Proving Group Elements & Inverse Have Same Order

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In any group, each element has a unique inverse, and both the element and its inverse share the same order. The order of an element g is defined as the smallest positive integer n such that g^n equals the identity element. To demonstrate that an element and its inverse have the same order, one can show that if g^n equals the identity, then (g^-1)^n also equals the identity. This relationship confirms that the orders of the element and its inverse are indeed equal. The discussion emphasizes the importance of accurately defining terms related to group theory.
semidevil
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so I need to show that in any group, it's elements and inverse has the same order.

so can I say that since it is a group, we know that there exists a unique inverse for each element. So each element would have 1 inverse. And then, that means we have the same number of elements as number of inverses?

does that work? or am I missing something?
 
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Can you state the definition of the order of an element in a group?
 
shmoe said:
Can you state the definition of the order of an element in a group?


so the order is the number of elements in a group...

is this suppose to be a hint? :)
 
semidevil...

The order of an element g in a group G is n such that g^n = identity where n is the smallest positive integer >= 1. To prove an element and its inverse have the same order you can say:

g^n = identity
(g^-1) ^ n = (g^n)^-1 = (identity)^-1 = identity
 
semidevil said:
so the order is the number of elements in a group...

That's the order of the group, not an element. I was hoping to get you to check the definition carefully;).
 
Heh, sorry shmoe...didn't mean to steal your thunder.
 
No thunder to be stolen, we're all here to help (or be helped) :smile:
 
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