# 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.

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.

• Thank you! I understand this, but I'm not getting anywhere with case 2. I know that $P_3=\langle y\rangle$, $y^3=e$ and $P_2=\{e,x_1,x_2,x_3\}$, $x_i^2=e$. $P_3\vartriangleleft G$ so $xyx^{-1}\in P_3$ and G non-abelian so $xyx^{-1}=xyx=y^{-1}$. Then I have $y=xy^{-1}x$. And now I'm lost. How can I find $a$? – Eva May 26 '16 at 19:01
• If,say $x_1yx_1=y^{-1}$ and $x_2yx_2=y^{-1}$, then $x_3yx_3=y$. So in any case one of $x_1,x_2,x_3$ - say $x_3$ must satisfiy $x_3yx_3=y$. Now $a:=x_3y$ has order $6$. – Derek Holt May 26 '16 at 20:38
• $xax=x(xy)x=yx$ and $a^{-1}=y^{-1}x$. But if they are the same then $y=y^{-1}$ which is not true? – Eva May 27 '16 at 6:12
• Put $x=x_1$ and $a=x_3y$. Then $xax=x_3y^{-1}=a^{-1}$. – Derek Holt May 27 '16 at 9:29

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).

• Yes, I know that. But I want to derive $D_6$ and $Dic_{12}$ from their subgroups. – Eva May 26 '16 at 9:53
• as you said, the 2_sylow of $D_6$ is $\Bbb{Z} / 2\Bbb{Z}\times\Bbb{Z} / 2\Bbb{Z}$, while the 2-Sylow of $Disc_{12}$ is $\Bbb{Z} / 4\Bbb{Z}$ So they are different and they are characterized by have order 12, not abelian and do not contains subgroups of order 6. Maybe I misunderstood the question. – m.idaya May 26 '16 at 10:18
• Sorry,i fix a typo in my above comment: D_6 and D_12 are characterized by the order 12, non abelian and containing a subgroup of order 6. – m.idaya May 26 '16 at 13:33