Using dihedral group in Lagrange theorem

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Discussion Overview

The discussion revolves around the properties of the dihedral group D4, particularly in relation to Lagrange's theorem. Participants explore the order of elements, the relationship between group elements, and the implications of these properties within the context of group theory.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants assert that the order of the element c should be 4, leading to the conclusion that \(c^4 = e\).
  • It is noted that the group contains \(c^3\) but not \(c^4\), suggesting that \(c^4\) must equal \(e\) based on group axioms.
  • One participant questions the sufficiency of the provided information about D4 to distinguish it from the quaternion group Q8.
  • There is a discussion about the inverse of c, where it is stated that \(c^{-1} = c^3\) because \(c \cdot c^3 = e\).
  • Some participants express confusion regarding the relationship between \(c^{-1}\) and \(c^3\), seeking clarification on how this equivalence is established.

Areas of Agreement / Disagreement

Participants generally agree on the order of the element c being 4 and the resulting conclusion that \(c^4 = e\). However, there is disagreement regarding the clarity of the information provided about D4 and its distinction from other groups, such as Q8. The discussion remains unresolved on some points of clarification regarding the group elements.

Contextual Notes

Some limitations include the lack of detailed information about D4 and the potential for confusion regarding the definitions and relationships of group elements, particularly concerning inverses.

onie mti
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i was given that
D4=[e,c,c2,c3,d,cd,c2d,c3d]
therfore D4=<c,d> is the subgroup of itself generated by c,d

then they defined properties of D4 as follows
ord(c)=d, ord(d)=2, dc=c-1d

i am strugging to understand how they got that c4=e=d2
 
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onie mti said:
i was given that
D4=[e,c,c2,c3,d,cd,c2d,c3d]
therfore D4=<c,d> is the subgroup of itself generated by c,d

then they defined properties of D4 as follows
ord(c)=d, ord(d)=2, dc=c-1d

i am strugging to understand how they got that c4=e=d2

You seem to have a typo.
It should be ord(c)=4.
That literally means that $c^4=e$ (and that all lower powers are different from $e$).

This can also be deduced from the group elements.
It contains $c^3$, but it does not contains $c^4$. Since as a group all powers of $c$ must be contained, $c^4$ must be one of the other elements. The only element that the axioms will allow is $e$.
 
Basically, the answer is "because the dihedral group is defined that way" :D
 
Is that ALL the information you are given about $D_4$? Because given ONLY that, I do not see how to distinguish it from $Q_8$ the group of quaternion units.
 
I like Serena said:
You seem to have a typo.
It should be ord(c)=4.
That literally means that $c^4=e$ (and that all lower powers are different from $e$).

This can also be deduced from the group elements.
It contains $c^3$, but it does not contains $c^4$. Since as a group all powers of $c$ must be contained, $c^4$ must be one of the other elements. The only element that the axioms will allow is $e$.

what you are saying kind of makes sense. you see what is confusing me the the group elements do not contain c-1 yet they sayn c-1 = c3 how so?

- - - Updated - - -

Deveno said:
Is that ALL the information you are given about $D_4$? Because given ONLY that, I do not see how to distinguish it from $Q_8$ the group of quaternion units.

yes that is all i was given
 
onie mti said:
what you are saying kind of makes sense. you see what is confusing me the the group elements do not contain c-1 yet they sayn c-1 = c3 how so?

The inverse $c^{-1}$ is defined as the (unique) element that satisfies $c \cdot c^{-1} = e$.

Since $c \cdot c^3=c^4=e$ it follows that $c^{-1}=c^3$.
 

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