Isomorphism Question

Apr 2010
51
0
[FONT=&quot]Let n be an even integer. Prove that Dn/Z(Dn)[/FONT] is isomorphic to
[FONT=&quot]D(n/2).

[/FONT]
 
Last edited:
Oct 2009
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[FONT=&quot]Let n be an even integer. Prove that Dn/Z(Dn)[/FONT] is isomorphic to [FONT=&quot]D(n/2). [/FONT]


Hints: if \(\displaystyle D_{2n}=\left\{a,b\;;\;a^2=b^n=1\,,\,aba=b^{-1}=b^{n-1}\right\}\) , then:

1) \(\displaystyle Z\left(D_{2n}\right)=\{1,b^{n/2}\}\)

2)\(\displaystyle D_{2n}/Z\left(D_{2n}\right)=\left\{\overline{a}\,,\,\overline{b}\;;\;\overline{a}^2=\overline{b}^{n/2}=\overline{1}\,,\,\overline{a}\overline{b}\overline{a}=\overline{b}^{-1}\right\}\) , with \(\displaystyle \overline{x}:=xZ\left(D_{2n}\right)\in D_{2n}/Z\left(D_{2n}\right)\,,\,x\in D_{2n}\)

Tonio
 
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Apr 2010
51
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It should say that Dn/Z(Dn) is isomorphic to D(n/2). I understand your definition for Dn/Z(Dn), but I don't get how to set up the isomorphism. I think I should use the fact that any group generated by a pair of elements of order 2 is dihedral to get the isomorphism from Dn/Z(Dn) to D(n/2) ?
 
Apr 2010
51
0
never mind, I think I just have to play with the elements in Dn/Z(Dn) until I get it to look that D(n/2).