Proof of Uniqueness of Non-Identity Commuting Element in D_2n

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SUMMARY

The discussion focuses on proving that for even integers n = 2k (where n ≥ 4), the element z = r^k in the dihedral group D_{2n} is the only non-identity element that commutes with all elements of D_{2n}. The proof relies on the established result that if x is an element of finite order n in a group G, then x^i = x^{-i} holds if and only if i = k. The participant is seeking assistance in demonstrating the uniqueness of z by assuming another commuting element exists and deriving a contradiction.

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  • Understanding of group theory, specifically dihedral groups D_{2n}
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  • Knowledge of commutativity in algebraic structures
  • Ability to manipulate group elements and equations
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Mathematics students, particularly those studying abstract algebra, group theory enthusiasts, and anyone interested in the properties of dihedral groups and their elements.

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Homework Statement


If ##n = 2k## is even and ##n \ge 4##, show that ##z = r^k## is an element of order 2 which commutes with all elements of ##D_{2n}##. Show also that ##z## is the only nonidentity element of ##D_{2n}## which commutes with all elements of ##D_{2n}##.


Homework Equations


The question also says to use this previously-proven result: If ##x## is an element of finite order ##n## in a group ##G, n = 2k##, and ##1 \le i < n##, then ##x^i = x^{-i}## if and only if ##i = k.##


The Attempt at a Solution


I have managed to show everything except the uniqueness of such an element (which is normally the easy part!). I know that I need to assume that another such element exists, and use this assumption to show that this element is in fact ##z##, giving a contradiction. However, I am struggling to generate such a contradiction; I always seem to end up with trivial equations of the type ##1 = 1## (where ##1## is the identity). I think I just need a quick hint as to how to generate my contradiction.

Thanks!
 
Last edited:
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Take an element ##r^a s^b## in ##D_{2n}##. Thus, we have ##1\leq a\leq n## and ##b = 0,1## that commutes with every element.

So take an arbitrary element ##r^x s^y##. Then we have

r^a s^b r^x s^y = r^x s^y r^a s^b

Try to write both sides of the above equation in the form ##r^q s^p##.
 

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