Maximal Real Subfields (14.5.7,9)

Dummit and Foote Abstract Algebra, section 14.5, exercises 7 and 9:

7. Show that complex conjugation restricts to the automorphism $σ_{-1}∈\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$. Show that the field $K^+=\mathbb{Q}(\zeta_n+\zeta_n^{-1})$ is the subfield of real elements in $K=\mathbb{Q}(\zeta_n)$, called the maximal real subfield of $K$.

9. Let $K_n=\mathbb{Q}(\zeta_{2^{n+2}})$ for $n ≥ 0$.
(a) Prove that $K_n^+/\mathbb{Q}$ is cyclic of degree $2^n$.
(b) Prove $K_n/K_{n-1}^+$ is biquadratic and that two of the three intermediate subfields are $K_{n-1}$ and $K^+$. Prove the remaining intermediate field is a cyclic extension of $\mathbb{Q}$ of degree $2^n$.

Proof: (7) Note that complex conjugation is an automorphism of $\mathbb{C}$, and also restricts to an automorphism of $K$ since $\overline{\zeta_n}=\zeta_n^{-1}$ ($|\zeta_n^n|=|\zeta_n|^n=1$ implies for $\zeta_n=a+bi$ that $|\zeta_n|=a^2+b^2=\zeta_n\overline{\zeta_n}=1$). Now, when $n > 2$ we note $\zeta_n∉\mathbb{R}$ so $K_n/K^+$ is of degree greater than or equal to $2$. As well, $\dfrac{1}{2}(\zeta_n+\zeta_n^{-1})=a$ and $1-a^2=b^2$, so when $α$ is a root of the equation $x^2+b^2$ we observe $K^+(α)=K_n$ and hence $K_n/K^+$ is of degree exactly $2$. Since the maximal real subfield is necessarily properly contained in $K_n$, we must have $K^+$ is the maximal real subfield.

(9)(a) The degree of the extension follows from above. Here we may observe the isomorphism $\text{Gal}(K/ℚ)≅(ℤ/2^{n+2}ℤ)^×≅ℤ/2ℤ×ℤ/2^nℤ$ shown in Exercises 2.3.22-23. We show $(ℤ/2^{n+2}ℤ)^×$ is generated by $5$ and $-1$. This is clearly the case when $n=0$ so assume $n ≥ 1$. Now $5$ is of order $2^n$ by the mentioned exercises and it suffices to show $5$ doesn't generate $-1$; we observe this by demonstrating $k_n$ such that $5^{k_n}≡2^{n+1}+1~\text{mod }2^{n+2}$, a different element of order $2$. First, we see $k_1=1$. Now, given $k_n$, we have $k_{n+1}=2k_n$ since $(5^{k_1})^2=(2^{n+1}+1+m2^{n+2})^2≡2^{(n+1)+1}+1~\text{mod }2^{(n+1)+2}$ when $n ≥ 1$. Thus $-1$ and $5$ generate $(ℤ/2^{n+2}ℤ)^×$, and since we notice the quotient group of $(ℤ/2^{n+2}ℤ)^×$ over the subgroup generated by $-1$ is the cyclic group generated by the image of $5$, we conclude $\text{Gal}(K^+/ℚ)≅\text{Gal}(K/ℚ)/\text{Gal}(K/K^+)$ is cyclic.

(b) We note that since $K_n/ℚ$ is Galois necessarily $K_n/K_{n-1}^+$ is Galois. This Galois extension by degree must be either cyclic or biquadratic, and since it cannot be cyclic by the presence of two distinct subfields of degree two, it must be biquadratic. We see that $K_{n-1}$, $K_n^+$, and the third subfield manifest as the fixed fields of the (isomorphic pullbacks) of the elements of order $2$ in $(ℤ/2^{n+2}ℤ)^×$, and viewing their Galois groups as the quotient of this group with the subgroups generated by those elements, we see as a result of only one of them intersecting nontrivialy with the order-$2^n$ subgroup generated by $5$ (that is, $2^{n+1}+1$ as shown above, which corresponds to $K_{n-1}/ℚ$ which we see is not cyclic) that the third subfield's Galois group's representation as a quotient is cyclic.$~\square$

Galois Traces and the Möbius Function (14.5.6,11)

Dummit and Foote Abstract Algebra, section 14.5, exercise 6:

(6) Show $$\text{Tr}_{\mathbb{Q}(\zeta_n)/\mathbb{Q}}(\zeta_n)=\mu(n)$$ (11) Show the primitive $n^{th}$ roots of unity form a basis of $\mathbb{Q}(\zeta_n)$ if and only if $n$ is squarefree.

Proof: (6) Let $n=p_1^{α_1}...p_k^{α_k}$. Since for any choice of primitives we have $\zeta_n=(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})^z$ for some $z$, by distributing powers we may simply assume $\zeta_n=\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}}$. Since $\mu(ab)=\mu(a)\mu(b)$ for $(a,b)=1$, it suffices to show $\text{Tr}(\zeta_a\zeta_b)=\text{Tr}(\zeta_a)\text{Tr}(\zeta_b)$ for $(a,b)=1$ and that the proposition holds for $n$ a prime power.

Note that for $1≤m≤n$ relatively prime to $n$, we have $(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})^m=\zeta_{p_1^{α_1}}^{r_1}...\zeta_{p_k^{α_k}}^{r_k}$ for $0 ≤ r_i < p_i^{α_i}$ and $(r_i,p_i^{α_i})=1$, and as well for such a selection of $r_i$ we have $\zeta_{p_1^{α_1}}^{r_1}...\zeta_{p_k^{α_k}}^{r_k}=(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})^m$ for some $(m,n)=1$ by the Chinese Remainder Theorem is the two-sided inverse of this operation. Since $$\text{Tr}(\zeta_n)=\sum_{(m,n)=1, 1≤m < n} \zeta_n^m$$ we therefore have $\text{Tr}(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})=\text{Tr}(\zeta_{p_1^{α_1}})...\text{Tr}(\zeta_{p_k^{α_k}})$.

Now, we may observe $\text{Tr}(\zeta_p)=\zeta_p+\zeta_p^2+...+\zeta_p^{p-1}=-1$. As well, assume $q=p^a$ for $a≥2$. Then $$\text{Tr}(\zeta_q)=\zeta_q+\zeta_q^2+...+\zeta_q^{p-1}+\zeta_q^{p+1}+...+\zeta_q^{2p-1}+\zeta_q^{2p+1}+...+\zeta_q^{p^a-1}=$$ $$(\zeta_q+...+\zeta_q^{p-1})+\zeta_q^p(\zeta_q+...+\zeta_q^{p-1})+\zeta_q^{2p}(\zeta_q+...+\zeta_q^{p-1})+...+$$ $$\zeta_q^{p(p^{a-1}-1)}(\zeta_q+...+\zeta_q^{p-1})=$$ $$(1+\zeta_q^p+\zeta_q^{2p}+...+\zeta_q^{p(p^{a-1}-1)})(\zeta_q+\zeta_q^2+...+\zeta_q^{p-1})=f(\zeta_q^p)(\zeta_q+\zeta_q^2+...+\zeta_q^{p-1})$$ where $f(x)=\dfrac{x^{p^{a-1}}-1}{x-1}$ which is valid since $\zeta_q^p≠1$. Thus we note $f(\zeta_q^p)=0$ as $(\zeta_q^p)^{p^{a-1}}-1=\zeta_q^{p^a}-1=0$ and so $\text{Tr}(\zeta_q)=0$ and $\text{Tr}(\zeta_q)$ agrees with $\mu(q)$ on prime powers, and since both are relatively multiplicative the proposition is complete.

(11) ($⇒$) By the above, we see $\text{Tr}(\zeta_n)=0$ if $n$ is not squarefree. ($⇐$) Let $n=p_1p_2...p_k$. Then we see $\mathbb{Q}(\zeta_n)=\mathbb{Q}(\zeta_{p_1}\zeta_{p_2}...\zeta_{p_k})$ is the composite of $\mathbb{Q}(p_k)$ and $\mathbb{Q}(\zeta_{p_1...p_{k-1}})$. By induction on prime width, we see the latter field has for basis over $\mathbb{Q}$ the elements $\zeta_{p_1}^{α_1}...\zeta_{p_{k-1}}^{α_{k-1}}$ for $1≤α_i < p_i$. Since the fields have relatively prime degree, we see the basis $\zeta_{p_k}^{α_k}$ for $1≤α_k < p_k$ for $\mathbb{Q}(\zeta_{p_k})$ over $\mathbb{Q}$ remains linearly independent over $\mathbb{Q}(\zeta_{p_1...p_{k-1}})$ (Corollary 13.2.22). Hence a basis for $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$ is $\zeta_{p_1}^{α_1}...\zeta_{p_k}^{α_k}$ for $1≤α_i < p_i$, which by the Chinese Remainder Theorem are all the primitive $n^{th}$ roots.$~\square$

p-Extensions and Galois Closures (14.4.5)

Dummit and Foote Abstract Algebra, section 14.4, exercise 5:

Let $p$ be a prime and let $F$ be a field. Let $K$ be a Galois extension of $F$ whose Galois group is a $p$-group. Such an extension is called a $p$-extension.

Let $L/K$ be a $p$-extension. Prove the Galois closure of $L$ over $F$ is a $p$-extension of $F$.

Proof: Since $L/K$ and $K/F$ are Galois of $p$-power degree, we have $L/F$ is separable and finite (and of $p$-power degree), thus simple. Let $L=F(α)$. We note that the Galois closure of $L$ must contain all the roots of $m_α(x)$ and also that the splitting field of $m_α(x)$ over $F$ is Galois, hence the latter is precisely the closure. Since $K/F$ is Galois, by 14.4.4 we see that in particular $m_α(x)∈F[x]$ splits over $K$ into a product of $n$ irreducibles of same degree $d$, and since $dn=p^a$ (the degree of $L/F$ and $m_α(x)$) we must have $d$ is a $p$-power thus $K(β)/K$ is of $p$-power degree for any root $β$ of $m_α(x)$.

Since the isomorphism $φ~:~K(α)→K(β)$ from mapping $α↦β$ induces an isomorphism $\text{Aut}(K(α)/K)≅\text{Aut}(K(β)/K)$ given by $σ↦φσφ^{-1}$, we see $K(β)/K$ is Galois. By observing degrees in Proposition 14.4.21 we see that composites of $p$-extensions are themselves $p$-extensions ($\text{Gal}(K_1/F)×\text{Gal}(K_2/F)$ is a $p$-group, and so too are its subgroups), hence it follows that the composite of the extensions $K(β_i)$ for roots $β_i$ of $m_α(x)$ (i.e. the splitting field of $m_α(x)$ over $F$, the Galois closure of $L$ over $F$) is in fact of $p$-power degree over $K$ and a $p$-extension of $F$.$~\square$

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 $2,048·322,560=660,602,880$.

Reductions in Finite Fields (14.3.6-7,11)

Dummit and Foote Abstract Algebra, section 14.3, exercises 6-7, 11:

6. Suppose $K=\mathbb{Q}(θ)=\mathbb{Q}(\sqrt{D_1},\sqrt{D_2})$ with $D_1,D_2∈\mathbb{Z}$ and that $θ=a+b\sqrt{D_1}+c\sqrt{D_2}+d\sqrt{D_1D_2}$ for integers $a$, $b$, $c$, $d$. Prove $m_θ(x)$ is reducible modulo every prime $p$. In particular show the polynomial $x^4-10x^2+1$ is irreducible over $\mathbb{Z}[x]$ but reducible modulo every prime.
7. Prove that one of $2$, $3$, or $6$ is a square in $\mathbb{F}_p$. Conclude $$x^6-11x^4+36x^2-36=(x^2-2)(x^2-3)(x^2-6)$$ has a root modulo $p$ for every prime $p$ but has no root in $\mathbb{Z}$.
11. Prove that $x^{p^n}-x+1$ is irreducible over $\mathbb{F}_p$ only when $n=1$ or $n=p=2$.

Proof: (6) Lemma 1: Let $f(x),g_i(x)∈\mathbb{Z}[x]$ be irreducible with finite indexing set $I$. If $θ∈\mathbb{Q}(α_1,...)$ for some roots $g_i(α_i)=0$ and the coefficients of $θ$ are given in integers, then if $Φ$ is reduction modulo $p$ and $β_i$ is a root of $Φ(g_i(x))$ we have $θ'$ (the corresponding integers having been reduced mod $p$, and $α_i$ replaced with $β_i$) is a root of $Φ(f)$. Proof: $Φ$ is an additive homomorphism $\mathbb{Z}(α_i)→\mathbb{F}_p(β_i)$ by its construction on a basis, and further it is multiplicative between two basis elements $Φ(\prod α_i)Φ(\prod α_j)=Φ(\prod α_i \prod α_j)$. Thus it is a ring homomorphism and we see $Φ(f)(θ')=Φ(f(θ))=Φ(0)=0$.$~\square$

Now, since a biquadratic extension of $\mathbb{F}_p$ would entail only the four automorphisms negating $\sqrt{D_1}$ and $\sqrt{D_2}$, none of which having order $4$ and thus none representing Frobenius $σ_p$, we must have $\mathbb{F}_p(\sqrt{D_1},\sqrt{D_2})$ is contained in $\mathbb{F}_{p^2}$ and now $m_θ(x)$ of degree $4$ cannot be irreducible mod $p$. With some work we find $\pm \sqrt{2} \pm \sqrt{3}$ are the solutions to the irreducible $x^4-10x^2+1$ and generate $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Hence $m_{\sqrt{2}+\sqrt{3}}(x)=x^4-10x^2+1$ generates only a second degree extension of $\mathbb{F}_p$ and is not irreducible.

(7) Lemma 2: One of $a,b,ab∈\mathbb{F}_{p^n}$ is a square in $\mathbb{F}_{p^n}$. Proof: Since $\_^2$ is a multiplicative endomorphism of $\mathbb{F}_{p^n}^×$ and for each square $α^2$ we have only two possible elements in the fiber by $x^2-α^2=(x-α)(x+α)$, we must have $[\mathbb{F}_{p^n}^×~:~\text{img }\_^2]≤2$. Now if $a$, $b$ are not squares then $a,b \neq 1$ in $\mathbb{F}_{p^n}^×/\text{img }\_^2$. Hence $ab=1$, i.e. $ab$ is a square.$~\square$

Thus we have proved either $2$, $3$, or $6$ is a square in $\text{F}_p$ and thus the polynomial above has a root in $\text{F}_p$.

(11) We note that if $α$ is a root, then so is $α+a$ for any $a∈\mathbb{F}_{p^n}$. Thus $\mathbb{F}_{p^n}⊆\mathbb{F}_p(α)$ as else the latter would contain all the roots of $x^{p^n}-x+1$ and then a nontrivial extension would yet more roots, a contradiction. Now we observe any automorphism of $\text{Gal}(\mathbb{F}_{p}(α)/\mathbb{F}_{p^n})$ must be defined by $α↦α+a$ for some $a∈\mathbb{F}_{p^n}$ and since these automorphisms fix $\mathbb{F}_{p^n}$ they all have order $p$. Since a Galois group of degree $k$ is always cyclic over $\mathbb{F}_p$ with generator $σ_p$ and hence always cyclic over $\mathbb{F}_{p^n}$, so $[\mathbb{F}_p(α)~:~\mathbb{F}_{p^n}]=p$.

Thus we must have $pn=p^n$. We write $n=p^k$ for some $k≥0$. If $k > 1$ then we notice $k+1=p^k$ and since $p≥2$ implies $k+1 < 2^k ≤ p^k$ when $k=2$ and inductively $(k+1)+1 < 2^k+1 ≤ 2^k+(2^{k+1}-2^k) = 2^{k+1} ≤ p^{k+1}$ we must have $k=0$ (and $n=1$) or $k=1$ (in which case $n=p$ and $p^2=p^n$ and $n=p=2$). We have previously verified $x^p-x+1$ is irreducible ($γ$ implies $γ+1$ a root, thus all roots are in the same extension, thus there is the same degree $d$ among $k$ irreducibles implying $kd=p$ and $d=1$, but there are no roots in $\mathbb{F}_p$) and one may check $x^4-x+1∈\mathbb{F}_2[x]$ has no roots and is not the square of the only irreducible quadratic $(x^2+x+1)^2=x^4+x^2+1$.$~\square$

Scratch Ideas (Research)

Let $R$ be a ring, $I$ an indexing set for which each $M_i$ is an $R$-module. We set $M=\bigoplus_{i∈I}M_i$. We say an element $s∈M$ is condensed if all but one of the coordinates of $s$ is zero (including $s=0$). We say a submodule $N⊆M$ is condensing if for every $m∈M$ we have $\overline{m}=\overline{s}∈M/N$ for some condensed $s∈M$. $N$ is further said to be halt-condensing if for condensed $s_1 \neq s_2$ we have $\overline{s_1} \neq \overline{s_2}$.

Let $N$ be a condensing submodule. We have a homomorphism of the modules $M_i$ into $M/N$, and the union of their images is all of $M/N$. Conversely, if there is such a homomorphism of the $M_i$ into a module $K$ with the union property, then we have a surjective homomorphism $φ : M → K$ given by evaluation of the sum of the nonzero components in $K$, giving rise to a condensing submodule as the kernel. The equivalence regarding halt-condensing submodules is modified by requiring the mapped modules to inject and be disjoint at nonzero values.
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Let $R$ be a UFD with group of units $V$. Choose a list of primes $p_i$ of cardinality $|S|=c$ to represent the associativity classes of prime elements. Letting $A=\text{Aut}(V)$ (group automorphisms), $S_c$ be the symmetric group on $c$ objects, and $V^*$ be the group of maps $S→V$ where the operation is multiplication $φ_1φ_2(s)=φ_1(s)φ_2(s)$, we claim $A × S_c × V^*$ contains (a copy of) $\text{Aut}(R)$ (ring automorphisms) where the former is under a custom operation elucidated below.

Let $φ∈\text{Aut}(R)$. Since automorphisms map units to units, we must have $φ|_V∈\text{Aut}(V)$. Now automorphisms also map primes to primes, and more generally associativity classes of primes to other associativity classes, so $φ$'s action on $p_i$ might be completely represented by an element $σ$ from $S_c$ together with a map $ψ:S→V$ given by $ψ(s)=φ(p_s)/p_{σ(s)}$ (i.e. so we may construct $φ$ from $ψ$ and $σ$ by $φ(p_s)=ψ(s)p_{σ(s)}$). Now, $φ$ is uniquely determined by this action on the units and primes, so we have an injection $\text{Aut}(R)$ into $A × S_c × V^*$. In order to claim an algebraic embedding the structure imposed on the latter is not simply componentwise multiplication but is rather modified to imitate composition of automorphisms of $R$. The operation is defined as such:

Let $x_1,x_2=(φ_1,σ_1,ψ_1),(φ_2,σ_2,ψ_2)∈A × S_c × V^*$. Then $x_1x_2=(φ_1∘φ_2,σ_1∘σ_2,(φ_1∘ψ_2)·(ψ_1∘σ_2))$, where multiplication and composition of maps are here distinct.
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Let $A,B$ be groups. Consider $\text{Aut}_A(A \times B)$, the set of automorphisms of $A \times B$ mapping $A \times 1$ to $A \times 1$. This is seen to be a subgroup of $\text{Aut}(A \times B)$ which we shall now classify. Let $\Phi \in \text{Aut}_A(A \times B)$. By definition $\Phi$ restricts to an automorphism $φ$ on $A \times 1$, and by simultaneously defining $(\psi(b), \sigma(b))=\Phi(1,b)$ we obtain homomorphisms $\psi : B \rightarrow A$ and $\sigma : B \rightarrow B$. Further, we see $\sigma$ is injective because if $\sigma(b)=1$ then $\Phi(1,b) \in A \times 1$ and the automorphism $φ$ leads to an element $\Phi(a,1)=\Phi(1,b)$ implying $a,b=1$. Further, $σ$ is surjective seeing as $Φ$ is surjective. Assume $\text{img }ψ \not ∈ Z(A)$; then $aψ(b) \neq ψ(b)a$ for some $a∈A$, $b∈B$. Then we have $$Φ((1,b)(φ^{-1}(a),1))=Φ(1,b)Φ(φ^{-1}(a),1)=$$ $$(ψ(b),σ(b))(a,1) \neq (a,1)(ψ(b),σ(b)) = Φ((φ^{-1}(a),1)(1,b))$$ a contradiction. Since $φ,ψ,σ$ uniquely determine $Φ$, we have an injection of sets $\text{Aut}_A(A × B) → \text{Aut}(A) × \text{Hom}(B,Z(A)) × \text{Aut}(B)$.

Conversely, let $φ∈\text{Aut}(A)$, $ψ∈\text{Hom}(B,Z(A))$, and $σ∈\text{Aut}(B)$. Define $Φ: A×B → A×B$ by $Φ(a,b)=(φ(a)ψ(b),σ(b))$. We observe $Φ((a_1,b_1)(a_2,b_2))=Φ(a_1,b_1)Φ(a_2,b_2)$, so it is homomorphic. As well, it is injective as $Φ(a,b)=(1,1)$ implies $σ(b)=1$ so $b=ψ(b)=1$ and consequently $a=1$. Finally, we observe surjectivity $(a,b)=Φ(φ^{-1}(aψ(b)^{-1}),σ^{-1}(b))$, so $Φ$ is an automorphism. Thus we have a bijection of sets $\text{Aut}_A(A × B) → \text{Aut}(A) × \text{Hom}(B,Z(A)) × \text{Aut}(B)$.

Suppose $Φ_1$ associates to $(φ_1,ψ_1,σ_1)$ and $Φ_2$ associates to $(φ_2,ψ_2,σ_2)$. Then we observe $Φ_1Φ_2$ associates to $(φ_1φ_2,φ_1ψ_2+ψ_1σ_2,σ_1σ_2)$. Thus, defining such a binary operation on $\text{Aut}(A) × \text{Hom}(B,Z(A)) × \text{Aut}(B)$ induces a group isomorphism between the two.

Footnotes: This machinery works especially well when $Z(B)$ is manageable, and in particular when $Z(B)=1$ and $A$ is abelian we have $\text{Aut}_A(A × B) = \text{Aut}(A × B)$, as then $Z(A × B) = A × 1$ and so $A × 1$ is characteristic. Otherwise, one may possibly discover $A × 1~\text{char}~A × Z(B)~\text{char}~A × B$ and thus $A × 1~\text{char}~A × B$ to the same effect.
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For groups $A≤B≤G$ let $\text{Aut}^G(A)$ be the subgroup of automorphisms of $A$ extending to automorphisms of $G$. Let $\text{Aut}_A^{G}(B)$ be the subgroup of automorphisms of $B$ mapping $A$ to $A$ and extending to automorphisms of $G$.

Let $H ≤ N \text{ char } G$. Then we have a bijection $ψ$ of coset representatives $[\text{Aut}(G)~:~\text{Aut}_H(G)]=[\text{Aut}^G(N)~:~\text{Aut}_H^G(N)]$ given by restriction $ψ(σ)=σ|_N$. This is injective as $σ_1|_N\text{Aut}_H^G(N)=σ_2|_N\text{Aut}_H^G(N)$ implies $σ_2^{-1}|_Nσ_1|_N∈\text{Aut}_H^G(N)$ so $σ_2^{-1}σ_1∈\text{Aut}_H(G)$, and surjective seeing as the representative $σ|_N$ by definition has an extension $σ$ and letting $σ'$ represent $σ$ we see $σ'|_N\text{Aut}_H^G(N)=σ|_N\text{Aut}_H^G(N)$.
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Let $(X,d)$ be a metric space, and let $\mathcal{C}$ be the set of all nonempty compact subspaces of $X$. Let $A∈\mathcal{C}$. Then given $x∈X$, define $d(x,A)=d(A,x)=\min_{a∈A}d(x,a)$. Note that since $A$ is compact this is well defined. Note also that fixing $A$ this function is continuous $X→ℝ$. Hence given $A,B∈\mathcal{C}$ $$D(A,B)=\max \{\max_{a∈A} d(a,B), \max_{b∈B} d(b,A)\}$$ is well defined. It is clear $D(A,A)=0$, and $D(A,B)=0$ quickly implies $A⊆B$ and $B⊆A$. As well, $D(A,B)=D(B,A)$. Now, we show it satisfies the triangle inequality.

Let $A,B,C∈\mathcal{C}$ and assume $D(A,C) > D(A,B) + D(B,C)$. We may assume $\max_{a∈A} d(a,C) ≥ \max_{c∈C} d(c,A)$, for otherwise interchanging $A,C$ would result in such a situation.

Let $a∈A$ and $β∈B$ be such that $d(a,β)=d(a,B) ≥ d(x,B)$ for all $x∈A$, let $b∈B$ and $γ∈C$ be such that $d(b,γ)=d(b,C)≥d(x,C)$ for all $x∈B$, and let $a'∈A$ and $γ'∈C$ be such that $d(a',γ')=d(a',C)≥d(x,C)$ for all $x∈A$. Since $d(a',B)≤d(a,B)$, let $x∈B$ be such that $d(a',x)≤d(a,β)$. Since $d(x,C)≤d(b,C)$, let $y∈C$ be such that $d(x,y)≤d(b,γ)$. Now we exhibit $$D(A,C) = d(a',γ') ≤ d(a',y) ≤ d(a',x)+d(x,y)$$ $$≤ d(a,β)+d(b,γ) ≤ D(A,B)+D(B,C)$$ Hence $(\mathcal{C},D)$ is a metric space. We garner some facts about this metric space.

Let $A⊆X$, and for each $a∈A$ let $f_a$ be a path from $f_a(0)=a$ to $f_a(1)$. The family $\{f_a\}$ is said to be uniform if for each $ε > 0$ and $r∈[0,1]$ there exists a neighborhood $U$ of $r$ in $[0,1]$ such that $f_a(U)⊆B_d(f_a(r),ε)$ for each $a∈A$. In other words, when $X^A$ is given the uniform topology, if the map $f : [0,1]→X^A$ given by $f(r)=\prod f_a(r)$ is continuous. When $B=∪f_a(1)$, we say $\{f_a\}$ is a uniform path from $A$ to $B$.
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Let $σ : ℕ→I^2$ be such that $\overline{σ(ℕ)}=I^2$. Given $ε > 0$ and a function $g : ℕ→\{0,1\}$, we may define $$h_g : I^2→I$$ $$h_g(x)=\lim_{N→∞}\sum_{σ(i)∈B(x,ε)}^N \dfrac{g(i)}{N}$$ when the above limit exists for all $x∈I^2$ (also: consider weighting the sum for guaranteed convergence). When $σ$ and $ε$ are fixed, which continuous functions $f:I^2→I$ are constructible as $h_g$ for some $g∈ℕ^{\{0,1\}}$?

Let $p : I→I^2$ be a path with only finitely many self-intersection points, i.e. $|p^{-1}(x)|=1$ for all but finitely many $x∈I^2$. Choose a countable dense subset $Q⊆[0,1]$ ordered by $φ : ℕ→Q$ with the property that, for all open $U⊆[0,1]$, the sequence $$\dfrac{|\{φ(1),φ(2),...,φ(N)\}∩U|}{N}$$ converges as $N→∞$ (the dyadic rationals under the usual ordering suffice). Let $V⊆[0,1]$ be a countable dense subset disjoint from $p(I)$ with ordering $ψ : ℕ→V$. When $σ : ℕ→I^2$ is defined $σ(1)=p∘φ(1)$, $σ(2)=ψ(1)$, $σ(3)=ψ(2)$, $σ(4)=p∘φ(2)$, $...$ (alternating between primes and non-primes), then $σ$ is an ordering of the countable dense subset $Q∪V$ (with the exception of finitely many points; modify without loss). Define $g : ℕ→\{0,1\}$ by $g(i)=1$ if $σ(i)∈Q$, and $g(i)=0$ otherwise. Then $h_g(x)$ denotes a well-defined function, which when the dyadic rationals as above are chosen and the limit of $$\dfrac{|\{φ(1),φ(2),...,φ(N)\}∩U|}{N}$$ as $N→∞$ thus coincides with the Lebesgue measure, and measures the "length" of the segment contained the $ε$-ball about $x$. For example, when $ε=1/2$ and $p(x)=0×x$, we observe $$h_g(x,y)=\text{min }\{1,y+\sqrt{1-4x^2}/2\}-\text{max }\{0,y-\sqrt{1-4x^2}/2\}$$ for $x≤1/2$ and $h_g(x,y)=0$ elsewhere.