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Math Help - intersection of subgroups

  1. #1
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    intersection of subgroups

    It is clear that intersection of two normal subgroups of finite index will result a normal subgroup of finite index.

    How about intersection of two normal subgroups with prime power index?
    Can we get a normal subgroup of some prime power index?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by deniselim17 View Post
    It is clear that intersection of two normal subgroups of finite index will result a normal subgroup of finite index.

    How about intersection of two normal subgroups with prime power index?
    Can we get a normal subgroup of some prime power index?
    Clearly, the indexes must be a power of the same prime!

    However, I believe that that is all that is needed.

    Let N_1 and N_2 be the subgroups of finite index, of index p^{n_1} and p^{n_2} respectively. Let N=N_1\cap N_2.

    Then, if a^m\in N then a^m\in N_1 and a^m\in N_2. So, clearly p divides m. So, let m=p^nm^{\prime} where p does not divide m^{\prime}. As N_1 and N_2 are both of prime power index then a^{p^n}\in N_1 and a^{p^n}\in N_2. Thus, a^{p^n} \in N=N_1\cap N_2 as required.

    EDIT: There is a much neater way. One can use the second and third isomorphism theorems!

    If N=N_1\cap N_2 then N_2N_1/N_2\cong N_2/N by the second isomorphism theorem. This means that N_2 is of index p^m for some m (why?).

    By the third isomorphism theorem, \frac{G/N}{N_1/N}\cong G/N_1. Everything is of finite index, so [G:N] = [G:N_1]\cdot[N_1:N] = p^{n_1}p^m=p^{n_1+m} as required.
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  3. #3
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    Quote Originally Posted by Swlabr View Post
    If N=N_1\cap N_2 then N_2N_1/N_2\cong N_2/N by the second isomorphism theorem. This means that N_2 is of index p^m for some m (why?).

    By the third isomorphism theorem, \frac{G/N}{N_1/N}\cong G/N_1. Everything is of finite index, so [G:N] = [G:N_1]\cdot[N_1:N] = p^{n_1}p^m=p^{n_1+m} as required.
    2nd Isomorphism Theorem can be used in this case?
    I thought this theorem can only be used when one of the subgroup is normal, not both normal.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by deniselim17 View Post
    2nd Isomorphism Theorem can be used in this case?
    I thought this theorem can only be used when one of the subgroup is normal, not both normal.
    For the second isomorphism theorem you need two subgroups, one normal and the other just has to be a subgroup. It doesn't matter if it is normal or not, so long as it is a subgroup. If something holds for a subgroup it will also hold for a normal subgroup, as normal subgroups are just subgroups but with a few more conditions thrown in.

    So yes, the second isomorphism theorem can be used in this case.
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