4. Descibe the symmetries of a thombus and prove that the set of symmetries forms a group. Give Cayley fables for both the symmetries of a rectanige and the symmetries of a thombus, Are the symmetries of a rectangle and those of a monbus the same?

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Consider a quadrilateral in the plane. We wish to describe its symmetry (isometry) transformations that preserve its geometric structure. In our discussion the two cases are:

• A rhombus (sometimes “thombus” is used to mean “rhombus”). In the general case of a non‐square rhombus the only symmetries are the identity and one reflection.  
• A (non‐square) rectangle. Its symmetries include the identity, two reflections (across the lines joining the midpoints of opposite sides), and a $$180^\circ$$ rotation.

We now explain these cases step by step.

---

### 1. Symmetries of a Rhombus

A general rhombus has all four sides equal. However, unless it is a square the angles are not all equal so that very few isometries preserve the vertex‐ordering. In particular, one may show that:

• There is always the identity transformation, denoted by $$e$$.  
• There is exactly one nontrivial isometry: a reflection, say $$f$$, about the line joining the vertices where the acute angles lie (or equivalently about the longer diagonal). This line is the unique axis of symmetry.

Thus the set of symmetries of a non‐square rhombus is  
$$
\{e, f\}.
$$

#### Proving that $$\{e,f\}$$ Forms a Group

To show that a set with a binary operation (here, the composition of functions) is a group we must check the following axioms:

1. **Closure:**  
   Compose any two symmetries. We have:
   - $$e \circ e = e,$$
   - $$e \circ f = f,$$
   - $$f \circ e = f,$$
   - $$f \circ f = e$$ (since reflecting twice across the same line gives the identity).  
   Hence, the composition of any two elements is in $$\{e,f\}$$.

2. **Associativity:**  
   Composition of functions is always associative.

3. **Identity:**  
   The identity transformation $$e$$ clearly satisfies $$e \circ x = x \circ e = x$$ for $$x \in \{e,f\}$$.

4. **Inverses:**  
   Both $$e$$ and $$f$$ are their own inverses:
   - $$e^{-1} = e,$$
   - $$f^{-1} = f,$$ since $$f \circ f = e$$.

Since all the group axioms are satisfied, the set $$\{e, f\}$$ forms a group. In fact, it is isomorphic to the cyclic group $$\mathbb{Z}/2\mathbb{Z}$$.

#### Cayley Table for the Rhombus

We can record the operation (composition) in the following Cayley table:

$$
\begin{array}{c|cc}
\circ & e & f \
\hline
e & e & f \
f & f & e \
\end{array}
$$

---

### 2. Symmetries of a Rectangle

A rectangle (which is not a square) has two pairs of equal but distinct side lengths. Its symmetries are:

1. The identity $$e$$.  
2. A rotation by $$180^\circ$$, denoted $$r$$.  
3. A reflection across the vertical line (the midline that joins the midpoints of the top and bottom edges), denoted $$v$$.  
4. A reflection across the horizontal line (the midline that joins the midpoints of the left and right edges), denoted $$h$$.

Let us verify briefly that these are all symmetries that preserve the rectangle. (Notice that the reflections across the diagonals do not preserve a non‐square rectangle.)

We check closure and inverses similarly as before. (Again the associativity is given by composition of functions.) For example:
- $$r \circ r = e$$ because a $$180^\circ$$ rotation done twice gives the identity.  
- Reflecting twice over the same line (whether vertical or horizontal) gives the identity.  
- Compositions like $$r \circ h = v$$ (and one may check other products) are in the set.

Thus, the set of symmetries of the rectangle,  
$$
\{e, r, h, v\},
$$
forms a group.

#### Cayley Table for the Rectangle

We now present the Cayley table. (The order in which elements are listed is a convention; here we choose the order $$e, r, h, v$$.) One valid Cayley table is:

$$
\begin{array}{c|cccc}
\circ & e & r & h & v \
\hline
e & e & r & h & v \
r & r & e & v & h \
h & h & v & e & r \
v & v & h & r & e \
\end{array}
$$

A brief check:  
- Since $$r$$ represents a $$180^\circ$$ rotation, $$r \circ h = v$$ and $$r \circ v = h$$.  
- Each element is its own inverse except that $$r^{-1} = r$$ as well.

---

### 3. Comparing the Two Groups

The symmetry group of a non‐square rhombus has two elements, while that of a non‐square rectangle has four elements. Thus, in these cases the groups are different (they are not isomorphic).

In summary:

- **Rhombus (non‐square):** Group of order 2, isomorphic to $$\mathbb{Z}_2$$.  
- **Rectangle (non‐square):** Group of order 4, the Klein four‐group often denoted by $$V_4$$ (or equivalently $$\mathbb{Z}_2 \times \mathbb{Z}_2$$).

(Note that if the rhombus is a square, its symmetry group becomes larger—namely, the dihedral group $$D_4$$ of order 8—but here we assume the general non‐square case.)

Thus, the symmetries of a rectangle and those of a (non–square) rhombus are not the same.
A rhombus has two symmetries: the identity and one reflection. Its symmetry group is isomorphic to $$\mathbb{Z}_2$$. A rectangle has four symmetries: the identity, two reflections, and a $$180^\circ$$ rotation. Its symmetry group is the Klein four-group $$V_4$$. Therefore, the symmetries of a rectangle and a rhombus are different.

Descibe the symmetries of a thombus and prove that the set of symmetries
forms a group. Give Cayley fables for both the symmetries of a rectanige and the
symmetries of a thombus, Are the symmetries of a rectangle and those of a monbus
the same?

Upstudy AI Solution

Answer

Solution

1. Symmetries of a Rhombus

Proving that Forms a Group

Cayley Table for the Rhombus

2. Symmetries of a Rectangle

Cayley Table for the Rectangle

3. Comparing the Two Groups

Beyond the Answer

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