Is G a Subgroup of GL[SUB]2[/SUB(Z) Isomorphic to {1,-1,i,-i}?

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Homework Help Overview

The problem involves determining whether the set G, consisting of specific 2x2 matrices, is a subgroup of GL₂(Z) and if it is isomorphic to the group {1, -1, i, -i}. The context is rooted in group theory and matrix algebra.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of isomorphisms and the requirements for a bijection to also be a group isomorphism. Questions arise about how to demonstrate the necessary properties of the proposed bijection.

Discussion Status

The discussion is ongoing, with participants exploring the conditions under which the bijection can be considered a group isomorphism. There is an acknowledgment of the need for careful selection of the bijection to satisfy group properties.

Contextual Notes

One participant notes that not every bijection qualifies as a group isomorphism, highlighting the importance of the operation compatibility condition.

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Homework Statement


Show that G = {[1 0 [-1 0 [0 -1 [0 1
0 1], 0 -1], 1 0], -1 0]} is a subgroup of GL2[/SUB(Z) isomorphic to {1,-1,i,-i}.

The Attempt at a Solution



I am clearly sure each element in G can be denoted as {1,-1,i,-i}.
(I can explain why {1,-1,i,-i}, but I will not explain at here.)
so G -> {1,-1,i,-i} is a bijection, so isomorphism.

Is it too simple?
 
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Not every bijection is a group isomorphism. For a map f:G->H, G and H are groups, to be a group isomorphism it needs to be a bijection and have the property that f(ab)=f(a)f(b) for all a,b in G.
You will need to choose your bijection carefully so that this property is satisfied.
 
How do I show that f(ab) = f(a)f(b)..?
Shows everything such that f(1*-1) = f(1)f(-1), f(i*-i)=f(i)*f(-i).. ?
 
Yes, there are 16 different pairs of elements in G.
 

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