All groups of order 12; Dic12 and D6 I know this has been asked in various forms before, but so far I have failed to understand those answers properly. I've also read several papers discussing this, but I don't really get it. 
I have an assignment in my algebraic structures class to find all groups of order 12. I know that there are 5; $Z_{12}, Z_2\times Z_2\times Z_3, A_4, D_6, Dic_{12}$. I have found the first three and shown that they are not isomorphic. So that leaves me $D_6\cong Z_2\times S_3$ (or $D_{12}$), the symmetries of a regular hexagon, and $Dic_{12}$, which I'm not familiar with.
Using Sylow's theorem I have that these two have 1 3-sylow subgroup and 3 2-sylow subgroups. The 3-sylow subgroup $P_3\cong Z_3$. The 2-sylow subgroups are either isomorphic to $Z_4$ or $Z_2\times Z_2$.
Case 1:
$P_3\cong Z_3$ and 2-sylow subgroups $\cong Z_4$ should give me $Dic_{12}=\{a,x|a^6=e, x^2=a^3, xax^{-1}=a^{-1}\}$ but I don't know how to show this.
Case 2:
$P_3\cong Z_3$ and 2-sylow subgroups $\cong Z_2\times Z_2$ should give me $D_6=\{a,x|a^6=x^2=e, xax=a^{-1}\}$ but I don't know how to show this either.
 A: in cas 2, is by definition of the group $D_6$. The second $Disc_{12}$ is a group not abelian and deferent to $D_6 $ and $A_4$,  the deferent  between the structure of $Disc_{12}$ and $D_6$ that the group $Disc_{12}$
contain an element $x$  of order $4$, and the second group $D_6$ no contain element fo order  $4$, of course this two groups are derent to $A_4$ (this is a classification of  non abelian groupes of order 12).
A: For Case 1, you have $P_3=\langle y \rangle$ with $|y|=3$ and $P_2 = \langle x \rangle$ with $|x|=4$.
Since $P_3 \lhd G$, we have $xyx^{-1} \in P$, so it must equal $y$ or $y^{-1}$. But we are assuming that $G$ is nonabelian, so $xyx^{-1}=y^{-1}$.
Hence $x^2yx^{-2}=xy^{-1}x^{-1} = y$, so $x^2$ (which has order $2$) commutes with $y$, and hence $a:=x^2y$ has order $6$.
Now $a^3=x^6y^3=x^2$ and $xax^{-1}=x(x^2y)x^{-1}=x^2y^{-1}=a^{-1}$. So $G$ is isomorphic to the group $\langle a,x \mid a^6=1,x^2=a^3,xax^{-1}=a^{-1}\rangle$, which is your definition of ${\rm Dic}_{12}$.
Maybe you could have another go at Case 2 yourself.
