MTH3003 Weekly Problems 3

Original Documents: Problem Sheet / [[mth3003 weekly problem sheet 3 handwritten solutions.pdf|My Handwritten Solutions]] / Provided Solutions

Vibes: Medium difficulty, ramp up from Week 1 permutations - now geometric intuition for dihedral group, cycle computations on shapes, first proofs using orders/inverses. Drawing diagrams essential for $D_{10}$/shapes.

Used Techniques:

• Product of Permutations for equations, verifying forms like $\sigma {\rho }^i$.
• Quick Inverse Proposition implicitly via $o(\sigma )=2$.
• Cycle notation for symmetries (rotations/reflections in $D_{2n}$).
• Relation $\rho \sigma =\sigma {\rho }^{-1}$ (L5 – Dihedral groups).

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3.1. Solving a Permutation Equation

Question

Find a permutation $\sigma \in S_9$ that satisfies the following equation:

$$(123)(45678)=(1578)\sigma (134)$$

Right-multiply both sides by $(134{)}^{-1}=(143)$: $(123)(45678)(143)=(1578)\sigma$.

Then left-multiply by $(1578{)}^{-1}=(8751)$: $\sigma =(8751)(123)(45678)(143)=\boxed{(23)(48)(56)}$.

Verified: $(1578)(23)(48)(56)(134)=(123)(45678)$.

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3.2. Identities in the Dihedral Group $D_{2n}$

Question

Consider the dihedral group $D_{2n}$. In the usual notation for dihedral groups, $\rho$ is an anticlockwise rotation by $\frac{2\pi }{n}$ and $\sigma$ is a reflection.

Prove that $\rho \sigma =\sigma {\rho }^{-1}=\sigma {\rho }^{n-1}$.

Hint: What is the order of $\rho \sigma$? What is the inverse of $\rho \sigma$?

Geometrically, $\rho \sigma$ is a reflection ($o(\rho \sigma )=2$), so $(\rho \sigma {)}^2=e$ implies $\rho \sigma \rho \sigma =e$.

Left-multiply by ${\rho }^{-1}{\sigma }^{-1}$: $\rho \sigma ={\sigma }^{-1}{\rho }^{-1}$. Since $o(\sigma )=2$, ${\sigma }^{-1}=\sigma$, so $\rho \sigma =\sigma {\rho }^{-1}$.

As $\rho {\rho }^{n-1}={\rho }^n=e$, ${\rho }^{-1}={\rho }^{n-1}$, hence $\rho \sigma =\sigma {\rho }^{n-1}$.

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3.3. Structure of the Dihedral Group $D_{10}$

Question

Consider the dihedral group $D_{10}$. It is the collection of symmetries of a pentagon.

In the usual notation for dihedral groups, label the corners $1\text{–}5$ anticlockwise and take $\rho$ to be an anticlockwise rotation by $\frac{2\pi }{5}$, and $\sigma$ to be a reflection through the line of symmetry that passes through the corner labelled $1$.

1. Write down, using cycle notation, the elements in $D_{10}$. Hint: draw a diagram
2. Verify that the elements of $D_{10}$ can be written as $\{e,\rho ,{\rho }^2,{\rho }^3,{\rho }^4,\sigma ,\sigma \rho ,\sigma {\rho }^2,\sigma {\rho }^3,\sigma {\rho }^4\}$, where all the rotations are of the form ${\rho }^i$ and all the reflections are of the form $\sigma {\rho }^i$.

Remark: In the near future, we will use Lagrange’s Theorem to quickly prove that every dihedral group $D_{2n}$ can be written as $\{e,\rho ,{\rho }^2,\dots ,{\rho }^{n-1},\sigma ,\sigma \rho ,\sigma {\rho }^2,\dots ,\sigma {\rho }^{n-1}\}$, where the rotations are of the form ${\rho }^i$ and the reflections are of the form $\sigma {\rho }^i$.

1. Rotations: $e$, $\rho =(12345)$, ${\rho }^2=(13524)$, ${\rho }^3=(14253)$, ${\rho }^4=(15432)$. Reflections: $\sigma =(25)(34)$, $(13)(45)$, $(15)(24)$, $(12)(35)$, $(14)(23)$.
2. Compute: $\sigma \rho =(15)(24)$, $\sigma {\rho }^2=(14)(23)$, $\sigma {\rho }^3=(13)(45)$, $\sigma {\rho }^4=(12)(35)$. Matches reflections; rotations match powers. Thus, $D_{10}=\{e,\rho ,{\rho }^2,{\rho }^3,{\rho }^4,\sigma ,\sigma \rho ,\sigma {\rho }^2,\sigma {\rho }^3,\sigma {\rho }^4\}$.

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3.4. Symmetry Group of a Labelled Shape

Question

Consider the following shape $S$ (drawn in the Euclidean plane with centre at the origin $O$), whose corners have been labelled $1,2,\dots ,8$ as shown (on original problem sheet).

The symmetry group of isometries of $S$ can be thought of as a subgroup of $S_8$. Write down (in cycle notation) all elements in the symmetry group of $S$.

Rotations (90° $\rho =(1357)(2468)$): $e$, $\rho$, ${\rho }^2=(15)(37)(26)(48)$, ${\rho }^3=(1753)(2864)$.

Reflections (through corners/edges): $(12)(38)(47)(56)$, $(14)(23)(67)(58)$, $(16)(25)(34)(78)$, $(18)(27)(36)(45)$.

$\boxed{\{e,(1357)(2468),(15)(37)(26)(48),(1753)(2864),(12)(38)(47)(56),(14)(23)(67)(58),(16)(25)(34)(78),(18)(27)(36)(45)\}}$ (preserves central square).

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3.5. Constructing a Shape with $D_{12}$ Symmetry

Question

Draw an example of a shape $S$ that is not a hexagon, whose symmetry group of isometries is isomorphic to $D_{12}$.

$\boxed{\text{A regular hexagon with a square attached outwardly to each of its 6 edges}}$ (12 symmetries: 6 rotations, 6 reflections; not a plain hexagon).

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3.6. Generating the Group $D_{10}$

Question

Let $\rho$ and $\sigma$ be (respectively) a nontrivial rotation and reflection in $D_{10}$. In lectures, we will soon see that $⟨\rho ,\sigma ⟩$ means the smallest subgroup of $D_{10}$ containing both $\rho$ and $\sigma$.

Using the following two facts, show that $⟨\rho ,\sigma ⟩=D_{10}$.

1. $D_{10}=\{e,\rho ,{\rho }^2,{\rho }^3,{\rho }^4,\sigma ,\sigma \rho ,\sigma {\rho }^2,\sigma {\rho }^3,\sigma {\rho }^4\}$
2. $\rho \sigma =\sigma {\rho }^{-1}$

Any $g\in ⟨\rho ,\sigma ⟩$ is ${\rho }^{n_1}{\sigma }^{k_1}\cdots {\rho }^{n_m}{\sigma }^{k_m}$. Using (2), commute $\rho$‘s past $\sigma$‘s: each $\sigma {\rho }^n={\rho }^{-n}\sigma$ (iteratively), reducing to ${\sigma }^k{\rho }^n$.

Since $o(\rho )=5$, $o(\sigma )=2$, $k\equiv 0,1\mathrm{mod}2$, $n\equiv 0.\mathrm{mod}5$, so $⟨\rho ,\sigma ⟩=\{e,\rho ,{\rho }^2,{\rho }^3,{\rho }^4,\sigma ,\sigma \rho ,\sigma {\rho }^2,\sigma {\rho }^3,\sigma {\rho }^4\}=D_{10}$ by (1).