Dih(8) is isomorphic to its own automorphism group

Use the fact that D_8 \vartriangleleft D_{16} to prove that \mathsf{Aut}(D_8) \cong D_8.

By Corollary 15, we have D_{16}/C_{D_{16}}(D_8) \leq \mathsf{Aut}(D_8).

Let s be in (the isomorphic copy in D_{16} of) D_8. Note that, for 0 \leq a < 8, r^a s r^{-a} = sr^{-2a}. Thus r^a \in C_{D_{16}}(s) if and only if 2a = 0 mod 8, if and only if a \in \{0,4\}. Similarly, sr^a s sr^a = sr^{2a}, so that sr^a \in C_{D_{16}}(s) if and only if 2a = 0 mod 8, if and only if a \in \{0,4\}. Thus C_{D_{16}}(s) = \{1, r^4, s, sr^4 \}. Similarly, \langle r^2 \rangle \leq C_{D_{16}}(r^2) and sr^a r^2 sr^a = sr^{-2} = sr^6, so that C_{D_{16}}(r^2) = \langle r^2 \rangle. Since Z(D_{16}) \leq C_{D_{16}}(D_8) = \bigcap_{x \in D_8} C_{D_{16}}(x), we have C_{D_{16}}(D_8) = Z(D_{16}) = \langle r^4 \rangle (using this previous exercise). Thus D_{16}/Z(D_{16}) \cong D_8 \leq \mathsf{Aut}(D_8) (using this previous exercise). Since |\mathsf{Aut}(D_8)| = 8, we have \mathsf{Aut}(D_8) \cong D_8.

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  • Steven  On December 5, 2011 at 6:55 pm

    Hi, I’m not sure if you’re using previously proved results, but I don’t find this proof rigorous. In particular, you skip right to the conclusion that D_{16}/Z(D_{16} \cong D_{8} without any reference to prior work, or a demonstrated isomorphism. Not to mention, the exposition that C_{D_{16}}(D_{8}) = Z(D_{16}) is brief, to say the least. I’ll try writing up my own version of this tonight and send you a link of copy.

    I don’t want it to sound like I’m not appreciative of your blog, I think it’s great you’ve attempted every problem. It’s just that I found this particular solution to be uncharacteristic of your previous work.

    • nbloomf  On December 5, 2011 at 10:38 pm

      I’ve added links to the prior results used.

      As for the centralizer exposition, we have two elements whose centralizers intersect in Z(D_{16}), so the centralizer of D_8 is bounded below and above by Z(D_{16}). Thus the two subgroups are equal.

  • shosho  On December 13, 2011 at 1:50 pm

    is D_8 really and normal subgroup of D_16 ?
    r in D_8 but r^7 is not in D_8
    can you explain ??

    • nbloomf  On December 13, 2011 at 2:15 pm

      D_8 (as a subgroup of D_{16}) is generated by s and r^2– that is, the element r^2 in D_{16} plays the role of r in the usual presentation of D_8.

      This subgroup has index 2, and so is normal.

      It’s not the group D_8 which is normal in D_{16}. There is a subgroup of D_{16} which is normal and happens to be isomorphic to D_8.

  • shosho  On December 13, 2011 at 11:09 pm

    so it is isomorphic to a normal subgroup of D_16 but itself is not a subgroup
    By the way I like your work very much it is very great thank you very much
    I hope you will get what you want

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