I Symmetry involving prime multiplication modulo 8

[Kevin McHugh]

I was reading Armstrong's Groups and Symmetry the other day and saw this table. It has beautiful symmetry. It is the the prime numbers multiplied modulo 8. It creates one of the most elegant things I've ever seen. What is so special about modulo 8 that creates such a symmetric matrix of primes?

 

[fresh_42]

Could you be a bit more specific? Which primes are you talking about? All primes in the natural numbers, or only those less than eight? Or the primes in $\mathbb{Z}_8$?

 

[Kevin McHugh]

I expect this is true for all primes, since it works for primes > 8.

PS Stupid me, I can't get the matrix lined up correctly.

 

[mfb]

What does that array represent, and where is the symmetry?
How can something mod 8 be 9?

I expect this is true for all primes, since it works for primes > 8.

That's not how mathematics works.

 

[Kevin McHugh]

What does that array represent, and where is the symmetry?
How can something mod 8 be 9?

That's not how mathematics works.

The matrix is supposed to be the primes 1-7 multiplied modulo 8. There is beautiful symmetry there.

 

[fresh_42]

I guess you have to leave out $2$ for the symmetry. Then you simply have a $(3 \times 3)-$matrix which is of course symmetric, because multiplication in $\mathbb{Z}_n$ is commutative for all $n$, not just eight. So it boils down to all diagonal entries being $p\cdot p = 1 \mod(8)$. But for every odd number $2n+1$ holds (not only primes): $(2n+1)^2=4n^2+4n+1 = 1 \mod(8)$ because either $n$ is even, then $4n$ and $4n^2$ are divisible by eight, or $n$ is odd and each term $4n$ and $4n^2$ provides a $4$ that adds up to eight.

I guess I missed your point.

 

[TeethWhitener]

Are you maybe referring to the fact that the set {1,3,5,7} under multiplication mod 8 is isomorphic with the Klein four-group?

 

[Kevin McHugh]

Are you maybe referring to the fact that the set {1,3,5,7} under multiplication mod 8 is isomorphic with the Klein four-group?

Are those the r s rotations of the dihedral group?

 

[Kevin McHugh]

I guess you have to leave out $2$ for the symmetry. Then you simply have a $(3 \times 3)-$matrix which is of course symmetric, because multiplication in $\mathbb{Z}_n$ is commutative for all $n$, not just eight. So it boils down to all diagonal entries being $p\cdot p = 1 \mod(8)$. But for every odd number $2n+1$ holds (not only primes): $(2n+1)^2=4n^2+4n+1 = 1 \mod(8)$ because either $n$ is even, then $4n$ and $4n^2$ are divisible by eight, or $n$ is odd and each term $4n$ and $4n^2$ provides a $4$ that adds up to eight.

I guess I missed your point.

No, it's a 4x4 matrix.

 

[fresh_42]

Are those the r s rotations of the dihedral group?

In this case, yes.

 

[TeethWhitener]

Are those the r s rotations of the dihedral group?

$C_2 \times C_2$ is isomorphic to the dihedral group $D_2$, if that's what you mean.

 

[fresh_42]

No, it's a 4x4 matrix.

But one isn't prime, but two is. And two disturbs the permutation.

 

[Kevin McHugh]

But one isn't prime, but two is. And two disturbs the permutation.

How come one isn't prime? It's factors are 1 and itself. Forgive my ignorance in all things mathematical, I'm just a dumb chemist.

 

[fresh_42]

How come one isn't prime? It's factors are 1 and itself. Forgive my ignorance in all things mathematical, I'm just a dumb chemist.

It's useful to exclude units (elements, that can be inverted; so in case of integers $\pm 1$). It makes theories and theorems more elegant. In addition the proper definition of a prime is:

"If $p$ divides a product $a \cdot b$ forces that it has to divide either $a$ or $b$ (or both), then we call $p$ a prime."

And with this definition it wouldn't make sense to allow units, because they simply divide anything and the condition above became self-evident.

Edit: As an example: "Every natural number can be written as a product of primes."
If we allow $\pm 1$ to be primes, how would this short sentence end up? Because we can add as many $1$ as we like, so we would have to add "... of course without $1$." Ugly.

 

[Kevin McHugh]

$C_2 \times C_2$ is isomorphic to the dihedral group $D_2$, if that's what you mean.

In chemistry, C2 is an axis of symmetry. Are these rotations about an axis of symmetry? What does C2 X C2 mean, are they successive rotations about different axes?

 

[fresh_42]

In chemistry, C2 is an axis of symmetry. Are these rotations about an axis of symmetry? What does C2 X C2 mean, are they successive rotations about different axes?

One is a reflection (s / symmetry), and one a rotation (r). In this case by $180°$.
https://en.wikipedia.org/wiki/Dihedral_group

 

[TeethWhitener]

In chemistry, C2 is an axis of symmetry. Are these rotations about an axis of symmetry? What does C2 X C2 mean, are they successive rotations about different axes?

In this case, I used $C_2$ to refer to the cyclic group of order 2. $C_2 \times C_2$ is the direct product of two copies of this group:
https://en.wikipedia.org/wiki/Direct_product_of_groups

 

[Kevin McHugh]

Thank you both gentlemen, I am learning something new everyday. Now back to my original question. Why is the set {1,3,5,7) multiplied by modulo 8 so symmetric? Is there something special about modulo 8 versus modulo Z? Did Fresh 42 answer this question is post # 6?

 

[TeethWhitener]

It can be shown that the set of natural numbers less than n that are relatively prime to n form a group under multiplication mod n. Therefore, the set {1,3,5,7} forms a group under multiplication mod 8. There are 4 members in this group, and it can also be shown that any 4-member group is isomorphic to either $C_4$, the cyclic 4-member group (with a single generator for all elements in the group) or the Klein group (isomorphic to $C_2\times C_2$ or $D_2$). The set {1,3,5,7} under multiplication mod 8 can't be isomorphic to the cyclic 4-group because $1^2 \equiv 3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \mod 8$, so no single element generates the rest of the elements in the group. Therefore, it has to be isomorphic to the Klein group (the $D_2$ dihedral group, or if you like, the symmetry group of a non-square rectangle). (You could also just compare the multiplication tables of the two groups directly).

Geometrically, we can see the relationship between the group above and the symmetry of a (non-square) rectangle in 2 dimensions. The 4 elements of the symmetry group for the rectangle are 1) the identity, 2) 180 degree rotation about the symmetry axis, 3 and 4) the two reflections perpendicular to the plane of the rectangle. The congruence $1^2 \equiv 3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \mod 8$ corresponds to the fact that 1) if you apply the identity twice, you get the identity, 2) if you rotate by 180 degrees twice, you get the identity, 3 and 4) if you reflect along the same mirror plane twice, you get the identity.

 

[Kevin McHugh]

Wow, thanks for that TW, I'm going to digest this tonight. Cheers!

 

[zinq]

"the Klein group . . . or if you like the symmetry group of a non-square rectangle"

I will make this more precise:

* This "Klein group" is known as the "Klein four-group". Other groups are almost never called a four-group so as to avoid confusion with this one.

* Instead of a "rectangle" think of a 3-dimensional rectangular solid with all three sides unequal. We will call this a "shoebox".

* Instead of "the symmetry group" — which often means the group of all self-mappings that preserve distance (a.k.a. the isometry group), think of the set of all isometries of the shoebox obtained solely from rotations. These are exactly the 180° rotations about each of its three axes of symmetry (normally the coordinate axes), as well as doing nothing (the identity rotation).

 

[Kevin McHugh]

Thank you Zinq, that is good info. I always like to hear/see different perspectives on the same topic. It usually leads to a deeper understanding of the underlying principle.