# Inverses of elements of a group

1. Mar 5, 2009

### duki

1. The problem statement, all variables and given/known data

Find the inverse for each element in the group {1, 9, 16, 22, 29, 53, 74, 79, 81}, which is under modulo91

2. Relevant equations

3. The attempt at a solution

From notes (again):
1 = 1
9 = 81
16 = 74
22 = 29
53 = 79

How were these numbers found? Sorry for so many questions, I'm just realy lost on this group stuff. Thanks :)

2. Mar 5, 2009

### sutupidmath

Well, again, an element of V_91, call it a, is invertible if and only if gcd(a,91)=1.

Find such elements first.

3. Mar 5, 2009

### duki

Does that hold true for all the elements? All of them have gcd(a,91) = 1

4. Mar 5, 2009

### sutupidmath

I ndeed to correct myself on my previous post. It should have read: an element of Z_91, call it a, is invertible iff gcd(a,91)=1. Because it doesn't make sense to say an elment of V_91, since we know that V_91 is the set of all invertibles of Z_91.

Now the task of findiing their inverses is another issue. What you need to do is find two elements a,b of V_91 such that

[a]=[1] that is, two elements such that when you multiply them and then take mod 91 they should give u 1.

Say 9 and 81=> 9*81=729=> 729=8*91+1=> so the remainder is 1, which means that

729=1(mod 91) so 9, and 81 are multiplicative inverses of each other.

5. Mar 5, 2009

### duki

Excellent, thanks! That makes perfect sense now.

If it's not too much trouble, there's another (3rd) post I made that has worked its way down the board, also on group theory. Could you maybe give me a hand on that one?