$D_3=\big<a,b| a^3=b^2=(ab)^2=1\big>$ and $|D_3|=6 $
$D_3=\{ 1, a, a^2, b, ab, a^2b\} $
Are these correct?
Then so that I can compute the orders of all the elements of the group $D_3$
$\endgroup$ 52 Answers
$\begingroup$Assuming your title reads 'What are the elements of the dihedral group $D_3$ (which has order $6$)?', rather than 'what are the order $6$ elements in $D_3$'.
Then you must be careful. Your presentation reads $\langle a,b\mid a^3=b^2=1,(ab)^2=1\rangle$, so $a^3=1$, and so your $6$ elements are not correct. One idea here is to consider elements that you do know are in the group, and then use the group axioms to work out the rest (of the representatives, up to relations). It's clear that $\{1,a,a^2,b\}\subset D_3$.
Then you can consider the closure axiom. Each of these multiplied by $a$, $a^2$ and $b$ must still be in the group. So consider the elements $a^2,a^3,ab$. The first is already in our, the second is the identity and hence is already in our set, and the third wasn't, so let us now consider $\{1,a,a^2,b,ab\}$ what about multiplying by $b$ now? $b,ab,a^2b,b^2,ab^2$ all of these are in our group but for $a^2b$. So you have $\{1,a,a^2,b,ab,a^2b\}\subseteq D_3$.
Then what are the orders? $a^3=1, (a^2)^3=1,b^2=1,(ab)^2=1,(a^2b)^2=1$.
$$(a^2b)^2=a^2ba^2b=aabaab=a(abab)b^{-1}ab=ab^{-1}ab=abab=1$$
$\endgroup$ 4 $\begingroup$$<1>=1$, $<a>=<a^2>=3$, $<b>=<ab>=<ba^2>=2$.
$$(ba^2)^2=ba^2ba^2=ba(abab)ba=baba=b(abab)b=b^2=1.$$
$\endgroup$ 5
