# Prove that $Gl_{n}(Z/pZ)$ forms a group under multiplication [closed]

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)

• 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, $\det (AB)=\det(A)\det(B)$. Feb 24, 2020 at 7:00
• 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? Feb 24, 2020 at 7:20
• 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 $GL_n(k)$? Feb 24, 2020 at 7:46
• @freakish I want to show that the inverse of any matrix in the Gln(Z/pZ) belongs to the same group Feb 24, 2020 at 9:00
• @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. Feb 24, 2020 at 9:52

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 $$G$$.