Prove an element and its inverse have same order


by sheepishlion
Tags: element, inverse, order, prove
sheepishlion
sheepishlion is offline
#1
Feb14-11, 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|.
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
Dick
Dick is offline
#2
Feb14-11, 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.
Robert1986
Robert1986 is offline
#3
Feb14-11, 09:54 PM
P: 828
Quote Quote by sheepishlion View Post
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|.

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 m|n (that means m divides n.) Now, all you need to do is to show that n|m (n divides m) to show that m=n. To do that, you have to do something really similar to what I did.


Register to reply

Related Discussions
To prove right inverse implies left inverse for square matrices. Calculus & Beyond Homework 15
Group of order 100 with no element of order 4? Calculus & Beyond Homework 3
Prove that the additive inverse -v of an element v in a vector space is unique. Precalculus Mathematics Homework 7
A group of order 2n conatins an element of order 2 Calculus & Beyond Homework 11
prove every even ordered group has an element of order 2 Calculus & Beyond Homework 7