need help with a couple proofs showing all steps (cannot us many theorems mostly just the 6 properties of a group)

1) let G be a group and let a be any element of G then (a) is a subgroup of G

(where (a) is any element from a group to denote the set a^n such that n is an integer a^0 is the identity)


2) let t={a+bi such that a,b are real numbers and a^2 + b^2 = 1} (you are allowed to use that t is a subgroup of the non zero complex numbers under multiplication)

prove 3/5 + 4/5i is an element of t

prove (3/5 + 4/5i)^-1 is an element of t


id just like some well done examples to compare and review to

post or pm them I always pay for help!

not sure about 1) but for 2) i think:

3/5 + 4/5i is an element of t because 3/5 and 4/5 are reals and (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 1

(3/5 + 4/5i)^-1 is also an element of t because, using the conjugate expression on the denominator: 1/(3/5 + 4/5i) = (3/5 - 4/5i)/[(3/5)^2 - (4/5i)^2] = (3/5 - 4/5i)/(9/25 + 16/25) = 3/5 - 4/5i, and again, (3/5)^2 + (-4/5)^2 = 1

Looking at your post again, i'm not sure this is what you were asking, sorry if it's not

This post was edited by m0hawk on Feb 15 2014 01:51am

1) Are you sure G is not supposed to be a finite group ?

Otherwise : let G = Z (set of integers) with operation + (addition) : (Z,+) is a group with identity element zero (0).
Let a = 1.
The set { a^n, n positive integer } = (a) = { 0,1,2,3,4,...} is not a subgroup.

Or maybe your notation (a) stands for {a^n, n positive or negative integer}, where a^(-1) is the inverse element of a, and consequently, a^(-n) = (a^(-1))^n for every positive integer n.
In this case, to prove that (a) is a subgroup, you just have to check that :
- (a) is not empty
- (a) is included into G
- for every x, y in (a), x * y ^ (-1) is an element of (a), where the power (-1) denotes the inverse element.

The first 2 points are obvious, for the third :
x in (a) means that x = a^m for a given integer m,
y in (a) means that y = a^n for a given integer n,
in turns, x*y^(-1) = a^m * (a^n)^(-1) = a^m * a^(-n) = a^(m-n) belongs to (a).

Notice that a^(-n) is the inverse element of a^n because a^n * a^(-n) = a^(-n) * a^n = e (the identity element of G).



2) (3/5)² + (4/5)² = 1, hence 3/5 + 4/5i is in t.

(3/5 + 4/5i)^(-1) = 3/5 - 4/5i

and (3/5)² + (-4/5)² = 1.

Or : Since t is a subgroup of (C*,multiplication), then the inverse of any element of t still belongs to t : if a is in t, then a^(-1) is in t.

yup this is what I was looking for thanks