Im not sure if this is against the rules so if it is just let me know and i wont do it again. I am a math major and have taken alot of math classes and have never had a teacher grade hw so difficult in my life. Im in first semester upper division abstract algebra and just wanted to check some of my solutions. If im wrong im not looking for answers just warning and maybe hints as to where i went wrong. Thanks in advance...

1. For n element integers defind n(integers) as all t * n such that n is an integer prove n is subgroup of integers.

Proof. Let a,b be elements in integers then there exists c,d element integers such that a = c * n and b = d *n then ab = c * n and b = d * n and ab = c*n + d*n (addition is operation since mult is not defined on integers) = n(c+d) is an element of n(integer) so its closed

We know 0 is identity in integers, 0 = and integer * 0 so 0 element n(integer)

Let a element integers then c element integers such that a = c * n a(inverse) = - (c * n) which is element of n(integers) therefore n(integers) is subgroup of integers.

2.If H and K are subgroups of G H intersection K is subgroup.

Proof. Let a,b element H intersect K then a element H and k and b element h and k, since a,b element H and H is subgroup of G ab element G, similarly a,b element K ab element K ab element H intersect K

We know H and K are subgroups so identity element of H and element of K therefore element of H intersect K

Inverse similar to identity therefore H intersect K is subgroup of G

Last one. Function f from real to real is linear if f(x) = ax + b show L = {f|f is linear function} is a group under composition

Proof. Show associativity. Let f,g,h element L then a,b,c,d,e,f Real numbers such that f(x) = ax + b g(x) = cx + d h(x)= ex+f. (F composed G) composed H = a(cx + d) + b composed h(x) = acx + da + b composed h(x) = ac(dx + e) + da + b = acdx + ace + da +b work it out the other way it works associativity proven

let e = F(x) |---> x that is identity under composition and it is element L

Show inverse. Let f(inverse) = x/a -b/a and f element L then f(x) = ax +b

ffinverse = e and finverse is element of L then L is a group

Thanks again for any feedback