How many necklaces can we create?

[I like Serena]

The second bead and the last but one should also be of the same colour, right?

Right.
So we have AB.BA

 

[mathmari]

Right.
So we have AB.BA

So for that reflection there are $k^{\lceil \frac{m}{2}\rceil}$ necklaces that stay unchanged, or not? (Wondering)

 

[I like Serena]

So for that reflection there are $k^{\lceil \frac{m}{2}\rceil}$ necklaces that stay unchanged, or not? (Wondering)

Yeap.
The axis of symmetry always goes through either 0, 1, or 2 beads.
In all those cases we get $\lceil \frac{m}{2}\rceil$ choices for the color. (Nod)

 

[mathmari]

The axis of symmetry always goes through either 0, 1, or 2 beads.

When the number of beads is odd the axis of symmetry goes through $1$ bead and when the number of beads is even the axis of symmetry goes through $2$ beads, right? (Wondering)

When goes the axis of symmetry through $0$ beads? (Wondering)

 

[I like Serena]

When the number of beads is odd the axis of symmetry goes through $1$ bead and when the number of beads is even the axis of symmetry goes through $2$ beads, right? (Wondering)

When goes the axis of symmetry through $0$ beads? (Wondering)

Take a look at $D_6$:
View attachment 5402

The axis of symmetry of $s$ in this picture goes through $2$ beads.
The axis of $rs$ goes through $0$ beads. (Thinking)

 

[mathmari]

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)

 

[I like Serena]

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.

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

Right. (Nod)