How many necklaces can we create?

  • MHB
  • Thread starter mathmari
  • Start date
In summary: Thinking )Yes.So the necklace will be $BABBA$.Suppose we pick the first bead to be $A$.If we rotate it to the right by one bead, that means the second bead also has to be $A$ doesn't... ( Thinking )Yes.So the necklace will be $BABBA$.
  • #36
I like Serena said:
Take a look at $D_6$:The axis of symmetry of $s$ in this picture goes through $2$ beads.
The axis of $rs$ goes through $0$ beads. (Thinking)

I got stuck right now... Isn't $m$ constant in this example, $m=6$ ? (Wondering)
How can we take cases if $m$ is odd or even? (Wondering)
 
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  • #37
mathmari said:
I got stuck right now... Isn't $m$ constant in this example, $m=6$ ? (Wondering)
How can we take cases if $m$ is odd or even? (Wondering)

I'm afraid $m$ has a different meaning in the picture. (Worried)
In the picture the number of beads is $6$.
And $m$ represents what we called $i$, the number of rotations that we apply.
 
  • #38
I like Serena said:
I'm afraid $m$ has a different meaning in the picture. (Worried)
In the picture the number of beads is $6$.
And $m$ represents what we called $i$, the number of rotations that we apply.

When we have an odd number of beads, would we have again that when the number of reflections is even then the symmetry axis goes through two vertices and when it is odd then the symmetry axis goes through sides? (Wondering)
 
  • #39
mathmari said:
When we have an odd number of beads, would we have again that when the number of reflections is even then the symmetry axis goes through two vertices and when it is odd then the symmetry axis goes through sides? (Wondering)

Suppose we take a look at $D_5$, which is represented by a pentagon.
What would it look like? (Wondering)
 
  • #40
I like Serena said:
Suppose we take a look at $D_5$, which is represented by a pentagon.
What would it look like? (Wondering)

It is of the following form:

View attachment 5403

right? (Wondering)
 

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  • #41
mathmari said:
It is of the following form:



right? (Wondering)

The vertical line is an axis of symmetry with 2 vertices to the left and 2 to the right.
However, the horizontal line has 1 above and 2 below. (Worried)
 
  • #42
I like Serena said:
The vertical line is an axis of symmetry with 2 vertices to the left and 2 to the right.
However, the horizontal line has 1 above and 2 below. (Worried)

So, when we have an odd number of beads the necklace is unchanged only under the reflection with the vertical line as the axis of symmetry, or not? (Wondering)
Or do we consider also the other two lines as the axis of symmetry of a reflection? (Wondering)
 
  • #43
mathmari said:
So, when we have an odd number of beads the necklace is unchanged only under the reflection with the vertical line as the axis of symmetry, or not? (Wondering)
Or do we consider also the other two lines as the axis of symmetry of a reflection? (Wondering)

The picture should be like:
View attachment 5410

There are 5 axes of symmetry, each intersecting with 1 bead. (Thinking)
 

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  • #44
I like Serena said:
The picture should be like:There are 5 axes of symmetry, each intersecting with 1 bead. (Thinking)

Ah ok... I see... (Thinking)

So, when we have an odd number of beads, there are $k^{\lfloor \frac{m}{2}\rfloor+1}=k^{\lceil \frac{m}{2}\rceil}$ necklaces that stay unchanged under a reflection, or not? (Wondering)

And when we have an even number of beads, there are either $k^{\frac{m}{2}}$ necklaces, when we rotate by an odd $i$, or $k^{\frac{m-2}{2}+2}=k^{\frac{m+2}{2}}=k^{\frac{m}{2}+1}$ necklaces, when we rotate by an even $i$ that stay unchanged under a reflection, right? (Wondering)
 
  • #45
Right. (Nod)
 

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