## 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 $\mathsf{Aut}(D_8) \cong D_8$.

Post a comment or leave a trackback: Trackback URL.

### Comments

• 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