MHB Using dihedral group in Lagrange theorem

[onie mti]

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

 

[I like Serena]

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 $e$.

 

[Nono713]

Basically, the answer is "because the dihedral group is defined that way" :D

 

[Deveno]

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.

 

[onie mti]

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?

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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

 

[I like Serena]

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 $c^{-1}=c^3$.