12 Dihedral Groups

#todo/school : Do problems 1, 3, 10 📅 2025-01-10 ✅ 2025-01-10

1

Question

Compute the order of each of the elements in the following groups:
a. $D_{6}$
b. $D_{8}$
c. $D_{10}$

Proof

a. Here $n = 3$ so then we have:

D_{6} = ⟨ r, s | r^{3} = s^{2} = 1, r s = s r^{- 1} ⟩

Then we can compute the orders of each of the operations:

$| 1 | = 1$
$| r | = 3$
$| r^{2} | = 3$ since $(r^{2})^{3} = (r^{3})^{2} = 1^{2} = 1$
$| s | = 2$
$| r s | = 2$ since $(r s)^{2} = (r s) (s r^{- 1}) = r (s^{2}) r^{- 1} = r r^{- 1} = 1$
$| r^{2} s | = 2$ since $(r^{2} s)^{2} = (r^{2} s) (s r^{- 2}) = r^{2} (s^{2}) r^{- 2} = r^{2} r^{- 2} = 1$ .

b. Here $n = 4$ so then we have:

D_{8} = ⟨ r, s | r^{4} = s^{2} = 1, r s = s r^{- 1} ⟩

Then we can compute the orders of each of the operators:

$| 1 | = 1$
$| r | = 4$
$| r^{2} | = 2$ since $r^{4} = (r^{2})^{2} = 1^{2} = 1$
$| r^{3} | = 4$ since $r^{3}, r^{6}, r^{9} \neq 1$
$| s | = 2$
$| r s | = 2$ since $(r s)^{2} = (r s) (s r^{- 1}) = r s^{2} r^{- 1} = 1$ like in the previous part.
$| r^{2} s | = 2$ by similar reasoning.
$| r^{3} s | = 2$ as well.

c. We can just directly compute the orders pretty easily:

$| 1 | = 1$
$| r | = 5$
$| r^{2} | = 5$ since $gcd (2, 5) = 5$
$| r^{3} | = 5$ since $gcd (3, 5) = 5$
$| r^{4} | = 5$ since $gcd (4, 5) = 5$
$| s | = 2$
$| r s | = 2$ by the same argument as the previous part.
$| r^{2} s | = 2$
$| r^{3} s | = 2$
$| r^{4} s | = 2$

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3

Question

Use the generators and relations above to show that every element of $D_{2 n}$ which is not a power of $r$ has order 2. Deduce that $D_{2 n}$ is generated by the two elements $s$ and $s r$ both of which have order 2.

Proof

The only elements that are in $D_{2 n}$ are all the $r^{i} s$ where $0 \leq i \leq n$ so let's show that $r^{i} s$ is always order 2, for all $i$ . Notice that for any of these $i$ 's that the order is 2 since:

(r^{i} s)^{2} = (r^{i} s) (s r^{- i}) = r^{i} s^{2} r^{- i} = r^{i} r^{- i} = 1

Thus showing the order here.

Notice to show that $D_{2 n} = ⟨ s, s r ⟩$ that we just need to show that we can construct $r$ from these generators themselves. Notice we can with:

(s) (s r) = s^{2} r = r

Thus then these generators work, and all the generators have order 2!

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10

Question

Let $G$ be the Group of rigid motions in $R^{3}$ of a cube. Show that $| G | = 24$ .

Proof

Here we have rigid motions, so we cannot flip. We can only rotate around the $x$ -axis ( $r_{x}$ ), the $y$ -axis ( $r_{y}$ ), and the $z$ -axis ( $r_{z}$ ). Thus:

G = ⟨ r_{x}, r_{y}, r_{z} | r_{x}^{4} = r_{y}^{4} = r_{z}^{4} = 1 ⟩

Now notice that the number of unique permutations of the faces can depend on our choices of rotations. This gets complicated so instead lets consider numbering the 6 faces:

If we color one face, there are 6 possibilities for where it ends up
If we draw a face on the face (:)) you can consider the 4 rotations of the face for where it ends up
Thus there are $6 \times 4 = 24$ possible permutations.

Therefore $| G | = 24$ .

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