example, $3 + 4 = 7 \equiv 7 \pmod {12}$, and $2 \cdot 5 = 10 \equiv 10 \pmod {12}$, but when we encounter something like $8 + 10 = 18$, it's customary to "reduce modulo $12$", which just means subtracting $12$ from our result until we land between $0$ and $11$ (inclusive). So for example, $8 + 10 = 18 \equiv 18 - 12 = 6 \pmod {12}$: 8 hours after 10 o'clock is of course 6 o'clock.

Also note that writing $8 + 10 \equiv 18 \pmod {12}$, or even $8 + 10 \equiv 30 \pmod {12}$, is perfectly OK, as $18 \equiv 6 \equiv 30 \pmod {12}$; it's just customary (and handy) to reduce and use only the numbers $0$ through $11$.

It's the same with multiplication. Take $5 \cdot 9 = 45 \equiv 45 - 3\cdot 12 = 45 - 36 = 9 \pmod {12}$: If you do five tasks that take 9 hours each, and start at 1 o'clock, you'll be done at 10 o'clock (several days later, perhaps!)

Main idea: zero divisors.

If $x$ and $y$ are real numbers, then $xy = 0$ implies that at least one of $x$ or $y$ must be $0$. This is what lets us factor quadratics to find solutions, for example (e.g., if $x^2 + 2x - 15 = 0$, then $(x + 5)(x - 3) = 0$ and we set our factors equal to $0$ for solutions).

But this needn't be the case working modulo $12$! For example, $4 \cdot 3 = 12 \equiv 0 \pmod {12}$! Or, slightly more interestingly, $8 \cdot 6 = 48 \equiv 0 \pmod {12}$. How can this happen? Factoring $12$ into a product of powers of primes, we see that $12 = 2^2 \cdot 3^1$. So if our pair of numbers have two factors of $2$ and a factor of $3$ between them, their product will be $0 \pmod {12}$. This was the case for $4$ and $3$, and for $8$ and $6$ as well. This wasn't the case for $5$ and $9$, so their product isn't congruent to $0 \pmod {12}$.

For a fixed number $k$, we say that $k$ is a zero divisor modulo $12$ if there's a nonzero solution to the equation $k \cdot x \equiv 0 \pmod {12}$. Above we saw that $3, 4, 6,$ and $8$ are all zero divisors modulo $12$ (in fact, the only numbers that aren't zero divisors, modulo $12$, are those that are coprime to $12$).

Modular arithmetic is covered in great detail in any book on elementary number theory, for example Rosen's Number Theory that I used as a student.

Interlude Over

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Now let's think more about these $\{12/k\}$ star polygons. The idea is that we start at $0$ and keep taking steps of size $k$ until we get back to where we started; until our path closes up. We'll hit all the vertices exactly when this takes $12$ steps (think about steps of size $3$: After $4$ steps, we're back at $0$ already, since $4 \cdot 3 \equiv 0 \pmod {12}$). Hitting all of the vertices is exactly what it means for $\{12/k\}$ to be connected.

So the question, then: When will it take the full $12$ steps of size $k$ before we get back to $0$? Precisely when $k$ and $12$ are coprime and share only $1$ as a common divisor! Suppose that $k$ and $12$ aren't coprime, so that there's some prime $p$ dividing both $k$ and $12$ (if concrete numbers help, think about $k = 2 = p$, or $k = 3 = p$). Then, after taking $\frac{12}{p}$ steps, we'll be at vertex

$$k \cdot \frac{12}{p} \pmod {12},$$

and we ask: Is this vertex $0$? That is, have we closed our path after only $\frac{12}{p} < 12$ steps? This is exactly equivalent to asking whether $k \cdot \frac{12}{p} \equiv 0 \pmod {12}$.

But we saw earlier that the product of two things is congruent to $0$ modulo $12$ exactly when the two things combined have all prime divisors of $12$, counting multiplicity, as factors. The only prime that could possibly be missing from $\frac{12k}{p}$ is $p$, but remember: We said that $p$ divides both $12$ and $k$, so $p^2$ must divide $12k$, meaning $p$ must still divide $\frac{12k}{p}$: Our path must have closed early.

Whenever $k$ and $12$ are coprime, then in order to have $k \cdot x \equiv 0 \pmod {12}$, all prime factors (counting multiplicity) of $12$ must come from $x$, our number of steps, alone: In other words, we must have $12 \mid x$, and must take a multiple of $12$ steps, in order for our path to close. In this case, we hit all vertices before winding up where we started, and $\{12/k\}$ is connected.

Our argument really didn't use $12$ in any specific way, and could have been any arbitrary $m$.