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):