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$D_4 = \{R_0,R_{90},R_{180},R_{270},H,V,D,D^{'}\}$

$|R_0|=1 ,\, |R_{90}|=4 ,\, |R_{180}|=2 ,\, |R_{270}|=4$

$|D|=|V|=|D|=|D^{'}|= 2$

So Does that mean order of $D_4$ is $4 = \text{lcm}\, (4,2,1).$

But answer is $8$.

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2  
The order of a group is not necessarily the $\mbox{lcm}$ of the orders of its elements. $D_4$ is a good example of that, and so is $C_2^2$. –  Alfonso Fernandez Mar 23 '13 at 14:14
6  
The notation used for dihedral groups is unfortunately inconsistent. About half of the world calls the dihedral group of order $2n$ $D_{2n}$, because it has $2n$ elements, and the other half calls it $D_n$, because it is the symmetry group of a regular $n$-gon. –  Derek Holt Mar 23 '13 at 16:15

3 Answers 3

up vote 11 down vote accepted

The order of $D_4$ is 8. This is because that list of elements that you gave, $$ \{R_0,R_{90},R_{180},R_{270},H,V,D,D^{'} \}$$

contains 8 items.

The orders of the various elements, $|R_0|, |V|$, and so on, must divide the order of the whole group. This follows from Lagrange's theorem. Here for example, each of 1, 2, and 4 divides the order of the group, 8.

But in general, there is no way to compute the order of a group just from the orders of its elements. For example, there is a group whose elements (except the identity) all have order 2 but which is infinite. Also for every $n$ there is a finite group of order $2^n$ whose elements (except the identity) all have order 2.

The Klein $V$-group has order 4, but its elements all have order 1 or 2:

$$\begin{array}{c|cccc} \oplus & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ 1 & 1 & 0 & 3 & 2 \\ 2 & 2 & 3 & 0 & 1 \\ 3 & 3 & 2 & 1 & 0 \\ \end{array}$$

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... assuming all of the items in the list of 8 items are distinct. (which they are) –  Hurkyl Mar 23 '13 at 14:23
    
@MJD: what is the operator in Klein V-group.I did not understand how $(1)^{2}$ = 0 –  TLE Mar 23 '13 at 14:29
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@stranger001: You can either take MJD's table as the definition of the operation, or you can do bitwise XOR. –  Jyrki Lahtonen Mar 23 '13 at 14:38
    
@stranger001: It isn't multiplication as you'd expect, but the special operation $\oplus$ as defined by the Cayley table. –  Cameron Buie Mar 23 '13 at 14:40
    
@stranger001 The operator in the Klein $V$-group is the one I showed in the table! –  MJD Mar 23 '13 at 15:45

I guess you are conflating two different meanings of the term order, plus the term exponent.

The order of a (sub)group is the number of its elements.

The order of an element $a$ is the least $k > 0$ such that $a^{k} = 1$. (If it exists, otherwise $\infty$.)

The link is that the order of the element $a$ is the order of the subgroup $\langle a \rangle$.

The least common multiple of the orders of the elements in a finite group $G$ is called the exponent of $G$. As hinted in other answers, the exponent of $G$ divides the order of $G$, but the two might be different.

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Besides to other answers which contain complete aspects of the answer to your question; I suggest you to consider another presentation of $D_4$ (or $D_8$). I am sure via this equivalent presentation, you can find the elements easier and find where you are confusing at. That is: $$D_4=\langle a,b\mid a^4=b^2=1, ba=a^{-1}b\rangle=\{1,a,a^2,a^3,b,ab,a^2b,a^3b\}$$ which has $8$ elements. Try this one too!

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+ 1 and $> 20$K!!! :-D –  amWhy Mar 24 '13 at 0:08
    
Congratulations! $\ddot\smile$ –  amWhy Mar 24 '13 at 2:01

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