# Subgroups

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• May 11th 2010, 11:16 PM
Primenumber
Subgroups
1. Draw the lattice of subgroups (up to conjugacy) of $\displaystyle S_{3} \times \mathbb{Z}_{2}$.

The symmetry of a system of balancing balls is given by
$\displaystyle S_{3} \times \mathbb{Z}_{2}$ are $\displaystyle S_{3} \times \mathbb{Z}_{2}$ which can be thought of as the group $\displaystyle \{id,(123),(321),(12),(13),(23),-id,-(123),-(321),-(12),-(13),-(23)\}$

Now I know that, up to conjugacy, the subgroups of $\displaystyle S_{3} \times \mathbb{Z}_{2}$ are $\displaystyle S_{3} \times \mathbb{Z}_{2}$, $\displaystyle \mathbb{Z}_{3} \times \mathbb{Z}_{2}=\{id,-id,(123),(321),-(123),-(321)\}$, $\displaystyle \mathbb{Z}_{2} \times \mathbb{Z}_{2}=\{id,-id,(12),-(12)\}$, $\displaystyle S_{3}=\{id,(123),(321),(12),(13),(23)\}$, $\displaystyle \mathbb{Z}_{3}=\{id,(123),(321)\}$, $\displaystyle \mathbb{Z}_{2}^{'}=\{id,(12)\}$, $\displaystyle \mathbb{Z}_{2}=\{id,-id\}$, $\displaystyle \widetilde{\mathbb{Z}_{2}}=\{id,-(12)\}$ and $\displaystyle I={\id\}$.

However, I don't know how to construct a lattice of subgroups from this. How should it be structured?

2. For each of the subgroups write down, and justify, its Fixed Point Subspace when acting on the two dimensional surface given by $\displaystyle \{\mathbb{R}^{3}: x_{1}+x_{2}+x_{3}=0 \}$.

What is a Fixed Point Subspace? How do you find it?

3. Draw the lattice of isotropy subgroups up to conjugacy.

Same problem as in 1.

If someone could enlighten me on this that would be great.
Thanks!
• May 12th 2010, 07:04 AM
TheArtofSymmetry
Quote:

Originally Posted by Primenumber
1. Draw the lattice of subgroups (up to conjugacy) of $\displaystyle S_{3} \times \mathbb{Z}_{2}$.

The symmetry of a system of balancing balls is given by
$\displaystyle S_{3} \times \mathbb{Z}_{2}$ are $\displaystyle S_{3} \times \mathbb{Z}_{2}$ which can be thought of as the group $\displaystyle \{id,(123),(321),(12),(13),(23),-id,-(123),-(321),-(12),-(13),-(23)\}$

Now I know that, up to conjugacy, the subgroups of $\displaystyle S_{3} \times \mathbb{Z}_{2}$ are $\displaystyle S_{3} \times \mathbb{Z}_{2}$, $\displaystyle \mathbb{Z}_{3} \times \mathbb{Z}_{2}=\{id,-id,(123),(321),-(123),-(321)\}$, $\displaystyle \mathbb{Z}_{2} \times \mathbb{Z}_{2}=\{id,-id,(12),-(12)\}$, $\displaystyle S_{3}=\{id,(123),(321),(12),(13),(23)\}$, $\displaystyle \mathbb{Z}_{3}=\{id,(123),(321)\}$, $\displaystyle \mathbb{Z}_{2}^{'}=\{id,(12)\}$, $\displaystyle \mathbb{Z}_{2}=\{id,-id\}$, $\displaystyle \widetilde{\mathbb{Z}_{2}}=\{id,-(12)\}$ and $\displaystyle I={\id\}$.

However, I don't know how to construct a lattice of subgroups from this. How should it be structured?

The action of $\displaystyle S_3$ alone in the hyperplane seems to be straightforward. On the contrary, is it true that $\displaystyle -(1 2)x_1=-x_2$?

I think you simply need to draw the subgroup lattice with a single bottom element $\displaystyle \{id\}$ and the next level elements $\displaystyle \{id, -id\}, \{id, (1 2)\}$, and $\displaystyle \{id, -(1 2)\}$, and the top element $\displaystyle S_3 \times \mathbb{Z}_2$. You need to fill the intermediate elements between them.

Quote:

2. For each of the subgroups write down, and justify, its Fixed Point Subspace when acting on the two dimensional surface given by $\displaystyle \{\mathbb{R}^{3}: x_{1}+x_{2}+x_{3}=0 \}$.

What is a Fixed Point Subspace? How do you find it?

For instance, every subgroup of $\displaystyle S_3$ fixes $\displaystyle \{\mathbb{R}^{3}: x_{1}+x_{2}+x_{3}=0 \}$.

Quote:

3. Draw the lattice of isotropy subgroups up to conjugacy.

Same problem as in 1.

If someone could enlighten me on this that would be great.
Thanks!

This part may need some lengthy computations and take every combination of $\displaystyle x_1, x_2,$ and $\displaystyle x_3$ into account, and their isotropy subgroups.