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Math Help - [SOLVED] Show that any subgroup of <a> is normal in a dihedral group D_n

  1. #1
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    [SOLVED] Show that any subgroup of <a> is normal in a dihedral group D_n

    Let the dihedral group D_n be given by elements a of order n and b of order 2, where ba=a^{-1}b. Show that any subgroup of <a> is normal in D_n.
    I have to say I'm not doing a good job of interpreting my textbook, here, which leaves me rather lost to prove what it asks. But here's a shot...

    If I read the previous chapter correctly, a dihedral group D_n=\{a^pb^q\}, for all integers p,q with 0\leq p<n and q\in\{0,1\}. Hopefully this much is correct.

    In order for the subgroup <a> to be normal, we need to show that for each element g\in D_n, it is true that g^{-1}a^rg=a^s for any integer r\in[0,n) and some other integer s\in[0,n). This clearly holds for all g=a^pb^0=a^p, since (a^p)^{-1}a^ra^p=a^{n-p}a^ra^p=a^r.

    So now we need to show that (a^pb)^{-1}a^r(a^pb)=a^s. Now, we know that since o(b)=2 then b^{-1}=b. That means (a^pb)^{-1}=ba^{n-p}. So now we evaluate ba^{n-p}a^ra^pb=ba^rb.

    Note that since ba=a^{-1}b=a^{n-1}b, then ba^rb=a^{n-1}ba^{r-1}b. And we can get by repeating this:

    ba^rb=a^{n-1}ba^{r-1}b

    =a^{2n-2}ba^{r-2}b

    =a^{3n-3}ba^{r-3}b

    =a^{4n-4}ba^{r-4}b

    ...and so on, up to:

    =a^{rn-r}ba^{r-r}b=a^{rn-r}\in D_n

    Q.E.D. ... ?

    Have I done this correctly? If not, how did I err? If so, is there a quicker/easier way to do it?

    Thanks for your help, guys.
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    I have to say I'm not doing a good job of interpreting my textbook, here, which leaves me rather lost to prove what it asks. But here's a shot...
    In general if G is a finite group and H is a subgrop with (G:H) = 2 then H is a normal subgroup.
    Try proving this instead.
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  3. #3
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    the problem is asking you to prove that every subgroup of <a> is normal. let <a^k>=H be any subgroup of <a> and g \in D_n. let h=a^{kr} \in H. we have ghg^{-1}=(gag^{-1})^{kr}.

    let g=a^ib^j. if j=0, then gag^{-1}=a and thus ghg^{-1}=a^{kr}=h \in H. if j=1, then g=a^ib=ba^{-i}, because ab=ba^{-1}. therefore gag^{-1}=a^{-1} and hence ghg^{-1}=h^{-1} \in H.
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