1. ## Group Theory

Hi anyone knows how to solve this?

i) Consider Z*15, the group of invertible element of Z15 under the operation of multiplication (mod 15). What is the size of Z*15?
ii) Consider Z*6, the group of invertible element of Z6 under the operation of multiplication (mod 6). What is the size of Z*6?

Appreciate if anyone can help me (:

2. ## 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.

3. ## 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.

4. ## Re: Group Theory

Guys, i still dun get it after reading the links will it be possible if u can enlighten me

5. ## 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 non-zero 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].

6. ## Re: Group Theory

Thanks! I worked out.
I got |Z*15| = {[1],[2],[4],[7],[8],[13],[14]}

(:

7. ## Re: Group Theory

Miss out 11.
I got |Z*15| = {[1],[2],[4],[7],[8],[11],[13],[14]}

Thanks you very much!

8. ## 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.