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The identity element will belong to the group.

Matrix multiplication is associative.

The closure property will be satisfied as when I multiply two matrices whose coefficients are elements from $Z/pZ$ , as $Z/pZ$ forms a group the elements will belong to the group(not very sure)

How do I show inverse? (I am clear about the idea of how I can calculate the inverse but how do I write it down)

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    $\begingroup$ You should recall that 1. a matrix with entries in a field is invertible if and only if its determinant is invertible in the field 2. for matrices $A, B$ with entries in a field, $\det (AB)=\det(A)\det(B)$. $\endgroup$
    – arnett
    Feb 24, 2020 at 7:00
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    $\begingroup$ It's not exactly clear where you're getting stuck. Would you be able to answer the question were about $GL_n(\Bbb R)$? If so, then how is the question about $GL_n(\Bbb Z/p \Bbb Z)$ different? $\endgroup$
    – Ben Grossmann
    Feb 24, 2020 at 7:20
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    $\begingroup$ Isn't $GL_n(k)$ the set of all $n\times n$ invertible matrices over $k$ by the definition? I'm not sure what you are trying to show. How do you define $GL_n(k)$? $\endgroup$
    – freakish
    Feb 24, 2020 at 7:46
  • $\begingroup$ @freakish I want to show that the inverse of any matrix in the Gln(Z/pZ) belongs to the same group $\endgroup$
    – Guria Sona
    Feb 24, 2020 at 9:00
  • $\begingroup$ @GuriaSona that doesn't answer my question at all. $GL_n$ is the set of all invertible matrices to begin with. Obviously the inverse of an invertible matrix is invertible. $\endgroup$
    – freakish
    Feb 24, 2020 at 9:52

1 Answer 1

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Let $p$ be a prime number. Then the ring of $n\times n$ matrices over $F_p$, which is $\mathcal{M_{n\times n}}(F_p)$ has $GL(n,F_p)$ as a subgroup: the general linear group of dimension $n$ over $F_p$.

Proving that $GL_n(F_p)$ satisfies the group axioms under multiplication can be done here.

  1. Every element of this group must be an unit, this is equal to $Det(A)\not \equiv_p 0$, so the determinant is an unit on the prime field.

  2. The previous argument proves the inverse lemma as matrices are non-singular.

  3. The product of two or more units is an unit, which implies closure as the result is an unit on $G$.

  4. Associativity works over matrices under multiplication as $A(BC) = (AB)C $.

  5. The group is non-commutative, so it's not abelian.

  6. The $n\times n$ Identity Matrix $I_n$ represents the identity element on $G$.

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