## Dih(8) is isomorphic to its own automorphism group

Use the fact that to prove that .

By Corollary 15, we have .

Let be in (the isomorphic copy in of) . Note that, for , . Thus if and only if mod 8, if and only if . Similarly, , so that if and only if mod 8, if and only if . Thus . Similarly, and , so that . Since , we have (using this previous exercise). Thus (using this previous exercise). Since , we have .

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## Comments

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 without any reference to prior work, or a demonstrated isomorphism. Not to mention, the exposition that 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.

I’ve added links to the prior results used.

As for the centralizer exposition, we have two elements whose centralizers intersect in , so the centralizer of is bounded below and above by . Thus the two subgroups are equal.

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 ??

(as a subgroup of ) is generated by and – that is, the element in plays the role of in the usual presentation of .

This subgroup has index 2, and so is normal.

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

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