
#1
Feb1411, 09:33 PM

P: 1

1. The problem statement, all variables and given/known data
Prove that in any group, an element and its inverse have the same order. 2. Relevant equations none 3. The attempt at a solution I know that I need to show g=g^1. This is how I approached it but I'm not sure I'm right, it seems like I'm missing something. Let g=m and g^1=n for g, g^1 elements of G. Then (g)^m=e and (g^1)^n=e. Therefore, either (g)^m <= (g^1)^n or (g)^m >= (g^1)^n. If (g)^m <= (g^1)^n then e<=e and m<=n. If (g)^m >= (g^1)^n then e>=e and m >= n. Therefore m = n and g=g^1. 



#2
Feb1411, 09:47 PM

Sci Advisor
HW Helper
Thanks
P: 25,174

It's right in spirit. But lacks in presentation. Sure, if x^k=e then (x^(1))^k=e^(1)=e. Now just use the definition of 'x'. Don't use '<' for the elements of the group. Groups generally aren't ordered.




#3
Feb1411, 09:54 PM

P: 828

I might be misreading your attempted solution, but I don't understand what you mean by f^m <= (g^1)^n. Here is a start: Let m = ord(g) and n = ord(g^1). Then g^m = (g^1)^n. Now, multiply on the left by g^n to get: (g^n)(g^m) = e. Since g^m=e, we must have that g^n = e, from which we see that mn (that means m divides n.) Now, all you need to do is to show that nm (n divides m) to show that m=n. To do that, you have to do something really similar to what I did. 


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