# Thread: Abstract algebra, short problems

1. ## Abstract algebra, short problems

I have a test Wednesday and I'm working on the practice test, just wanna know if these are right, I've done them all

1)
a)list all of the elements of z30that have order 10
b)let g be a group, x ϵ G, with order of x = 40. List all elements of <x> with order 10.

2)consider z25. show that the map ϕ: z25 -> z25 give by ϕ(x)=2x is an automorphism of z25

operation perserving
ϕ(x+y)
2(x+y) mod 25
2x mod 25 + 2y mod 25
ϕ(x)+ϕ(y)

As for the bijective step, I know I can show the mapping for every element but is this right also?
1:1
ϕ(x)=ϕ(y)
2x=2y
x=y
onto
let y = 2x
ϕ(x)=2x=y

Thanks!

2. ## Re: Abstract algebra, short problems

your answer for number 1 is correct. note that the first subset is the elements of <3> that are generators of <3>, <3> has order 10, so should have φ(10) = 4 generators. similarly <x4> has order 10, and will have 4 generators as well.

there is a problem with your "proof" ϕ is injective. you cannot conclude that 2x = 2y implies x = y. how do you get from step 1, to step 2? it looks like you divided by 2, but you cannot do that in Z25, unless 2 is a unit (not all elements of Z25 ARE units, for example 5 is NOT).

just for emphasis: Aut(Zn) ≅ U(n).

in this case, 2 IS a unit in Z25: (2)(13) = 26 = 1 (mod 25).

therefore, you should have proceeded like this:

2x = 2y
(13)(2x) = (13)(2y)
((13)(2))x = ((13)(2))y
x = y

there is also a problem with your "proof" that ϕ is surjective, when you say "let y = 2x". that is precisely what you want to PROVE, is that there is some x with y = 2x. so you can't ASSUME there is right from the start. for all we know the set of 2x's might be SMALLER than the set of y's. but...we already know that (2)(13) = 1 (mod 25). so choose x = 13y. this is clearly "some" element of Z25, and ϕ(x) = ϕ(13y) = 2(13y) = ((2)(13))y = 1y = y, as desired.