How would you write $D_4$ in cycle notations? When writing $D_4$ in cycle notation, I am struggling to find unique cycle notations for each of the rotations. I find that all the rotations would be written as the cycle:
$$( 1 2 3 4 )=( 2 3 4 1 )=( 3 4 1 2 )=(4 1 2 3)$$
However, since $D_4$ has order $8$, shouldn't $D_4$ contain $8$ unique cycle elements?
We have $D_4=\{ ( 1 2 3 4 ), ( 2 1 4 3), ( 3 4 2 1), (2 3), (1 4)\}$
 A: $D_4$ is the  group of symmetries on a square, and also is a subgroup of $S_4$ so let's label a square as thus:

A key point: the group of symmetries does not consist solely of rotations. It also consists of reflections about various axes. For even-sided polygons, $D_n$ includes reflections across the axes that go through opposite pairs of vertices, and those through the midpoints of opposing sides. For odd-sided polygons, the reflections consist of those across the axes formed by considering the line through each vertex and the midpoint of the side opposite it.
It might be clearer if I include the axes for each reflection. Each applicable axis is in blue below:

A regular $n$-gon will also have $n-1$ rotational symmetries, and $n$ reflection symmetries, plus a final "do nothing" symmetry that serves as the identity for the group. This means $D_n$ has $2n$ elements in total - so you are correct, $D_4$ should have eight elements.
With this all in mind, it becomes easy to figure out the elements of $D_4$ in cycle notation, since it will just be the elements of $S_4$ interchanging the corresponding numbers that align with the vertices in the symmetry of the polygon. For example: the symmetry corresponding to a reflection about the main diagonal fixes $2$ and $4$, but swaps $1$ and $3$. Thus we could write this as $(2) \; (4) \; (1 \; 3)$, or more compactly by eliminating the $1$-cycles and just having $(1 \; 3)$.
In summary: the idea is to determine all of the symmetries of the square, and then consider where each symmetry maps each vertex/number. Write that result in cycle notation.
Namely, $D_4$ consists of:

*

*Identity / "Do Nothing:" $e$ (or $0$ or $id$ or whatever notation you prefer)

*Rotation $90^\circ$ counterclockwise: $(1 \; 2 \; 3 \; 4)$

*Rotation $180^\circ$ counterclockwise: $(1 \; 3) \; (2 \; 4)$

*Rotation $270^\circ$ counterclockwise: $(1 \; 4 \; 3 \; 2)$

*Reflection about vertical axis: $(1 \; 2) \; (3 \; 4)$

*Reflection about horizontal axis: $(1 \; 4) \; (2 \; 3)$

*Reflection about main diagonal (the one through $2,4$): $(1 \; 3)$

*Reflection about anti-diagonal (the one through $1,3$) $(2 \; 4)$


I find that all the rotations would be written as the cycle:
$( 1 2 3 4 )=( 2 3 4 1 )=( 3 4 1 2 )=(4 1 2 3)$

That would only correspond to a $90^\circ$ rotation counterclockwise. It does not encompass the other two rotational symmetries.

We have $D_4=\{ ( 1 2 3 4 ), ( 2 1 4 3), ( 3 4 2 1), (2 3), (1 4)\}$

Based on this (namely the last two elements), I imagine you noticed the diagonal symmetries, and thus have your square labeled as below:

On this premise, then:

*

*The first and second symmetry make no sense to me (maybe I guessed your numbering wrong?)

*The third symmetry is indeed a $90^\circ$ counterclockwise rotation

*The fourth and fifth symmetries are about the diagonals

*The $5$ symmetries you lack (if we throw away the first two that don't make sense) are the identity, $180^\circ$ and $270^\circ$ rotations, and reflections about opposing midpoints

A: $D_4$ is generated by $(1234)$, a rotation,  and $(14)(23)$, a reflection. 
So we get $D_4=\{e,(1234),(13)(24),(1432),(14)(23),(42),(13),(12)(34)\}$. 
