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I'm studying group theory and now I'm analyzing non-abelian groups of order 12. I see that the dihedral group $D_6$ can be expressed by $$D_6=\langle a,b : a^6=1, b^2=1, aba=b\rangle =\langle 1,a,a^2,a^3,a^4,a^5, b, ba, ba^2, ba^3, ba^4, ba^5\rangle,$$ and the alternating group $A_4$ by $$A_4=\langle 1, (123),(132),(124),(142),(134),(143),(234),(243),(12)(34),(13)(24), (14)(23)\rangle.$$ I want to know what are the elements of the dicyclic group $\text{Dic}_3$ defined as $$\text{Dic}_3=\langle a,b : a^6=1, b^2=a^3, bab^{-1}a=1\rangle.$$ Any help would be appreciate.

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  • $\begingroup$ In the second displayed equation you mean $A_4 = $, not $D_6 =$. Also, you ask "what are the elements" of that last group, but what do you mean? Some people would say they just are what they are. After all, it is not essential that every element of a group isomorphic to $A_4$ has to be defined as an even permutation of 4 elements. And you wrote out the elements of $D_6$ in an abstract way, so why are you not satisfied doing the same thing with the last group of order 12 (after you know it exists)? $\endgroup$ – KCd Jul 1 '18 at 20:41
  • $\begingroup$ I've corrected the mistake. I want to write the elements in an abstract way to find some properties easier (I'm a beginner at this so it helps me) $\endgroup$ – user326159 Jul 1 '18 at 20:58
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    $\begingroup$ Notice that the subgroup $\langle a\rangle$ is normal in ${\rm Dic}_3$, so the elements can be labelled in the same way as in $D_6$, though obviously the binary operation will be different. $\endgroup$ – Robert Chamberlain Jul 1 '18 at 21:15
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Just to follow on from my comment:

Notice that the subgroup $\langle a\rangle$ is normal in ${\rm Dic}_3$, so the elements can be labelled in the same way as in $D_6$, though obviously the binary operation will be different.

In fact it is relatively easy to check that ${\rm Dic}_3$ is generated by $a^2$ and $b$ and that $\langle a^2$ is a normal subgroup of order $3$, so ${\rm Dic}_3\cong C_3\rtimes C_4=\langle x,y|x^3=y^4=1,yxy^{-1}=x^{-1}\rangle$. This is similar to the description $D_{6}\cong C_6\rtimes C_2$ that makes dihedral groups perhaps a little more tangible.

I noticed this by chance, but could have found it by looking for the Sylow $3$-subgroups. This is often a good place to start when trying to understand a new finite group.

Alternatively, for very tangible elements, one may identify ${\rm Dic}_3$ with the subgroup $\langle (1,2,3),(2,3)(4,5,6,7)\rangle$ of $S_7$.

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  • $\begingroup$ Then, I can express the group as $\text{Dic}_3=\{1, a^2, a^4, b, ba^2, ba^4, b^2,b^2a^2, b^2a^4, b^3, b^3a^2,b^3a^4 \}$, can't I? $\endgroup$ – user326159 Jul 2 '18 at 8:26
  • $\begingroup$ @mishu Yes, that is one way of expressing the elements. Another, to my mind easier, would be $Dic_3 = \{ 1, a, a^2, a^3, a^4, a^5, b, ba, ba^2, ba^3, ba^4, ba^5 \}$. $\endgroup$ – Leppala Jul 2 '18 at 11:23
  • $\begingroup$ But as the order of $b$ is $4$ and by the definition of the group, $b^2$ is not the identity and $b^2\in \text{Dic}_3$. What element of your expression corresponds to $b^2$? $\endgroup$ – user326159 Jul 2 '18 at 11:44
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    $\begingroup$ both of you are correct, if you substitute $b^2=a^3$ you go from mishu's set to Leppala's $\endgroup$ – Robert Chamberlain Jul 2 '18 at 17:48
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[Aside: I recently made use of this group in one of my other answers]

Many groups can be represented as 3-dimensional rotation groups. For instance, any cyclic group can be understood as a group of rotations. The same is also true of the dihedral groups, where the elements of order $2$ can be understood as $180^\circ$ rotations of 3d space.

3d rotations can be represented as quaternions. Except this introduces some redundancy, as every rotation can be expressed as two quaternions: some quaternion $q$ and its negation $-q$. For example, a $0^\circ$ rotation can be expressed by the quaternions $1$ and $-1$.

Therefore if one takes a 3D rotation group, like $D_3$ (the dihedral group of order $6$), and models its elements as quaternions, one gets a group of double the order. The resulting group can be denoted $2D_3$, where $2$ denotes the doubling. The resulting group is the dicyclic group of order $12$. In fact, this is how one generates all the dicyclic groups.

By the way, as an aside, every non-abelian group of order up to $15$ can be understood either as a 3D rotation group or a quaternionic double-covering of a 3D rotation group. Some abelian groups like $C_2\times C_2$ (the Klein 4-group) can also be understood this way (in that particular case, it's a group consisting of $180^\circ$ rotations about $3$ orthogonal axes and a single identity rotation; its quaternionic double-covering is the group $Q$ which is the smallest dicyclic group).

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