I am having trouble finding the order of each element in $D_4$.
I know $D_4 = \{R_0, R_{90}, R_{180}, R_{270}, V, H, D, D'\}$ and $|D_4| = 8$.
How would I go about finding the order of the elements?
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See this duplicate for its subgroups, hence the orders of elements. – Dietrich Burde Oct 05 '17 at 16:24
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Start with the definition of order, which loosely means in this context: "how many times can you iterate one of the operations before everything is back to how it started?" For instance, $R_{90}$ refers to a rotation by 90 degrees. If we do that 4 times, our object is back to how it started, so the order of that element is 4.
Aaron Montgomery
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so let me see if i have done this right: $|R_0| =1, |R_{90}|=4, |R_{180}|=2, |R_{270}|=4, |V|=1? |H|=1?, |D|=2, |D'|=2$ – rover2 Oct 05 '17 at 14:03
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Not quite. A general hint: the only element of any group that can ever have order 1 is the identity element, which you've called $R_0$ here. – Aaron Montgomery Oct 05 '17 at 14:04
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are my answers for the rotations right? i was having trouble with $V, H, D, D'$ – rover2 Oct 05 '17 at 14:05
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The Dihedral Group D4 has representation $ \{I,A,B,C,D,E,F,G\} $ which
$ I=
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} $,
$ A=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix} $,
$ B=
\begin{bmatrix}
-1 & 0 \\
0 & -1
\end{bmatrix} $,
$ C=
\begin{bmatrix}
0 & 1 \\
-1 & 0
\end{bmatrix} $,
$ D=
\begin{bmatrix}
-1 & 0 \\
0 & 1
\end{bmatrix} $,
$ E=
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix} $,
$ F=
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix} $,
$ G=
\begin{bmatrix}
0 & -1 \\
-1 & 0
\end{bmatrix} $
By multiplication of Matrices, the order of each element is
$ |I|=1 $, $ |A|=4 $, $ |B|=2 $, $ |C|=4 $, $ |D|=2 $, $ |E|=2 $, $ |F|=4 $, $ |G|=2 $.
Pieter Geerkens
- 923
Sotiris Z.
- 45