Monomorphism of a group automorphisms

Question

Let $\mathbb{A}$ and $\mathbb{B}$ be two algebras on the same language $\mathcal{L}.$ How can I prove that there exists a monomorphism of the group

${\rm Aut}\mathbb{A}\times{\rm Aut}\mathbb{B}$ to the group ${\rm Aut}(\mathbb{A}\times\mathbb{B})?$

Answer 1

Hint:

An element of $\text{Aut} (\mathbb{A}) \times \text{Aut}(\mathbb{B})$ looks like $(f,g)$ where $f \in \text{Aut}(\mathbb{A})$ and $g \in \text{Aut}(\mathbb{B})$.

Then consider the function $(f \times g)(a,b) = (f(a),g(b))$.

Can you show $f \times g \in \text{Aut}(\mathbb{A} \times \mathbb{B})$? If yes, then the map $(f,g) \mapsto f \times g$ is your desired group homomorphism.

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I hope this helps ^_^

Answer 2

Let $f\in$Aut$ \mathbb{A}$ and $g\in$ Aut$\mathbb{B}.$ Let us show that $f\times g\in$ Aut($\mathbb{A}\times\mathbb{B}$), where $(f\times g)(a,b)=(f(a),g(b)).$ $f\times g$ is 1-1 because

$(f\times g)(a,b)=(f\times g)(c,d)\Rightarrow (f(a),g(b))=(f(c),g(d))\Rightarrow f(a)=f(c), g(b)=g(d).$

Because $f$ and $g$ are automorphisms, we have $a=c$ and $b=d,$ so $(a,b)=(c,d).$

$f\times g$ is sirjection because for all $(c,d)\in \mathbb{A}\times\mathbb{B}$ we know there exists $a\in\mathbb{A}$ and $b\in\mathbb{B}$ such that $f(a)=c$ and $g(b)=d$ because $f$ and $g$ are sirjections. So, $(f\times g)(a,b)=(f(a),g(b))=(c,d).$

$f\times g$ is a homomorphism because

for $c\in$ Const$_{\mathcal{L}}$ we have $(f\times g)(c^{\mathbb{A}\times\mathbb{B}})=(f\times g)(c^{\mathbb{A}},c^{\mathbb{B}})=(f(c^{\mathbb{A}}),g(c^{\mathbb{B}}))=(c^{\mathbb{A}},c^{\mathbb{B}})=c^{\mathbb{A}\times \mathbb{B}},$

for $F\in$ Fun$_{\mathcal{L}},$ $ar(F)=n$ and $(a_1,b_1),\ldots, (a_n,b_n)\in A\times B$ we have

$(f\times g)(F^{\mathbb{A}\times \mathbb{B}}((a_1,b_1),\ldots, (a_n,b_n)))=(f\times g)(F^{\mathbb{A}}(a_1,\ldots,a_n),F^{\mathbb{B}}(b_1,\ldots,b_n))=(f(F^{\mathbb{A}}(a_1,\ldots,a_n)),g(F^{\mathbb{B}}(b_1,\ldots,b_n)))=(F^{\mathbb{A}}(f(a_1),\ldots,f(a_n)), F^{\mathbb{B}}(g(b_1),\ldots, g(b_n)))=F^{\mathbb{A}\times \mathbb{B}}((f(a_1),g(b_1)),\ldots,(f(a_n),g(b_n)))=F^{\mathbb{A}\times\mathbb{B}}((f\times g)(a_1,b_1),\ldots,(f\times g)(a_n,b_n)).$

Now, the map $h:$ Aut$\mathbb{A}\times$ Aut$\mathbb{B}\rightarrow$ Aut($\mathbb{A}\times \mathbb{B}$) defined with $h(f,g)=f\times g$ is a desired monomorphism.