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Dragonfall
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What is the size of [itex]GL_n(\mathbb{Z}_2)[/itex]?
HallsofIvy said:You could also think of this as n by n matrices whose entries must be either 1 or 0. How many entries are then in an n by n matrix? And if there are only two possible values for each?
morphism said:Think in terms of linear independence.
Yes!Dragonfall said:Is it true that an nxn matrix is invertible iff the column space has n dimensions?
HallsofIvy said:In fact, a matrix A, in Z2 is invertible if det(A)= 1 and not invertible if det(A)= 0. Does that imply that exactly half of all n by n matrices in Z2[/sup] are invertible?
morphism said:Yes!
GLn(Z2) is the special linear group of size n over the ring of integers modulo 2. It consists of all invertible n x n matrices with entries in Z2, where multiplication is defined as usual matrix multiplication.
The size of GLn(Z2) is given by (2n - 1)(2n - 2)(2n - 4)...(2n - 2n-1). This can be derived by counting the number of possible choices for each entry in an invertible n x n matrix over Z2.
GLn(Z2) is important in various areas of mathematics, including algebra, geometry, and number theory. It is a fundamental group in algebraic topology and plays a role in the classification of manifolds. It is also used in cryptography and coding theory.
Yes, the size of GLn(Z2) can be generalized to other rings. In general, the size of GLn(R) is given by the product of (|R|n - 1)(|R|n - |R|)(|R|n - |R|2)...(1 - |R|n-1), where R is any finite ring with identity and |R| denotes the number of elements in R.
The size of GLn(Zp) is (pn - 1)(pn - p)(pn - p2)...(pn - pn-1), which is a generalization of the formula for GLn(Z2). This means that the size of GLn(Zp) is larger than the size of GLn(Z2) for any prime number p other than 2.