Thread: 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 (:

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.

The key thing to remember is that " has multiplicative inverse in if and only if and 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.

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

Do you know what the definition is of an invertible element in ? If yes, then try to apply it, one element at a time, to . In other words, go through the non-zero elements of , 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 for you:

In , [1] [1] = [1], so [1] is invertible, and its inverse is [1].

In , [2] [3] = [0], so [2] is not invertible (if [2] [a] = [1], then [2] [a] [3]= [1] [3] = [3], so [0] = [3]. Impossible.)

In , [3] [2] = [0], so [3] is not invertible (same reasoning as with [2].)

In , [4] [3] = [12] = [0], so [4] is not invertible (same reasoning as with [2].)

In , [5] [5] = [25] = [24+1] = [(6)(4)+1] = [1], so [5] is invertible, with inverse [5].

Thus , and so .

Here are a few from :

In , [1] [1] = [1], so [1] is invertible, and its inverse is [1].

In , [2] [8] = [16] = [1], so [2] is invertible, and its inverse is [8].

In , [3] [5] = [15] = [0], so [3] is not invertible.

In , [4] [4] = [16] = [1], so [4] is invertible, and its inverse is [4].

... (try to fill these in) ...

In , [13] [7] = [91] = [(15)(6)+1] = [1], so [13] is invertible, and its inverse is [7].

In , [14] [14] = [196] = [195+1] = [(15)(13)+1] = [1], so [14] is invertible, and its inverse is [14].

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

(:

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

Thanks you very much!

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.