Inverses of elements of a group

[duki]

Homework Statement

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

Homework Equations

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 really lost on this group stuff. Thanks :)

 

[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.

 

[duki]

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

 

[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.

 

[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?