Showing that GL(F_p) is isomorphic to an automorphism group

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In summary, the conversation discusses how to show that ##\operatorname{Aut}(\mathbb{Z}_p^n)## is isomorphic to ##GL_n(\mathbb{F}_p)##, and suggests a possible isomorphism between the two sets. However, there is a difficulty in showing the map is well-defined and onto. Another approach could be to use induction along ##n## and consider the cases of ##n=1## and ##n=2## to find a way to translate between the two sets.
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Mr Davis 97
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Homework Statement


Let ##p## be a prime integer. Show that ##\operatorname{Aut}(\underbrace{Z_p\times \dots \times Z_p}_{n \text{ factors }})\cong GL_n(\mathbb{F}_p)##, where ##\mathbb{F}_p## is ##Z_p## viewed as a field.

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The Attempt at a Solution


First, note that ##Z_p = \{1,x,\dots,x^{p-1}\}##.= is the cyclic group of order ##p##. Also, let ##e_i## be the element in ##(Z_p)^n## with ##x## in the ##i##-th position and the identity everywhere else.

Here is my idea of what an isomorphism would look like:

##\Psi : GL_n(\mathbb{F}_p) \to \operatorname{Aut}((Z_p)^n)## given by ##(v_1~\dots~v_n) \mapsto \alpha## where ##\alpha(e_i) = v_i##.

So here are my questions. How can I show that this map is well-defined in the sense that the ##\alpha## being mapped to is actually an automorphism? Also, how would I show that this map is onto?
 
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Mr Davis 97 said:
Also, how would I show that this map is onto?
I see a fundamental difficulty here. To show all elements of ##\operatorname{Aut}(\mathbb{Z}_p^n)## are hit, you will have to know what they are. But if you know what they are, i.e. found a way to describe them, you probably will also have found a way to establish a bijection to ##GL_n(\mathbb{F}_p)##. So your idea looks to me as a circle argument: Solve the problem in order to be able to solve the problem.

Another approach would be an induction along ##n##. The situation ##n=1## and ##n=2## should already point out the way, i.e. how to translate ##\operatorname{Aut}(\mathbb{Z}_p)## to ##\mathbb{F}_p^*=GL_1(\mathbb{F}_p)## and how to deal with cross mappings, that is the anti-diagonal (##n=2##).
 

Related to Showing that GL(F_p) is isomorphic to an automorphism group

1. What is GL(F_p)?

GL(F_p) is the general linear group of degree p over the finite field F_p. It consists of all invertible p×p matrices with entries from the field F_p.

2. What does it mean for two groups to be isomorphic?

Two groups are isomorphic if there exists a bijective function between them that preserves the group operation. In other words, they have the same algebraic structure and can be considered equivalent.

3. How do you show that GL(F_p) is isomorphic to an automorphism group?

To show that GL(F_p) is isomorphic to an automorphism group, we can construct a mapping between the groups that is bijective and preserves the group operation. This can be done by defining an automorphism as a linear transformation on F_p and showing that it is a bijection between GL(F_p) and the set of automorphisms on F_p.

4. What is the significance of proving this isomorphism?

Proving that GL(F_p) is isomorphic to an automorphism group is significant because it helps us understand the algebraic structure of these groups and how they are related. It also allows us to use the tools and techniques from one group to study the other group, making it easier to solve problems and prove theorems.

5. Can this isomorphism be extended to other fields?

Yes, this isomorphism can be extended to other fields. In fact, it is a well-known result in abstract algebra that the general linear group GL(n,F) is isomorphic to the automorphism group Aut(F^n) for any field F and any positive integer n. However, the specific proof for GL(F_p) may differ from that of other fields.

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