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 z_{30}that have order 10

answer: {3, 9, 21, 27}

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

answer: x^{4} x^{12} x^{28} x^{36}

2)consider z_{25}. show that the map ϕ: z_{25 }-> z_{25} give by ϕ(x)=2x is an automorphism of z_{25 }

answer:

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!

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 <x^{4}> 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 Z_{25}, unless 2 is a unit (not all elements of Z_{25} ARE units, for example 5 is NOT).

just for emphasis: Aut(Z_{n}) ≅ U(n).

in this case, 2 IS a unit in Z_{25}: (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 Z_{25}, and ϕ(x) = ϕ(13y) = 2(13y) = ((2)(13))y = 1y = y, as desired.