Dihedral group

**vincisonfire** · November 12th 2008, 02:08 PM

I have to find the order of the elements of the dihedral group $D_4 = \{1,y,y^2,y^3,x,xy,xy^2,xy^3\}$ but I don't really understand what this group is even though I've read my notes 10 times. It must simply be the way it is explain that doesn't fit my mind. So how can we find this and why is it like this?

**whipflip15** · November 12th 2008, 02:28 PM

The group $D_4$ is the group of symmetries of a square. For your group, y corresponds to a clockwise (or anticlockwise) rotation of 90 degrees. So $y^4=1$ where 1 is the identity (not rotating). x can be thought of as 'flipping' the square over and x^2=1.

Wiki explains it better

Dihedral group - Wikipedia, the free encyclopedia

**vincisonfire** · November 12th 2008, 02:29 PM

Ok but what about the order of the elements?

**ThePerfectHacker** · November 12th 2008, 05:01 PM

Originally Posted by vincisonfire

I have to find the order of the elements of the dihedral group $D_4 = \{1,y,y^2,y^3,x,xy,xy^2,xy^3\}$ but I don't really understand what this group is even though I've read my notes 10 times. It must simply be the way it is explain that doesn't fit my mind. So how can we find this and why is it like this?

The order of $y$ is 4. Therefore the order of $y^3$ is 4 also since $\gcd(3,4)=1$ . While the order of $y^2$ is 2 since $\gcd(2,4)=2$ .

Now, $(xy^a)^2 = xy^axy^a = x(y^ax)y^a=x(xy^{-a})y^a = x^2 = 1$ .
Thus, $x,xy,xy^2.xy^3$ each has order 2.

**whipflip15** · November 12th 2008, 05:02 PM

If you label the corners of a square 1,2,3,4 how many times do you rotate to get back to the original position (the identity)?

If it helps you can label the corners 1,2,3,4 then your elements can be thought of as permutations on the corners of the square, y=(1234) and x=(13)(24)

Hence o(y)=4, o(x)=2

Try the rest

**vincisonfire** · November 12th 2008, 05:07 PM

Thanks for the answers but how can I deduce that the order of y is 4 and the order of y^2 is gcd(2,4) and so on?

**ThePerfectHacker** · November 12th 2008, 05:54 PM

Originally Posted by vincisonfire

Thanks for the answers but how can I deduce that the order of y is 4 and the order of y^2 is gcd(2,4) and so on?

It should be a know fact that the dihedral group is generated by two elements. One has order 4 (geometrically these are rotations) and the other has order 2 (geometrically these are reflections). Therefore $y$ has order 4, that is basically the definition of the dihedral group. If $y$ has order 4 then $y^2$ has order 2 because $(y^2)^2 = y^4 = 1$ . Similarly $y^3$ has order 4.

In general if $a\in G$ (any group) has order $n$ then $a^k$ has order $\frac{n}{\gcd(k,n)}$ .

**vincisonfire** · November 12th 2008, 05:57 PM

Thanks, I got it.
Sorry the definition we saw in class was similar to cardinality but I found the "right" definition on the web as you posted your answer.

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