What does a vertical bar mean in this context?

I'm doing a few exercises regarding Schnorr's identification scheme. I have the exercise starting off like this, with the values defined:

Let $$p = 311$$ and $$r = 31\ |\ (p - 1)$$. Let $$g = 169$$, which has order $$r$$.

I just really can't figure out what the vertical bar means here?

Sometimes in discrete maths, a vertical bar means absolute value, sometimes two them are cardinalty? Programming would suggest that it means a logical or?

One place I saw something indicating that it might be xor, but I really have no idea.

• Do you have more context? $|$ can also mean divides in that 31 divides 310, but it doesn't make sense to assign that to $r$ May 18 '21 at 15:57
• @AmanGrewal This is apparently a widely used problem, see books.google.com/… for one example. May 18 '21 at 16:52

The meaning of that $$\ \vert\$$ in this context is divides (as in evenly divides, or is a divisor of), and that's a standard usage of this sign. The quote should be read as:

let $$p=311$$ and $$r=31$$, which divides $$(p-1)$$. Let $$g=169$$

In other words: $$r$$ is a divisor of $$p-1$$. Or, exists integer $$q$$ with $$r\times q=p-1$$. Or, $$((p-1)\bmod r)=0$$, also writable as $$p-1\bmod r=0$$ or $$p-1\equiv0\pmod r$$ or $$p\equiv1\pmod r$$. In many common programing languages, (p-1)%r == 0. That's because $$31$$ (evenly) divides $$311-1$$, since $$31\times10=310$$.

That was correctly guessed by Aman Grewal in comment, but as noted, proximity with the assignment makes the notation confusing. Elision of the implied which is something I would try to avoid.

The end of the sentence says «Let $$g=169$$, which has order $$r$$». Does that mean that $$g$$ is really 169%31?

No. The term order is used in its meaning in group theory. In this context, it means that when we repeatedly multiply $$1$$ by $$g$$, reducing modulo $$p$$ after each multiplication, we'll first get back to $$1$$ after performing $$r$$ multiplications. That's related to $$r\,\vert\,(p-1)$$, because the order of any element in a finite group is a divisor of the order of the group, that is the number of elements in the group. Here the group is the multiplicative group modulo $$p$$, noted $$\mathbb Z_p^*$$ or $$(\mathbb Z/p\mathbb Z)^\times$$, which has $$p-1$$ elements since $$p$$ is prime. The powers of $$g$$ form a subgroup of order $$r$$, called a Schnorr group.

• Thanks, One more thing that I don't understand. The end of the sentence says Let G=169, which has order r does that mean that G is really 169%31? May 18 '21 at 17:37
• thanks a lot again. A quick addition though,. I have just received an answe that the actual value of r is actually 8? How does this make sense? May 18 '21 at 18:04
• @Garsty100: Asking Wolfram order of 169 modulo 311 yields 31, thus the quote is correct. I recommend studying some group theory. That serves in many fields (including the non-mathematical sense of that).
– fgrieu
May 18 '21 at 18:42
• “Let… $r$ be 31, which divides $(p-1)$…” May 19 '21 at 18:48