Show that o(x) = o(y^-1xy), where o(x) means order of x

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In summary, if G is a group and x, y are in G, then the order of x is equal to the order of y-1xy. This can be shown by computing the powers of y-1xy and using the definition of the order of an element. Additionally, conjugation is an isomorphism and isomorphisms preserve the order of elements, making it sufficient to show that (y-1xy)n = e when n is the order of x. However, it is also necessary to show that no smaller positive power of y-1xy equals e, by picking the smallest power (y-1xy)k that equals e and showing that k = |x|.
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If G is a group and x, y are in G. Show that o(x) = o(y^-1xy), where o(x) means order of x.

thanks
 
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  • #2
Which is your group operation?
 
  • #3
raj123 said:
If G is a group and x, y are in G. Show that o(x) = o(y^-1xy), where o(a) means order of a.

thanks
Why not try computing a few powers of y-1xy and see if that gives you any clues?
 
  • #4
0rthodontist said:
Why not try computing a few powers of y-1xy and see if that gives you any clues?

isn't y-1xy=x since y-1y=e. so o(y-1xy)=o(x).
 
  • #5
raj123 said:
isn't y-1xy=x since y-1y=e. so o(y-1xy)=o(x).
Why would you think that? Not all groups are commutative.

What is (y-1xy)2 equal to?
 
  • #6
Do not assume your group is commutative unless expressly told it is. yxy^-1 is most definitely not the same as x in general, and rule number 1 is that you should never assume otherwise unless told you may. I'm sure this was hammered home in your first lesson (and if not it should have been).

Now, please try to follow the hint given. What is (yxy^-1)^2? (yxy^-1)^n?

Remember that the order of z is the least positive n with z^n=1, so it obviously suffices to show that the order of z is less than the order of x, since by symmetry the order of x is then less than the order of z, so they are equal.

You should go through that argument a couple of times to see why it is true, as well.
 
  • #7
0rthodontist said:
Why would you think that? Not all groups are commutative.

What is (y-1xy)2 equal to?
(y-1xy)2 =y-2x2y2 =y-2+2x2=x2
 
  • #8
raj123 said:
(y-1xy)2 =y-2x2y2 =y-2+2x2=x2
I don't think you're getting it. Could you provide what you think the justification is for each of those steps? This means working directly from the group axioms. You have three equal signs, which means you'll need three justifications.

There's more than one problem here. First you say that (y-1xy)2 =y-2x2y2. That isn't necessarily true. Then you say that y-2x2y2 =y-2+2x2, which again is not necessarily true.

To get you started,
(y-1xy)2 = (y-1xy)(y-1xy). What does that equal? You can use the associative law and the definition of inverses.
 
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  • #9
raj123 said:
(y-1xy)2 =y-2x2y2 =y-2+2x2=x2


For the fourth (?) time: groups are not to be presumed commutative, you cannot do this. xy=/=yx. You cannot just swap round the order of x and y at will.
 
  • #10
i see. how about this..
(y-1xy)(y-1xy) = y-1xyy-1xy = y-1x2y
 
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  • #11
Yes. Now what about (yxy^-1) raised to the n'th power?
 
  • #12
that will be...
(y-1xy)n = y-1xny
 
  • #13
And, now what happens when [itex]n=o(x)[/itex]?
 
  • #14
1) conjugation is an isomorphism.
2) isomorphisms preserve order of elements.
 
  • #15
NateTG said:
And, now what happens when [itex]n=o(x)[/itex]?

xn= e.what are the next steps in this proof?
 
  • #16
Can you please try and put two and two together? You're asked to raise something to the power n, and then asked to consider what happens when n is the order of x. Now, please, try to think what that might mean.
 
  • #17
matt grime said:
Can you please try and put two and two together? You're asked to raise something to the power n, and then asked to consider what happens when n is the order of x. Now, please, try to think what that might mean.

thanks. that was easy.
 
  • #18
You're not done yet, though, because you also must show that no smaller positive power of y-1xy equals e. Pick the smallest power (y-1xy)k that equals e, and show that k = |x|. (the order of x)
 
  • #19
You just said "xn= e". Use that in your formula (y-1xy)n= y-1xny!

However, o(x) is defined as the smallest n such that xn= e. Just showing that (y-1xy)n= e is not enough to say that the order of y-1xy is n.
 

1. What does "order of x" mean in this context?

The order of x refers to the smallest positive integer n such that x^n = e, where e is the identity element. Essentially, it is the number of times you have to multiply x by itself to get back to the identity element.

2. How do you prove that o(x) = o(y^-1xy)?

To prove that o(x) = o(y^-1xy), we must show that the two expressions have the same value. This can be done by showing that x^n = e if and only if (y^-1xy)^n = e. This can be done using algebraic manipulations and the properties of group operations.

3. Can you provide an example to illustrate this concept?

Sure, let's say we have a group G with elements x and y. If o(x) = 4, this means that x^4 = e, where e is the identity element. Now, if we take y^-1xy, this is essentially the same as saying that we are replacing x with y in the expression x^4. If o(y^-1xy) is also equal to 4, then this means that (y^-1xy)^4 = e. This shows that o(x) = o(y^-1xy) in this specific example.

4. How does this relate to the concept of conjugacy in group theory?

The concept of conjugacy in group theory is closely related to this proof. Conjugacy essentially means that two elements are related to each other by a change of basis or perspective. In this case, we are showing that x and y^-1xy are conjugate to each other, which is why they have the same order.

5. Is this proof applicable to all groups?

Yes, this proof is applicable to all groups. The concept of order is a fundamental property of groups and the proof relies on the basic principles of group operations. Therefore, it can be applied to any group, regardless of its specific structure or elements.

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