Let p be a prime and let GL2(?p) be the group of 2 !! 2,...

Let p be a prime and let GL_2(?_p) be the group of 2 !! 2, invertible matrices over ?_p. Let G ? GL_2(?_p) be the subgroup of upper-triangular matrices: G = { ? a b ? : a, b, c ? ?_p, det = ac ? 0 ? . ? 0 c ? You may use without proof the fact that G is a subgroup. (a) Show that the function ? : ? a b ? ? (a, c) is a surjective group homomorphism ? : G ? (?/p?)* !! (?/p?)*. ? 0 c ? (b) Show that G is a solvable group. (It will help to examine ker ?.)

Transcript

00:01 Hello everyone we are going to solve a question and this question we are given a g is equals to a b 0c and this a b c belongs to field f p and where p is a prime.

00:13 So in a part we have to show that phi from a b zero c to a c to a c is subjective is subjective we have to show that it is a subjective group homomorphism homomorphism and here we are given that phi is a mapping from g to z star mod p z and here also z star mod p z where p is a prime so let's start with solution of this here firstly we show that it is a homomorphism so so for homomorphism for homomorphism to show that it is a homomorphism, we prove that, we prove that we show, we show that 5 of xy is equals to 5 of x multiplied with 5 of y.

01:29 So here we take elements x and y.

01:32 Let us take x as ab 0c and we take y as d e 0f so here firstly we find phi of x y so we have phi of ab 0c multiplied with d e 0 this is equals to phi of here we have by multiplying this is equal to five of here we have by multiplying this here we have ae plus bf, here we have 0 and here we have 0 plus cf that is cf.

02:14 Now by using the mapping this value is equals to ad and here we have cf.

02:21 Now we can write it as ac multiplied with df.

02:26 So this is equals to 5 of ab 0c and and phi of, here we have phi of d is 0f.

02:38 So that means we show that this is a homomorphism.

02:43 So hence, phi is a group homomorphism.

02:55 So let's, moving ahead, we have to prove that it is subjective.

03:00 So for subjective, for subjective, we show that for subjective, we show that for every a and for every a and c that belongs to z star p multiplied with z star p that is our range and domain so there exists a matrix a b 0 c in g such that such that phi of such that phi of ab 0c is equals to ac.

03:43 So that means phi is an subjective homomorphism.

03:48 So hence phi is subjective homomorphism.

03:56 So we proved our first part that is a part of the question.

04:00 In b part we have to show that g is solvable group.

04:09 G is solvable group...

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