
Group Theory
Hi anyone knows how to solve this?
i) Consider Z*_{15}, the group of invertible element of Z_{15} under the operation of multiplication (mod 15). What is the size of Z*_{15}?
ii) Consider Z*_{6}, the group of invertible element of Z_{6} under the operation of multiplication (mod 6). What is the size of Z*_{6}?
Appreciate if anyone can help me (:

Re: Group Theory
Recall that the multiplicative inverse modulo n is found using the Bézout's identity. Also recall that the number of positive integers less than or equal to n that are relatively prime to n is given by the Euler's totient function.
For more, see Multiplicative group of integers modulo n.

Re: Group Theory
The key thing to remember is that "$\displaystyle x$ has multiplicative inverse in $\displaystyle \mathbb{Z}_n^*$ if and only if $\displaystyle x$ and $\displaystyle n$ are relatively prime."
(One way is straightforward to prove, the other way is from Bezout's identity.)
The details are found in emakarov's post and its links.

Re: Group Theory
Guys, i still dun get it :( after reading the links :( will it be possible if u can enlighten me :)

Re: Group Theory
Do you know what the definition is of an invertible element in $\displaystyle \mathbb{Z}_n$? If yes, then try to apply it, one element at a time, to $\displaystyle \mathbb{Z}_{15}$. In other words, go through the nonzero elements of $\displaystyle \mathbb{Z}_{15}$, and for each, either find the inverse, or declare that it's not invertible. Then count the number of elements that have an inverse.
I'll do $\displaystyle \mathbb{Z}_{6}$ for you:
In $\displaystyle \mathbb{Z}_{6}$, [1] [1] = [1], so [1] is invertible, and its inverse is [1].
In $\displaystyle \mathbb{Z}_{6}$, [2] [3] = [0], so [2] is not invertible (if [2] [a] = [1], then [2] [a] [3]= [1] [3] = [3], so [0] = [3]. Impossible.)
In $\displaystyle \mathbb{Z}_{6}$, [3] [2] = [0], so [3] is not invertible (same reasoning as with [2].)
In $\displaystyle \mathbb{Z}_{6}$, [4] [3] = [12] = [0], so [4] is not invertible (same reasoning as with [2].)
In $\displaystyle \mathbb{Z}_{6}$, [5] [5] = [25] = [24+1] = [(6)(4)+1] = [1], so [5] is invertible, with inverse [5].
Thus $\displaystyle \mathbb{Z}_{6}^* = \{ [1], [5] \}$, and so $\displaystyle \mathbb{Z}_{6}^* = 2$.
Here are a few from $\displaystyle \mathbb{Z}_{15}$:
In $\displaystyle \mathbb{Z}_{15}$, [1] [1] = [1], so [1] is invertible, and its inverse is [1].
In $\displaystyle \mathbb{Z}_{15}$, [2] [8] = [16] = [1], so [2] is invertible, and its inverse is [8].
In $\displaystyle \mathbb{Z}_{15}$, [3] [5] = [15] = [0], so [3] is not invertible.
In $\displaystyle \mathbb{Z}_{15}$, [4] [4] = [16] = [1], so [4] is invertible, and its inverse is [4].
... (try to fill these in) ...
In $\displaystyle \mathbb{Z}_{15}$, [13] [7] = [91] = [(15)(6)+1] = [1], so [13] is invertible, and its inverse is [7].
In $\displaystyle \mathbb{Z}_{15}$, [14] [14] = [196] = [195+1] = [(15)(13)+1] = [1], so [14] is invertible, and its inverse is [14].

Re: Group Theory
Thanks! I worked out.
I got Z*15 = {[1],[2],[4],[7],[8],[13],[14]}
(:

Re: Group Theory
Miss out 11.
I got Z*15 = {[1],[2],[4],[7],[8],[11],[13],[14]}
Thanks you very much!

Re: Group Theory
If the problem makes sense to you now, you might want to go back and reread emakarov's post and its links. That now might make more sense.