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Let H is subgroup g… - QuestionCove
OpenStudy (loser66):

Let H is subgroup generated by e, r^k j, that is \(H=\) for k =0,1,....,n-1 When is H normal in \(D_n\)? Please, help

2 years ago
OpenStudy (loser66):

\(D_n =<r^k, r^k j>\) such that r^n = j^2=e

2 years ago
OpenStudy (perl):

ok e is the identity

2 years ago
OpenStudy (loser66):

elements of H

2 years ago
OpenStudy (loser66):

To have H is normal, then every elements in H must satisfy a r a^- for all a in D_n My attempt: let \(r^l j\) is an arbitrary element in H, then I need prove 2 cases: 1) \(r^k r^lj r^{k^{-1}}\in H\)

2 years ago
OpenStudy (loser66):

2) \((r^kj) r^lj (r^kj)^{-1}\in H\)

2 years ago
OpenStudy (loser66):

since \(D_n \) has only 2 forms of elements, then it contains just 2 cases as above.

2 years ago
OpenStudy (loser66):

@perl r: stands for rotation an angle 2pi/n in n-gon. j : stands for reflection about x-axis in n-gon

2 years ago
OpenStudy (loser66):

That definition gives us j^2 =e since j and then j again , we get original one.

2 years ago
OpenStudy (loser66):

Have to wait for the mood of "Someone" :)

2 years ago
OpenStudy (loser66):

oh, I got the first one since \(jr^{-k}= r^kj\), we have \(r^k r^lj r^{k^{-1}}= r^kr^lr^kj = r^{k+l+k}j\)

2 years ago
OpenStudy (loser66):

then, to get that is the form of r^l j, we must have \(r^k= r^{-k}\), that is the first condition.

2 years ago
OpenStudy (loser66):

oh, I got it too. :) \((r^kj) r^lj (r^kj)^{-1}=r^kjr^ljj^{-1}r^{-k}=r^kjr^lr^{-k}\) and then \(r^kj = jr^{-k}\) so, it is \(j(r^{-k}r^lr^{-k})= r^{-(-k+l-k)}\) so, if r^k = r^-k then the expression above becomes \(r^{-l}\) in H lalalala... hopefully it is not wrong. If I am wrong, please tell me what is wrong @Zarkon

2 years ago
OpenStudy (loser66):

I have class now, gtg. Please, leave your comment :)

2 years ago
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