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The permutation induces on the set

mathmari

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Apr 14, 2013
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Hey!! :eek:

I am looking at the following exercise:

Make a sketch of a regular tetrahedron and label the corners with the numbers $1, 2, 3, 4$. For $1\leq i\leq 5$ the permutations $\pi_i \in \text{Sym} (4)$ are defined as follows:
\begin{align*}&\pi_1:=\text{id} \\ &\pi_2:=(1 \ \ 2) \\ &\pi_3:=(1 \ \ 2 \ \ 3) \\ &\pi_4:=(1 \ \ 2)\ (3 \ \ 4) \\ &\pi_5:=(1 \ \ 2 \ \ 3 \ \ 4)\end{align*}
  1. For $1 \leq i \leq 5$, check whether there is a symmetry of the tetrahedron that the permutation $\pi_i$ induces on the set of vertices $\{1, 2, 3, 4\}$ and, if so, give a geometric description of this symmetry.
  2. Is there for every $\pi \in \text{Sym} (4)$ a symmetry of the tetrahedron, that induces the permutation $\pi$ on the set of vertices $\{1, 2, 3, 4\}$ ?


Could you explain to me the exercise and what I am supposed to do?

I got stuck what it means "induce on a set".

(Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Hey mathmari !!

Those permutations are of themselves merely reorderings of numbers.
At the next step we consider that those numbers represent the corners of the tetrahedron.
The question is then if for instance swapping corners 1 and 2 is actually a symmetry of the tetrahedron.
As it is, it corresponds to a reflection in the plane through corners 3 and 4 that transforms the tetrahedron to itself with corners 1 and 2 swapped.
So they say that the permutation (1 2) "induces" a reflection on the tetrahedron, which is a symmetry of the tetrahedron. (Nerd)

How about the others? What kind of geometric transformation do they represent if any?
Are they symmetries of the tetrahedron? (Thinking)
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,036
Those permutations are of themselves merely reorderings of numbers.
At the next step we consider that those numbers represent the corners of the tetrahedron.
The question is then if for instance swapping corners 1 and 2 is actually a symmetry of the tetrahedron.
As it is, it corresponds to a reflection in the plane through corners 3 and 4 that transforms the tetrahedron to itself with corners 1 and 2 swapped.
So they say that the permutation (1 2) "induces" a reflection on the tetrahedron, which is a symmetry of the tetrahedron. (Nerd)

How about the others? What kind of geometric transformation do they represent if any?
Are they symmetries of the tetrahedron? (Thinking)
As for $\pi_3$, the question is then if corner 1 is changed to corner 2 and corner 2 is changed to corner 3 and corner 3 changed to corner 1 is a symmetry of the tetrahedron.
As it is, it corresponds to a rotation around the axis of corner 4 and the middle of its opposite side.
So they say that the permutation (1 2 3) "induces" a rotation of the tetrahedron, which is a symmetry of the tetrahedron.

As for $\pi_4$, the question is then if swapping corners 1 and 2 and swapping corners 3 and 4 is a symmetry of the tetrahedron.
As it is, it corresponds to a rotation around the axis that passes through the midpoints of two opposite edges.
So they say that the permutation (1 2)(3 4) "induces" a rotation of the tetrahedron, which is a symmetry of the tetrahedron.

As for $\pi_5$, the question is then if corner 1 is changed to corner 2 and corner 2 is changed to corner 3 and corner 3 changed to corner 4 and corner 4 changed to corner 1 is a symmetry of the tetrahedron.
This is not a symmetry, is it?


Is everything correct so far? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Looks good.
Before we decide whether $\pi_5$ is a symmetry or not, perhaps we should list all symmetries of the tetrahedron? (Thinking)
 

mathmari

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Apr 14, 2013
4,036
Looks good.
Before we decide whether $\pi_5$ is a symmetry or not, perhaps we should list all symmetries of the tetrahedron? (Thinking)
We have the identity.
We have rotations when 1 vertex is fixed.
We have rotations when no vertex is fixed.
We have reflections.

(Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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We have the identity.
We have rotations when 1 vertex is fixed.
We have rotations when no vertex is fixed.
We have reflections.
How many of each? (Wondering)

There are $4!=24$ permutations of 4 numbers.
The symmetries of the tetrahedron must be a subset with a number of elements that must divide 24. (Thinking)
 

mathmari

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Apr 14, 2013
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How many of each? (Wondering)

There are $4!=24$ permutations of 4 numbers.
The symmetries of the tetrahedron must be a subset with a number of elements that must divide 24. (Thinking)
We have 1 identity.

Rotations when one vertex is fixed: 8

So far we have 9.

To find the others, do we have to find all permitations of 4 numbers?
 

Klaas van Aarsen

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Mar 5, 2012
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We have 1 identity.

Rotations when one vertex is fixed: 8

So far we have 9.

To find the others, do we have to find all permitations of 4 numbers?
We can use the permutations to find more symmetries.
After all we already know that (1 2) corresponds to a reflection. Just like all other cycles of 2 numbers.
They are the reflections in the planes that contain an edge of the polyhedron.
How many would there be? (Wondering)
 

mathmari

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Apr 14, 2013
4,036
We can use the permutations to find more symmetries.
After all we already know that (1 2) corresponds to a reflection. Just like all other cycles of 2 numbers.
They are the reflections in the planes that contain an edge of the polyhedron.
How many would there be? (Wondering)
We have (1 2), (1 3), (1 4), (2 3), (2 4), (3 4). So the number of such reflections is 6, right? (Wondering)
 

Klaas van Aarsen

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mathmari

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Apr 14, 2013
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Klaas van Aarsen

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Mar 5, 2012
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So we have 6 more symmetries, so 9+6 = 15.

So are just the rotations missing? Or have I forgotten something?
How about an axis of rotation through 2 opposing edges? (Wondering)

Which permutations of the 24 are left? (Wondering)
 

mathmari

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Apr 14, 2013
4,036
How about an axis of rotation through 2 opposing edges? (Wondering)

Which permutations of the 24 are left? (Wondering)
We have the following:
  • Identity. - 1
  • Rotation so that one vertex is fixed. For each vertex there are two such rotations. Therefore is no total there 2* 4 = 8 such rotations. - 8
  • Rotations where two pair of vertices are switched. This is a rotation of $180^{\circ}$ around the perpendicular bisector of the edges of the two pairs of vertices. These rotations are of the form: (1 2)(3 4), (1 3)(2 4), (1 4)(2 4). - 3
  • Reflections about the plane that goes through one edge and the midpoint of the opposite edge. These reflections switch two edges. These reflections are of the form: (1 2), (1 3), (1 4), (2 3), (2 4), (3 4). - 6

What am I missing? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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We have the following:
  • Identity. - 1
  • Rotation so that one vertex is fixed. For each vertex there are two such rotations. Therefore is no total there 2* 4 = 8 such rotations. - 8
  • Rotations where two pair of vertices are switched. This is a rotation of $180^{\circ}$ around the perpendicular bisector of the edges of the two pairs of vertices. These rotations are of the form: (1 2)(3 4), (1 3)(2 4), (1 4)(2 4). - 3
  • Reflections about the plane that goes through one edge and the midpoint of the opposite edge. These reflections switch two edges. These reflections are of the form: (1 2), (1 3), (1 4), (2 3), (2 4), (3 4). - 6

What am I missing?
Which symmetries did we have for a square? (Wondering)
 

mathmari

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Apr 14, 2013
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Which symmetries did we have for a square? (Wondering)
At a square we have the rotations by 0, 90, 180, 270 degrees.
The reflections along the perpendicular bisectors.
The reflections along the diagonal and the antidiagonal.

Have I missed something?
 

Klaas van Aarsen

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Mar 5, 2012
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At a square we have the rotations by 0, 90, 180, 270 degrees.
The reflections along the perpendicular bisectors.
The reflections along the diagonal and the antidiagonal.

Have I missed something?
Ah, I actually wanted to point out that we can have compositions of a rotation and a reflection.
However, in the case of a square, these are also reflections. (Blush)

Either way, in 3 dimensions we have something called improper rotations, also called rotation-reflections.
For instance the permutation (1 2 3)(3 4) = (1 2 3 4) is a reflection followed by a rotation, which is also a symmetry in 3 dimensions. (Thinking)
 

mathmari

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Apr 14, 2013
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Either way, in 3 dimensions we have something called improper rotations, also called rotation-reflections.
For instance the permutation (1 2 3)(3 4) = (1 2 3 4) is a reflection followed by a rotation, which is also a symmetry in 3 dimensions. (Thinking)
Ahhh (Thinking)

At at the first question in each there is a symmetry of the tetrahedron that the permutation πi induces on the set of vertices {1,2,3,4}, right?

At the second question do we have to proof that the symmetry group of the tetraeder is the permutation group of the 4 numbers? Or which is the difference between the two questions? (Wondering)
 
Last edited:

Klaas van Aarsen

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Mar 5, 2012
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Ahhh (Thinking)

At at the first question in each there is a symmetry of the tetrahedron that the permutation πi induces on the set of vertices {1,2,3,4}, right?
Yes. (Nod)
We found that $\pi_5 = (1\,2\,3\,4)$ is also a symmetry, and that it is geometrically a rotation-reflection.

At the second question do we have to proof that the symmetry group of the tetraeder is the permutation group of the 4 numbers? Or which is the difference between the two questions? (Wondering)
In the first question we verified that 5 specific permutations of 4 numbers induced symmetries of the tetrahedron.
And we also specified which geometrical transformations they were.

In the second question we have to verify that all 24 permutations correspond to a geometrical symmetry of the tetrahedron.
You've listed 18 of them. We could already tell that it couldn't be right, since 18 does not divide 24.
Do we have all 24 now? (Wondering)
 

mathmari

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Apr 14, 2013
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In the first question we verified that 5 specific permutations of 4 numbers induced symmetries of the tetrahedron.
And we also specified which geometrical transformations they were.

In the second question we have to verify that all 24 permutations correspond to a geometrical symmetry of the tetrahedron.
You've listed 18 of them. We could already tell that it couldn't be right, since 18 does not divide 24.
Do we have all 24 now? (Wondering)
We have the following permuatations:

  • Identity. - 1
  • Rotations that one vertex is fixed. Such rotations are of the form: $(3\ \ 4 \ \ 2)$, $(4\ \ 2 \ \ 3)$, $(3\ \ 4 \ \ 1)$, $(4\ \ 1 \ \ 3)$, $(2\ \ 4 \ \ 1)$, $(4\ \ 1 \ \ 2)$, $(2\ \ 3 \ \ 1)$, $(3\ \ 1 \ \ 2)$. - 8
  • Rotations that two pair of vertices are switched. Such rotations are of the form: $(1 2)(3 4)$, $(1 3)(2 4)$, $(1 4)(2 4)$. - 3
  • Reflections that switch two vertices. Such reflections are of the form: $(1 2)$, $(1 3)$, $(1 4)$, $(2 3)$, $(2 4)$, $(3 4)$. - 6
  • Rotations-reflections. For that do we take every product of a reflection with every rotation? Which order do we take? Or is it the same and we can consider both orders?
    We would have:
    $(3\ \ 4 \ \ 2)(1 2)$, $(4\ \ 2 \ \ 3)(1 2)$, $(3\ \ 4 \ \ 1)(1 2)$, $(4\ \ 1 \ \ 3)(1 2)$, $(2\ \ 4 \ \ 1)(1 2)$, $(4\ \ 1 \ \ 2)(1 2)$, $(2\ \ 3 \ \ 1)(1 2)$, $(3\ \ 1 \ \ 2)(1 2)$
    $(3\ \ 4 \ \ 2)(1 3)$, $(4\ \ 2 \ \ 3)(1 3)$, $(3\ \ 4 \ \ 1)(1 3)$, $(4\ \ 1 \ \ 3)(1 3)$, $(2\ \ 4 \ \ 1)(1 3)$, $(4\ \ 1 \ \ 2)(1 3)$, $(2\ \ 3 \ \ 1)(1 3)$, $(3\ \ 1 \ \ 2)(1 3)$
    $(3\ \ 4 \ \ 2)(1 4)$, $(4\ \ 2 \ \ 3)(1 4)$, $(3\ \ 4 \ \ 1)(1 4)$, $(4\ \ 1 \ \ 3)(1 4)$, $(2\ \ 4 \ \ 1)(1 4)$, $(4\ \ 1 \ \ 2)(1 4)$, $(2\ \ 3 \ \ 1)(1 4)$, $(3\ \ 1 \ \ 2)(1 4)$
    $(3\ \ 4 \ \ 2)(2 3)$, $(4\ \ 2 \ \ 3)(2 3)$, $(3\ \ 4 \ \ 1)(2 3)$, $(4\ \ 1 \ \ 3)(2 3)$, $(2\ \ 4 \ \ 1)(2 3)$, $(4\ \ 1 \ \ 2)(2 3)$, $(2\ \ 3 \ \ 1)(2 3)$, $(3\ \ 1 \ \ 2)(2 3)$
    $(3\ \ 4 \ \ 2)(2 4)$, $(4\ \ 2 \ \ 3)(2 4)$, $(3\ \ 4 \ \ 1)(2 4)$, $(4\ \ 1 \ \ 3)(2 4)$, $(2\ \ 4 \ \ 1)(2 4)$, $(4\ \ 1 \ \ 2)(2 4)$, $(2\ \ 3 \ \ 1)(2 4)$, $(3\ \ 1 \ \ 2)(2 4)$
    $(3\ \ 4 \ \ 2)(3 4)$, $(4\ \ 2 \ \ 3)(3 4)$, $(3\ \ 4 \ \ 1)(3 4)$, $(4\ \ 1 \ \ 3)(3 4)$, $(2\ \ 4 \ \ 1)(3 4)$, $(4\ \ 1 \ \ 2)(3 4)$, $(2\ \ 3 \ \ 1)(3 4)$, $(3\ \ 1 \ \ 2)(3 4)$
    If we calculate them we may see that we have some elements more than once? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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Each of those must map to one of the possible 24 permutations of 4 numbers.
And not all of them will actually be rotation-reflections.
For instance (123)(12)=(13), which is a reflection we already had.
Which permutations are still missing such as (1234)? (Wondering)
 

mathmari

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Apr 14, 2013
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Which permutations are still missing such as (1234)? (Wondering)
The permutations that are missing are:
$(1 \ \ 4 \ \ 3 \ \ 2)$
$(1 \ \ 3 \ \ 4 \ \ 2)$
$(1 \ \ 2 \ \ 4 \ \ 3)$
$(1 \ \ 3 \ \ 2 \ \ 4)$
$(1 \ \ 4 \ \ 2 \ \ 3)$

Which am I still missing? (Wondering)
 

Klaas van Aarsen

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Mar 5, 2012
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The permutations that are missing are:
$(1 \ \ 4 \ \ 3 \ \ 2)$
$(1 \ \ 3 \ \ 4 \ \ 2)$
$(1 \ \ 2 \ \ 4 \ \ 3)$
$(1 \ \ 3 \ \ 2 \ \ 4)$
$(1 \ \ 4 \ \ 2 \ \ 3)$

Which am I still missing?
So there are 6 of them.
That makes 24 with the 18 you already have.
So we have all of them! (Happy)
 

mathmari

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So there are 6 of them.
That makes 24 with the 18 you already have.
So we have all of them! (Happy)
Ahh yes!!

So to show that all these permutations correspond to geometric symmetries we have to show that these 6 are rotation-reflections, since we have already seen the others, right?

So show this do we have to write them as product of smaller cycles? (Wondering)
 

Klaas van Aarsen

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Ahh yes!!

So to show that all these permutations correspond to geometric symmetries we have to show that these 6 are rotation-reflections, since we have already seen the others, right?

So show this do we have to write them as product of smaller cycles?
Yes. (Nod)