Automorphism Computation (Research)

Compute $|\text{Aut}(Z_2^4 × D_8)|$.

Let $A=Z_2^4$, $B=D_8$, and $G=\text{Aut}(A×B)$. We note $A$ is abelian, so we consider $F=\text{Aut}_A(A × B)≤G$. We also note $Z(A×B)=Z_2^4 × \langle r^2 \rangle ≅Z_2^5$. Since $A×1 ≤ Z(A×B) \text{ char } A×B$ we observe $[G~:~F]=[\text{Aut}^{A×B}(Z(A×B))~:~\text{Aut}_A^{A×B}(Z(A×B))]$.

Let $Φ∈G$. Then we note $Φ(r^2)=Φ(r)^2$ so since $r^2$ and also $Φ(r^2)$ are in $Z(A×B)$ we have $Φ(r^2)=r^2$ ($r^2$ is the only nonidentity square in $Z(A×B)$). Conversely, let $σ∈\text{Aut}(Z(A×B))$ be such that $σ(r^2)=r^2$. We observe the relations of $A×B$ generated in $\langle a,b,c,d,s,r \rangle$: $$a^2=b^2=c^2=d^2=s^2=r^4=1$$ $$aba^{-1}b^{-1}=aca^{-1}c^{-1}=ada^{-1}d^{-1}=asa^{-1}s^{-1}=ara^{-1}r^{-1}=...=1$$ $$(rs)^2=1$$ where the second line lists all the commutation relations (all commute but $r$, $s$). Some extending map $σ'$ defined by its action on $a,...,r,s$ is an automorphism iff $σ'(a)$, $...$ , $σ'(r),σ'(s)$ generate $A×B$ and $σ'$ is one on these relations generators in the free group. If we let $σ'$ act on $a$, $b$, $c$, $d$ as $σ$ does and set $σ'(s)=s$ and $σ'(r)=r$ we observe $σ'(r^2)=σ(r^2)=r^2$ and it follows that $σ'$ extends $σ$, and is an automorphism since the orders line is fulfilled ($Z(A×B)$ has exponent $2$, so $σ'(a)^2=σ(a)^2=1$ and similar for $b,c,d$ and clearly $σ'(s)^2=σ'(r)^4=1$), the commutation line is fulfilled (all of these relations involve $a$, $b$, $c$, or $d$, and $σ'$ preserves their centricity), the last relation is fulfilled ($σ'$ fixes $D_8$), and the generation is established as $a,...,d$ is in the image of $σ$ on $Z(A×B)$, hence $σ'$ on $A×B$, plus $s=σ'(s)$ and $r=σ'(r)$. Thus $\text{Aut}^{A×B}(Z(A×B))$ is isomorphic to the group of isomorphisms fixing the last coordinate of $Z_2^5$, whose order we compute $(2^5-2^1)(2^5-2^2)(2^5-2^3)(2^5-2^4)=322,560$.

Utilizing the previous reasoning we note that $σ∈\text{Aut}_A(Z(A×B))$ extends iff $σ(r^2)=r^2$, so $|\text{Aut}_A^{A×B}(Z(A×B))|$ is the number of automorphisms of $Z_2^5$ fixing the last coordinate and restricting to an automorphism of $Z_2^4 × 1$, i.e. the number of automorphims of $Z_2^4$ which is computed to be $(2^4-2^0)(2^4-2^1)(2^4-2^2)(2^4-2^3)=20,160$. Thus we calculate $[\text{Aut}^{A×B}(Z(A×B))~:~\text{Aut}_A^{A×B}(Z(A×B))]=16$ distinct cosets of $\text{Aut}_A(A × B)$ in $\text{Aut}(A×B)$.

We have seen $|\text{Aut}_A(A × B)|=|\text{Aut}(A)|·|\text{Hom}(B,A)|·|\text{Aut}(B)|$. We have already computed $|\text{Aut}(Z_2^4)|=20,160$. As well, utilizing the relations for $D_8$ yields $|\text{Aut}(D_8)|=8$. We see $\text{Hom}(B,A)$ is the set of all maps defined freely on $r$ and $s$ in the free group factoring through the relations of $D_8$, i.e. mapping one on the relations of $D_8$, and since $Z_2^4$ has exponent $2$ we see every such mapping is a homomorphism, so $|\text{Hom}(B,A)|=16^2=256$. $$|\text{Aut}(A×B)|=|G|=[G~:~F]·|F|=$$ $$[G~:~F]·|\text{Aut}(A)|·|\text{Hom}(B,A)|·|\text{Aut}(B)|=$$ $$16·20,160·256·8=660,602,880$$ ~~~~~
Alternatively, we calculate the number of automorphisms of the characteristic center that extend $|\text{Aut}^{A×B}(Z_2^4×\langle r^2 \rangle)|=322,560$ (see above) which represent the cosets of the subgroup of automorphisms that fix the center. We see that sending $r$ to either $r$ or $r^3$ with any coordinates for $a$, $b$, $c$, $d$ together with any mapping of $s$ to $s$, $rs$, $r^2s$, or $r^3s$ and any coordinates of $a$, $b$, $c$, $d$ are all automorphisms that fix the center, so $2,048·322,560=660,602,880$.