Please show the details of your proof for my understanding. Thanks! Please do al

Question

Please show the details of your proof for my understanding. Thanks! Please do all parts to get the full credit. For part (a) I was given the hint of using the follwoing corollary since this is a finite set."" The corollary states: Let G be a group , and let H be a finite, nonempty subset of G. Then H is a subgroup of G if and only if abEH for all a,b EH."""

Let G be the vet of all matrices in GL2(Z5) of the form [m b 0 1]. Show that G is a subgroup of GL2(Z5). Show that the subset N of all matrices in G of the form [1 c 0 1], with c e Z5, is a normal subgroup of G. Show that the factor group G/N is cyclic of order 4.

Explanation / Answer

23)

You need to prove that G is stable by multiplication AND inverse

(a) Suppose we have two matrix A,B in G of the given form A=[[m1,b1],[0,1]] and B=[[m2,b2],[0,1]] for some m1,m2,b1,b2 in Z5

Then AB = [[m1m2,m1b2+b1],[0,1]]= [[ M , B],[0,1]] with M=m1m2 and B=m1b2+b1 both in Z_5 => so AB is in G

So G is stable by multiplication

Now since Z5 is a field (since 5 is prime) then det(A) = m and m^-1 exists in Z_5.

The inverse of a matrix A=[[m,b],[0,1]] is given by the formula A^( -1) = m^-1 * [[ 1 , -b ],[0 , m ]]=[[ m^-1 , -bm^-1],[0, 1]]

So A^(-1) = [M, B],[0,1]] ith M=m^-1,B=-bm^-1 which are in Z_5, so A^(-1) is in G

So G is stable by inverse

So G is a subgroup of GL_2(Z_5)

24)

N is a non-empty subset of G, we need to prove that N is stable by multiplication AND inverse.

If we look what we did in (a) with m1=m2=1 we see that :

AB = [[m1m2,m1b2+b1],[0,1]] = [[1*1,b2+b1],[0,1]] = [[1,C],[0,1]] is in N

A^-1 = [[m^-1, -bm^-1],[0,1]] = [[1,-b],[0,1]] = [[1,C],[0,1]] is in N (here m=m^-1=1)

So N is stable by multplication and inverse so N is subgroup of G

Now we have to prove that gN = Ng for all g in G

Suppose we have x in gN

so we can write x = gn with g = [[m,b],[0,1]] and n = [[1,c],[0,1]]

Computing the product gn = [[m,mc+b],[0,1]]

Now we can see that with n_2 = [[1,mc],[0,1]] (which is in N) we have n_2g = [[n,mc+b],[0,1]] = gn

So x is in Ng

Similarly if x in is Ng we can write g=[[m,b],[0,1]] and n=[[1,c],[0,1]]

Computing the product : ng = [[m,b+c],[0,1]]

Now we can see that with n_2 = [[1, m^-1c],[0,1]] we have gn_2 = [[m,b+c],[0,1]] = ng

So x is in gN

So Ng = gN and N is a normal subgroup of G

(c) Here Z_5* is the multiplicate group of units, which is a cylic group of order 4 as we know

Here we define the map phi : G -> Z_5* such that phi(g)= phi([[m,b],[0,1])) = m

Since with g1=[[m1,b1],[0,1]] and g2=[[m2,b2],[0,1]] we have g1g2=[[m1m2,m1b2+b1],[0,1]]

Then phi(g1*g2)=m1*m2=phi(g1)*phi(g2) , so phi is an homormorphim

phi is trivially onto by construction so Im(phi) = Z_5*

Now if we take g in Ker(phi) then m=1 <=> g has the form [[1,b],[0,1]] which is exactly N !

So Ker(phi) = N

Using the first isomorphism theorem we have G/Ker(phi) ~ Im(phi) <=> G/N ~ Z_5*

So G/N is a cylic group of order 4

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