How many distinct composition series does the group $D_{12}$ have? 
How many distinct composition series does the group $D_{12}$ have?

I know that $D_{12}  \trianglerighteq \mathbb{Z}_6  \trianglerighteq \mathbb{Z}_3 \trianglerighteq \{e\}$ is a composition series ( as the elemebts in the series are of index $2$ and therefore therefore maximal, so the quotients must be simple).
Moreover, I know that the Jordan-Holder Theorem states that any two composition series have the same composition length and the same composition factors, up to permutation (reordering) and isomorphism.
Does it mean that the answer is $1$? Or that could be composition series of other length?
 A: Let $G = \langle x,y \mid x^6 = y^2 = (xy)^2 = 1 \rangle \cong D_{12}.$ So $x$ is a rotation of order $6$, and $y$ is a reflection. Then the four distinct composition series of $G$ are:
$$1.\ \ \ \ G > \langle x \rangle > \langle x^3 \rangle > 1; \ \ \ (D_{12} > C_6 >C_2 > 1)$$
$$2.\ \ \ \ G > \langle x \rangle > \langle x^2 \rangle > 1; \ \ \ (D_{12} > C_6 >C_3 > 1)$$
$$3.\ \ \ \ G > \langle x^2,y \rangle > \langle x^2 \rangle > 1; \ \ \ (D_{12} > D_6 >C_3 > 1)$$
$$4.\ \ \ \ G > \langle x^2,xy \rangle > \langle x^2 \rangle > 1; \ \ \ (D_{12} > D_6 >C_3 > 1)$$
A: There are two others up to isomorphism.  I'm on my phone so I'm going to make typing easy.  In the series you give with $Z/6Z$ the top factor, you can also put $Z/2Z$ next.  There is also a series through $D_6$ and $Z/3Z$.  The top factor must have index 2, since an index 3 subgroup is 2-Sylow and not normal as there are 3 2-Sylows.  You can check that these are the only three possibilities.  However, since there are two $D_6$ subgroups, there are two series of that type related by an automorphism of $D_{12}$, as Derek Holt pointed out. So there are a total of four composition series.
