Prove that ord(a) = ord(bab^(-1))

  • #1
The_Iceflash
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Prove that ord(a) = ord(bab^(-1)) (Edited/Fixed)

Homework Statement


Prove that ord(a) = ord(bab-1)

Homework Equations


N/A


The Attempt at a Solution



Part 1: Show that ord(bab-1) = n
Suppose that ord(a) = n
Then an = e.
I then have to use the associative law to show (bab-1)n = e but that's where I'm having troublem.

Part 2: Show that ord(a) = n
Suppose ord(bab-1) = e
Then (bab-1)n = e
I then have to use the associative law to show that an = e but I'm stuck there as well.

Help is greatly appreciated.
 
Last edited:

Answers and Replies

  • #2
The_Iceflash
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I fixed the many typos I had.
 
  • #3
jbunniii
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Try writing out the multiplication:

[tex](bab^{-1})^n = (bab^{-1})(bab^{-1})\cdots (bab^{-1})[/tex] (product of n identical terms)

and simplifiy.
 
  • #4
The_Iceflash
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Try writing out the multiplication:

[tex](bab^{-1})^n = (bab^{-1})(bab^{-1})\cdots (bab^{-1})[/tex] (product of n identical terms)

and simplifiy.

So, doing that I get this:

[tex](bab^{-1})^n = (bab^{-1})(bab^{-1})\cdots (bab^{-1})[/tex] = [tex]ba^nb^{-1} = beb^{-1} = bb^{-1} = e[/tex]

Is that a valid thing to do?

On that thought, How would I show [tex]a^n = e[/tex] for part 2?
 
  • #5
jbunniii
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So, doing that I get this:

[tex](bab^{-1})^n = (bab^{-1})(bab^{-1})\cdots (bab^{-1})[/tex] = [tex]ba^nb^{-1} = beb^{-1} = bb^{-1} = e[/tex]

Is that a valid thing to do?

Yes, it's valid. So this shows that [itex](bab^{-1})^n = e[/itex]. It does NOT necessarily follow that [itex]o(bab^{-1}) = n[/itex], because there could be some smaller exponent [itex]k[/itex] such that [itex](bab^{-1})^k = e[/itex]. You have to do a little more work to prove that this can't happen.

On that thought, How would I show [tex]a^n = e[/tex] for part 2?

I'm not sure why you think this is a two-part proof. You want to prove that

[tex]o(a) = o(bab^{-1})[/tex]

So let [itex]n = o(a)[/itex], and then prove that [itex]o(bab^{-1}) = n[/itex] and you're done.
 
  • #6
The_Iceflash
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Yes, it's valid. So this shows that [itex](bab^{-1})^n = e[/itex]. It does NOT necessarily follow that [itex]o(bab^{-1}) = n[/itex], because there could be some smaller exponent [itex]k[/itex] such that [itex](bab^{-1})^k = e[/itex]. You have to do a little more work to prove that this can't happen.

Ok.


I'm not sure why you think this is a two-part proof. You want to prove that

[tex]o(a) = o(bab^{-1})[/tex]

So let [itex]n = o(a)[/itex], and then prove that [itex]o(bab^{-1}) = n[/itex] and you're done.

That's how my professor outlined the proof for me. They broke it up into two parts. The conclusion that is to come out of that as well is that doing both parts will prove that both orders are either infinite or both finite and equal. I believe that's why they want it broken up into two parts.
 
  • #7
jbunniii
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That's how my professor outlined the proof for me. They broke it up into two parts. The conclusion that is to come out of that as well is that doing both parts will prove that both orders are either infinite or both finite and equal. I believe that's why they want it broken up into two parts.

I'm not sure I understand the prof's rationale. Structuring the proof as

"Part 1: show that [itex]o(bab^{-1}) = n[/itex]
Part 2: show that [itex]o(a) = n[/itex]"

begs the question: what is [itex]n[/itex]? I don't see how this helps address the finite vs. infinite cases. More natural would seem:

"Part 1: Suppose that [itex]o(a) = n < \infty[/itex]. Show that [itex]o(bab^{-1}) = n[/itex]
Part 2: Suppose that [itex]o(a) = \infty[/itex]. Show that [itex]o(bab^{-1}) = \infty[/itex]"
 

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