I came across a question asking for me to find all the possible quotient groups for the dihedral group $D_6$. How must I go about this?


1 Answer 1


I assume that your definition of $D_6$ is $$D_6=\{1,r,r^2,s,sr,sr^2\}$$ where $r^3=s^2=rsrs=1$. A quotient set $D_6/N$ is a group if and only if $N\lhd D_6$. Hence, we want to find all normal subgroups of $D_6$. We always have the trivial ones $$D_6/D_6\cong\{1\}$$ and $$D_6/\{1\}\cong D_6.$$ The subgroup $\langle r\rangle = \langle r^2\rangle= \{1,r,r^2\}$ is of index $2$ and thus normal. So $$D_6/\langle r\rangle$$ is a quotient group also. If $N\lhd D_6$ is a different normal subgroup, then $|N|=2$, so either $N=\langle s \rangle$, $N=\langle sr\rangle$ or $N=\langle sr^2\rangle$. But none of them are normal. For example $\langle s \rangle$ is not normal since $(sr)s(sr)^{-1}=sr^2\notin\langle s \rangle$. Hence, the only non-trivial quotient group is $$D_6/\langle r\rangle\cong\mathbb{Z}/2\mathbb{Z}$$

  • $\begingroup$ How can we just assume $D_6$? $\endgroup$
    – Artemisia
    Nov 24, 2013 at 21:03
  • $\begingroup$ I meant: "I assume that your definition of $D_6$ is the following..." I will clarify. $\endgroup$
    – Spenser
    Nov 24, 2013 at 21:14

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