I'm now studying Fraleigh's Abstract algebra(7th). In section 8, there is a group table for $D_4$, with some strange notations that I can't compute it easily.

He uses $\rho_0=\left( \begin{array}{cccc} 1&2&3&4\\1&2&3&4 \end{array} \right)$, $\rho_1=\left( \begin{array}{cccc} 1&2&3&4\\2&3&4&1 \end{array} \right)$, $\rho_2=\left( \begin{array}{cccc} 1&2&3&4\\3&4&1&2 \end{array} \right)$, $\rho_3=\left( \begin{array}{cccc} 1&2&3&4\\4&1&2&3 \end{array} \right)$, $\mu_1=\left( \begin{array}{cccc} 1&2&3&4\\2&1&4&3 \end{array} \right)$, $\mu_2=\left( \begin{array}{cccc} 1&2&3&4\\4&3&2&1 \end{array} \right)$, $\delta_1=\left( \begin{array}{cccc} 1&2&3&4\\3&2&1&4 \end{array} \right)$, $\delta_2=\left( \begin{array}{cccc} 1&2&3&4\\1&4&3&2 \end{array} \right)$. ($\rho$ is for rotations, $\mu$ is for mirror images in perpendicular bisectors of sides, and $\delta$ is for diagonal flips)

Although he gave the full table of $D_4$ with this notation in the book, I want to compute any element operation without seeing the table. But to compute $\mu_1 \delta_1$, I have to recall what permuatations were the $\mu_1, \delta_1$ and then do the function composition, and then have to recall what's the name for the permuation($\rho_3$).

Is the above process unnecessary and you can compute it easily? Is there any way to understand the strange notation? Or can you recommend any other easy-computable notation for $D_n$(e.g., group presentation)?

  • 1
    $\begingroup$ Yes...and no. Perhaps the biggest problem with groups is that the most "natural" way of giving a group is by means of generators and relations (look up "group presentation"). However, this method is flawed (search for the word problem - basically, given an arbitrary presentation one cannot determine if two elements are different!). So people study group representations. What you are looking at is a nice representation for your group. But your right - it doesn't give you any real "feel" for $D_4$. A presentation for $D_4$ is $<a, b: a^4=1, b^2=1, bab=a^{-1}>$ with $a=\rho_1$, $b=\mu_1$. $\endgroup$
    – user1729
    Jul 11, 2011 at 9:06
  • 2
    $\begingroup$ Strange notation? That's the standard notation for permutations $\endgroup$ Jul 11, 2011 at 9:09
  • $\begingroup$ @Gobi: The group $S_3$ is isomorphic to $D_6$. Construct a cayley table and you will understand. Then try and understand this for $D_4$. $\endgroup$
    – user38268
    Jul 11, 2011 at 22:34

2 Answers 2


The dihedreal groups $D_n$ (or $D_{2n}$ for some authors) are meant to encode the symmetries of the regular $n$-gon. If we let $rho$ be the rotation by $2\pi/n$, and $\tau$ some reflection (any one will do), we have that $\rho$ and $\tau$ generate $D_n$. Half the time, when you think about the dihedral group, this is what you want to think about (not just the group, but the natural way the group acts).

We can give a nice presentation for $D_n$ in terms of $\rho$ and $\tau$ by noting that $\rho^n=\tau^2=e$, and $\tau \rho =\rho^{-1}\tau$. Once you have shown that all these relations hold, we notice that these relations allow us to simplify any expression in $\rho$ and $\tau$ to one of the form $\rho^k\tau^{\epsilon}$ where $k\in \{0,\ldots, n-1\}$ and $\epsilon \in \{0,1\}$. If you want to work with elements of $D_n$, this is usually the most convenient way.

The approach taken by Fraleigh is essentially an algebraic version of our first point of view. Namely, we take a square, and we label the vertices with the numbers $1,2,3,$ and $4$. Then every element of $D_4$ permutes the vertices, and specifying how the permutation acts on the vertices completely determines the action. Therefore, we get a concrete realization of the group by writing out how the elements permute the vertices.

While one can explicitly multiply out the permutations algebraically, it is easier to proceed as follows: We let $\rho$ act as the permutation $\rho(i)=i+1 \pmod n$ and $\tau(i)=-i\pmod n$. Note that $\tau$ is the reflection about the line through the vertex $n$. Then, using the group presentation, it is easy to multiply elements in the group. Once you express the various reflections in terms of $\rho$ and $\tau$, determining the permutations corresponding to a product is straight forward.

For example, when $n=4$, we can compute the action of $\rho^3 \tau$ by $\rho^3\tau(i)=\rho^3(-i) = -i+3 \pmod 4.$

Just to clarify his notation, given a matrix, the corresponding permutation is the one sending $i$ to the element that appears right underneath $i$. For example, the permutation $i\mapsto i+1 \pmod 4$ is expressed with the matrix $\pmatrix{1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1}.$


To understand this notation you need to draw a square and put the numbers $1,2,3,4$ at the vertices. Now the notation makes sense. For instance $\rho_1$ will be a rotation by $\frac{\pi}{2}$, i.e $1$ goes to $2$, $2$ goes to $3$, $3$ goes to $4$ and $4$ to $1$. Apply the same reasoning to all of them and compose them by drawing pictures of the square.

In other words you have to look at the columns and see to which number in the second line each number in the first line is going.

  • $\begingroup$ I misunderstood how elements of $D_4$ acts on a square(with the numbers 1,2,3,4 at the vertices). I didn't know exactly how to do the second action since the numbers 1,2,3,4 at the vertices have been changed by the first action. But now I got complete understanding of it due to this site: I should do "action" regardless of what the numbers are at vertices. $\endgroup$
    – Gobi
    Jul 13, 2011 at 8:21

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