Understanding SL\(_2(\mathbb{Z})\) and \(\Gamma(p)\)

The question asks us to evaluate several statements about the set \(\Gamma(p)\) in relation to the group SL\(_2(\mathbb{Z})\). Let's first define these sets:

• SL\(_2(\mathbb{Z})\) is the set of all 2x2 matrices with integer entries whose determinant is 1. This forms a group under matrix multiplication.
• \(\Gamma(p)\) is a subset of SL\(_2(\mathbb{Z})\) for a prime number p. A matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is in \(\Gamma(p)\) if \(a, b, c, d \in \mathbb{Z}\), \(ad-bc=1\), and the entries satisfy the following congruence conditions modulo p:
  • \(a \equiv 1 \pmod p\)
  • \(d \equiv 1 \pmod p\)
  • \(c \equiv 0 \pmod p\)
  • \(b \equiv 0 \pmod p\)

Analyzing Option 1: \(\Gamma(p)\) as a Subgroup

To check if \(\Gamma(p)\) is a subgroup of SL\(_2(\mathbb{Z})\), we need to verify three conditions:

1. Non-empty set: Check if the identity element of SL\(_2(\mathbb{Z})\) is in \(\Gamma(p)\).

   The identity matrix is \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\). Its determinant is \(1 \times 1 - 0 \times 0 = 1\), so it is in SL\(_2(\mathbb{Z})\). Let's check the congruence conditions modulo p:

   • \(1 \equiv 1 \pmod p\) (True)
   • \(1 \equiv 1 \pmod p\) (True)
   • \(0 \equiv 0 \pmod p\) (True)
   • \(0 \equiv 0 \pmod p\) (True)

   Since all conditions are met, the identity matrix \(I\) is in \(\Gamma(p)\). Thus, \(\Gamma(p)\) is not empty.

2. Closure under multiplication: If \(A, B \in \Gamma(p)\), is \(AB \in \Gamma(p)\)?

   Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) and \(B = \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}\) be in \(\Gamma(p)\). Since \(A, B \in SL_2(\mathbb{Z})\) and SL\(_2(\mathbb{Z})\) is a group, their product \(AB\) is also in SL\(_2(\mathbb{Z})\), which means \(AB\) has integer entries and determinant 1.

   Now, let's check the congruence conditions for \(AB = \begin{pmatrix} aa'+bc' & ab'+bd' \\ ca'+dc' & cb'+dd' \end{pmatrix}\) modulo p.

   We know \(a \equiv 1\), \(b \equiv 0\), \(c \equiv 0\), \(d \equiv 1 \pmod p\) and \(a' \equiv 1\), \(b' \equiv 0\), \(c' \equiv 0\), \(d' \equiv 1 \pmod p\). Let's look at the entries of \(AB\) modulo p:

   • \(aa'+bc' \equiv 1 \cdot 1 + 0 \cdot 0 \equiv 1 \pmod p\)
   • \(ab'+bd' \equiv 1 \cdot 0 + 0 \cdot 1 \equiv 0 \pmod p\)
   • \(ca'+dc' \equiv 0 \cdot 1 + 1 \cdot 0 \equiv 0 \pmod p\)
   • \(cb'+dd' \equiv 0 \cdot 0 + 1 \cdot 1 \equiv 1 \pmod p\)

   The product \(AB\) satisfies the congruence conditions and is in SL\(_2(\mathbb{Z})\). Thus, \(AB \in \Gamma(p)\). Closure under multiplication holds.

3. Closure under inverses: If \(A \in \Gamma(p)\), is \(A^{-1} \in \Gamma(p)\)?

   Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma(p)\). We know \(ad-bc=1\). The inverse of A is \(A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\). Since A has integer entries, \(A^{-1}\) also has integer entries. We already know its determinant is 1.

   Now, let's check the congruence conditions for \(A^{-1}\) modulo p:

   We know \(a \equiv 1\), \(b \equiv 0\), \(c \equiv 0\), \(d \equiv 1 \pmod p\). Let's look at the entries of \(A^{-1}\) modulo p:

   • \(d \equiv 1 \pmod p\)
   • \(-b \equiv -0 \equiv 0 \pmod p\)
   • \(-c \equiv -0 \equiv 0 \pmod p\)
   • \(a \equiv 1 \pmod p\)

   The inverse \(A^{-1}\) satisfies the congruence conditions and is in SL\(_2(\mathbb{Z})\). Thus, \(A^{-1} \in \Gamma(p)\). Closure under inverses holds.

Since all three subgroup conditions are met, \(\Gamma(p)\) is a subgroup of SL\(_2(\mathbb{Z})\). Option 1 is true.

Analyzing Option 3: \(\Gamma(p)\) Cardinality (at least two elements)

We already showed that the identity matrix \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) is in \(\Gamma(p)\). To show \(\Gamma(p)\) has at least two elements, we need to find at least one other element besides the identity.

Consider the matrix \(M = \begin{pmatrix} 1 & p \\ 0 & 1 \end{pmatrix}\). Let's check if \(M \in \Gamma(p)\):

• Entries are integers.
• Determinant: \(1 \times 1 - p \times 0 = 1\). So \(M \in SL_2(\mathbb{Z})\).
• Congruence conditions modulo p:
  • \(a = 1 \equiv 1 \pmod p\) (True)
  • \(d = 1 \equiv 1 \pmod p\) (True)
  • \(c = 0 \equiv 0 \pmod p\) (True)
  • \(b = p \equiv 0 \pmod p\) (True)

Since all conditions are met, the matrix \(M = \begin{pmatrix} 1 & p \\ 0 & 1 \end{pmatrix}\) is in \(\Gamma(p)\). Since p is a prime number, \(p \neq 0\), so \(M\) is different from the identity matrix \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).

Therefore, \(\Gamma(p)\) contains at least the identity matrix and the matrix \(\begin{pmatrix} 1 & p \\ 0 & 1 \end{pmatrix}\). This proves that \(\Gamma(p)\) has at least two elements. Option 3 is true.

Analyzing Option 2: \(\Gamma(p)\) as a Normal Subgroup

A subgroup H of a group G is normal if for every element \(g \in G\) and every element \(h \in H\), the conjugate \(ghg^{-1}\) is in H. Here, \(G = SL_2(\mathbb{Z})\) and \(H = \Gamma(p)\).

Let \(h = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma(p)\) and \(g = \begin{pmatrix} x & y \\ z & w \end{pmatrix} \in SL_2(\mathbb{Z})\). We know \(a \equiv 1, b \equiv 0, c \equiv 0, d \equiv 1 \pmod p\), which means \(h \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \pmod p\). Also, \(xw-yz=1\). The inverse of \(g\) is \(g^{-1} = \begin{pmatrix} w & -y \\ -z & x \end{pmatrix}\), and \(g^{-1} \in SL_2(\mathbb{Z})\).

The matrix \(ghg^{-1}\) is a product of matrices in SL\(_2(\mathbb{Z})\), so it has integer entries and its determinant is det(g)det(h)det(g\(^{-1}\)) = \(1 \cdot 1 \cdot 1 = 1\). Thus, \(ghg^{-1} \in SL_2(\mathbb{Z})\).

To check if \(ghg^{-1} \in \Gamma(p)\), we need to check its entries modulo p. Let's compute \(ghg^{-1} \pmod p\):

\[ ghg^{-1} \pmod p \equiv (g \pmod p) (h \pmod p) (g^{-1} \pmod p) \]

Since \(h \in \Gamma(p)\), \(h \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \pmod p\). Let \(\bar{g} = g \pmod p\). Then \(\bar{g} \in SL_2(\mathbb{Z}/p\mathbb{Z})\) and \(\bar{g}^{-1} = g^{-1} \pmod p\).

\[ ghg^{-1} \pmod p \equiv \bar{g} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \bar{g}^{-1} \equiv \bar{g} \bar{g}^{-1} \equiv \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \pmod p \]

This shows that the entries of \(ghg^{-1}\) satisfy the congruence conditions modulo p (top-left and bottom-right are \(1 \pmod p\), top-right and bottom-left are \(0 \pmod p\)). Since \(ghg^{-1}\) is also in SL\(_2(\mathbb{Z})\), it must be in \(\Gamma(p)\).

Therefore, \(\Gamma(p)\) is a normal subgroup of SL\(_2(\mathbb{Z})\). Option 2, which states that \(\Gamma(p)\) is *not* a normal subgroup, is false.

Analyzing Option 4: \(\Gamma(p)\) Countability

The elements of \(\Gamma(p)\) are 2x2 matrices with integer entries. The set of all 2x2 matrices with integer entries, \(M_2(\mathbb{Z})\), is equivalent to \(\mathbb{Z}^4\) (by listing the four entries), which is countable since the set of integers \(\mathbb{Z}\) is countable.

The set SL\(_2(\mathbb{Z})\) is a subset of \(M_2(\mathbb{Z})\). Any subset of a countable set is also countable. Therefore, SL\(_2(\mathbb{Z})\) is countable.

The set \(\Gamma(p)\) is defined as a subset of SL\(_2(\mathbb{Z})\). Since SL\(_2(\mathbb{Z})\) is countable, \(\Gamma(p)\) must also be countable.

Option 4, which states that \(\Gamma(p)\) is uncountable, is false.

Summary of Conclusions

• Option 1: \(\Gamma(p)\) is a subgroup of SL\(_2(\mathbb{Z})\). (True)
• Option 2: \(\Gamma(p)\) is not a normal subgroup of SL\(_2(\mathbb{Z})\). (False)
• Option 3: \(\Gamma(p)\) has at least two elements. (True)
• Option 4: \(\Gamma(p)\) is uncountable. (False)

Based on our analysis, options 1 and 3 are true.