Let G be the group (under matrix multiplication) of 2 × 2 invertible matrices with entries from /9ℤ. Let a be the order of G. Which of the following statements are true?
Order of GL2(ℤ9) Group
Let G be the group of 2 × 2 invertible matrices with entries from ℤ9 under matrix multiplication. This group is also known as the General Linear Group of degree 2 over the ring ℤ9, denoted as GL2(ℤ9). We are asked to find the order of this group, denoted by 'a', and determine which of the given statements about its divisibility are true.
The ring ℤ9 is a commutative ring with unity. Since $9 = 3^2$, ℤ9 is of the form ℤpk, where $p=3$ and $k=2$. The size of the matrices is 2 × 2, so $n=2$.
Calculating the Order of GL2(ℤ9)
The formula for the order of the General Linear Group GLn(ℤpk) is given by:
$$|GL_n(\mathbb{Z}_{p^k})| = p^{k n^2 - n} \prod_{i=0}^{n-1} (p^n - p^i)$$
We have $p=3$, $k=2$, and $n=2$. Substituting these values into the formula:
$$a = |GL_2(\mathbb{Z}_{3^2})| = 3^{2 \cdot 2^2 - 2} \prod_{i=0}^{2-1} (3^2 - 3^i)$$ $$a = 3^{2 \cdot 4 - 2} (3^2 - 3^0)(3^2 - 3^1)$$ $$a = 3^{8 - 2} (9 - 1)(9 - 3)$$ $$a = 3^6 \cdot 8 \cdot 6$$
Next, we find the prime factorization of 8 and 6:
$$8 = 2^3$$ $$6 = 2 \cdot 3$$
Substitute these prime factors back into the expression for 'a':
$$a = 3^6 \cdot (2^3) \cdot (2^1 \cdot 3^1)$$ $$a = 3^6 \cdot 2^3 \cdot 2^1 \cdot 3^1$$
Combine the terms with the same base by adding their exponents:
$$a = 3^{6+1} \cdot 2^{3+1}$$ $$a = 3^7 \cdot 2^4$$
So, the order of the group G is $a = 3^7 \cdot 2^4$.
Analyzing Divisibility Statements
We need to check the divisibility statements based on the calculated order $a = 3^7 \cdot 2^4$.
- Statement 1: a is divisible by $3^4$.
The prime factorization of $a$ is $3^7 \cdot 2^4$. For $a$ to be divisible by $3^4$, the exponent of 3 in the prime factorization of $a$ must be greater than or equal to 4. The exponent of 3 in $a$ is 7, and $7 \geq 4$. Thus, $a$ is divisible by $3^4$. This statement is true. - Statement 2: a is divisible by $2^4$.
The prime factorization of $a$ is $3^7 \cdot 2^4$. For $a$ to be divisible by $2^4$, the exponent of 2 in the prime factorization of $a$ must be greater than or equal to 4. The exponent of 2 in $a$ is 4, and $4 \geq 4$. Thus, $a$ is divisible by $2^4$. This statement is true.
The statements that are true are that $a$ is divisible by $3^4$ and $a$ is divisible by $2^4$.
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