---- > [!definition] Definition. ([[automorphism]]) > An **automorphism** of an object $A$ in a [[category]] $\mathsf{C}$ is an [[isomorphism]] from $A$ to itself. > > The set of automorphisms of $\mathsf{C}$ forms a [[group]] under composition, denoted $\text{Aut}_{\mathsf{C}}(A) \subset \text{End}_{\mathsf{C}}(A)$. It's called the **automorphism group** of $A$. > [!justification] > We need to show that the set $\text{Aut}_{\mathsf{C}}(A)$ has the structure of a [[group]] in a natural way. The [[binary operation|group operation]] is morphism composition. Since morphism composition is [[associative]], we just need to give $\text{Aut}_{\mathsf{C}}(A)$ an an identity and show closure under inverses. > \ > The identity is $1_{A}$. Let $f \in \text{Aut}_{\mathsf{C}}(A)$ Since the inverse of an [[group isomorphism|isomorphism]] is an [[group isomorphism|isomorphism]], $f ^{-1} \in \text{Aut}_{\mathsf{C}}(A)$. > [!basicexample] Most commonly we are considering $\mathsf{C}=\mathsf{Grp}$. - [[inner automorphism]] **Lemma. (Automorphisms and Generators)** A bit similar to how a [[linear maps and basis of domain|linear map is completely determined by its values on a basis]], an [[group isomorphism]] is completely determined by where it takes [[generating set of a group|generators]]: as long as we map [[generating set of a group|generators]] to [[generating set of a group|generators]], an [[group isomorphism|isomorphism]] can be obtained by extending our map using the [[binary operation|group operation]]. # $\text{Aut}(C_{5})$ Represent $C_{5}$ as $\langle a \rangle = \{ e, a, a^{2},a^{3},a^{4}\}$. Since every nontrival element of $C_{5}$ [[generating set of a group|generates]] $C_{5}$, there are four possible maps to define: 1. $a \mapsto a$ and extend using the group operation $(\phi_{e})$; 2. $a \mapsto a^{2}$ and extend using the group operation $(\phi_{1})$; 3. $a \mapsto a ^{3}$ and extend using the group operation $(\phi_{2})$; 4. $a \mapsto a^{4}$ and extend using the group operation $(\phi_{3})$. So $\text{Aut}(C_{5})$ has [[order of a group|order]] $4$. We will show it is [[group isomorphism|isomorphic]] to $C_{5}$ by demonstrating that it is [[generating set of a group|generated by]] the element $\phi_{2}$ defined above. Since each $\phi_{i}$ is characterized by where it sends $a$, it suffices to show that by raising $\phi_{2}$ to various powers (of composition) we obtain all of the $\phi_{i}$. We have $\begin{align} \phi_{2}^{1}(a) = & \phi_{2}^{}(a) = a^{3} \implies \phi_{2}^{1}= \phi_{2}\\ \phi_{2}^{2}(a^{}) = & (\phi_{2} \circ \phi_{2})(a^{})=\phi_{2}(a^{3})=a^{9}=a^{4} \implies \phi_{2}^{2}= \phi_{3}\\ \phi_{2}^{3}(a) = & \phi_{2} \circ \phi_{2}^{2}(a)=a^{12}=a^{2} \implies \phi_{2}^{3}=\phi_{1} \\ \phi_{2}^{4}(a)= & \phi_{2} \circ \phi_{2}^{3}(a) = a ^{6}=a \implies \phi_{2}^{4}= \phi_{e}, \end{align}$ as desired. FAILED Generalization attempt: ![[DEPRECTAED for prime p, automorphism group of p-ordered cyclic group is (p-1)-ordered cyclic#^a045bf]] Update 3 weeks later: we now have the machinery to show this. See [[automorphism group of cyclic group]] and [[finite subgroup of multiplicative group of a field is cyclic]]! As a corollary, we have the [[classification of groups of order pq]]. # $\text{Aut } C_{8}$ The [[generating set of a group|generators]] of $C^{8}$ are $\{ a,a^{3} , a^{5} ,a ^{7}\}$ (powers are [[relatively prime integers|coprime]] with $8$, see [[automorphism group of cyclic group]]). So there are four possible maps: 1. $a \mapsto a$ and extend using the group operation $(\phi_{e})$; 2. $a \mapsto a^{3}$ and extend using the group operation $(\phi_{1})$; 3. $a \mapsto a ^{5}$ and extend using the group operation $(\phi_{2})$; 4. $a \mapsto a^{7}$ and extend using the group operation $(\phi_{3})$. Now, $|\text{Aut }C_{8}|=4$, so it is [[group isomorphism|isomorphic]] to either $C_{4}$ or to the [[Klein 4-group]]. We claim the latter. It suffices to show that $\phi_{1}^{2}=\phi_{2}^{2}=\phi_{3}^{2}=\phi_{e}$. We have $\begin{align} \phi_{1}^{2}(a)= & \phi_{1} \circ \phi_{1}(a) = (a^{3})^{3}=a^{9 \text{ mod } 8}=a & \implies \phi_{1}^{2}=\phi_{e} \\ \phi_{2}^{2}(a)= & \phi_{2} \circ \phi_{2}(a) = (a^{5})^{5}=a ^{25 \text{ mod }8} = a & \implies \phi_{2}^{2}=\phi_{e} \\ \phi_{2}^{2}(a)= & \phi_{2} \circ \phi_{2}(a) = (a^{7})^{7}=a ^{49 \text{ mod }8} = a & \implies \phi_{3}^{2}=\phi_{e}, \end{align}$ as desired. # $\text{Aut } C_{11}$ Proceed analogously to the $C_{5}$ case above. Represent $C_{11}$ as $\langle a \rangle = \{ e, a, a^{2}, \ldots, a^{10} \}$. Since every nontrivial element of $C_{11}$ generates $C_{11}$, there are ten possible maps to define: - $a \mapsto a$ and extend using the group operation ($\phi_{e}$); - $a \mapsto a^{2}$ and extend using the group operation ($\phi_{1}$); - $a \mapsto a^{3}$ and extend using the group operation ($\phi_{2}$); - $\vdots$ - $a \mapsto a^{10}$ and extend using the group operation ($\phi_{9}$); So $\text{Aut}(C_{11})$ has order $10$. We will show it is isomorphic to $C_{10}$ by demonstrating that it is generated by the element $\phi_{2}$ defined above. Since each $\phi_{i}$ is characterized by where it sends $a$, it suffices to show that by raising $\phi_{2}$ to various powers (of composition) we obtain all of the $\phi_{i}$. We have: $\begin{align*} \phi_{2}^{1}(a) & = \phi_{2}(a) = a^{3} \implies \phi_{2}^{1} = \phi_{2} \\ \phi_{2}^{2}(a) & = (\phi_{2} \circ \phi_{2})(a) = \phi_{2}(a^{3}) = a^{9} \implies \phi_{2}^{2} = \phi_{8} \\ \phi_{2}^{3}(a) & = \phi_{2} \circ \phi_{2}^{2}(a) = a^{27} = a^{5} \implies \phi_{2}^{3} = \phi_{4} \\ & \vdots \\ \phi_{2}^{10}(a) & = \phi_{2} \circ \phi_{2}^{9}(a) = a^{1024} = a \implies \phi_{2}^{10} = \phi_{e} \end{align*}$ Thus, repeated compositions of $\phi_{2}$ yield all the automorphisms of $C_{11}$, showing that $\text{Aut}(C_{11})$ is cyclic of order $10$, and therefore isomorphic to $C_{10}$. # $\text{Aut }S_{3}$ [[symmetric group|Recall that]] $S_{3} \cong D_{3}$. From this we see that there are exactly $6$ [[group isomorphism|automorphisms]] of $S_{3}$:![[CleanShot 2023-09-09 at 20.06.34.jpg]] Denote by $\phi_{e},\phi_{2},\dots,\phi_{6}$ the [[automorphism]]s from left-to-right displayed above. We see that $|\text{Aut } S_3|=6$. [[classification of small groups|Is it isomorphic to]] $C_{6}$ or $S_{3}$? We claim $S_{3}$: we will show that $\text{Aut }S_{3}$ has one element, $\phi_{2}$, of order $3$ and three elements, $\{ \phi_{i} \}_{i=4}^{6}$, of order $2$. To begin, $\phi_{2}$ maps $e$, $\tau$, and $\tau^{2}$ to themselves, so we only need to check it on $\sigma, \sigma \tau$ and $\sigma \tau^{2}$. We have $\begin{align} \phi_{2}^{3}(\sigma) = & \phi_{2} (\phi_{2}(\phi_{2}(\sigma)))= \phi_{2}(\phi_{2}(\sigma \tau^{2}))=\phi_{2}(\sigma \tau)=\sigma \\ \phi_{2}^{3}(\sigma \tau) = & \phi_{2} (\phi_{2}(\phi_{2}(\sigma \tau)))= \phi_{2}(\phi_{2}(\sigma))=\phi_{2}(\sigma \tau^{2})=\sigma \tau, \end{align}$ and now to ensure $\phi_{2}$ is an [[injection]] we must finally have $\phi_{2}^{3}(\sigma \tau^{2})=\sigma \tau^{2}$. Thus $\phi_{2}^{3}=\phi_{e}$ and so $|\phi_{2}|=3$. It is clear from observing the picture above that $\phi_{i} \circ \phi_{i}=\phi_{e}$ for $i =4,5,6$; thus each of $\phi_{4},\phi_{5},\phi_{6}$ have order $2$. Hence $\text{Aut }S_{3} \cong S_{3}$. # $\text{Aut } C_{2} \times C_{2}$ The [[Klein 4-group]] is characterized as $\{ e,a,b,c \}$ with $a^{2}=b^{2}=c^{2}=e$. Any [[bijection]] of it [[permutation|permutes]] $a,b,c$; since there are only $6$ candidate maps we'll check them individually. Let $1,2,3$ represent $a,b,c$ respectively, and let $\phi_{\text{cycle}}$ represent the map that permutes these elements according to $\text{cycle}$. - $\phi_{()}$: The [[identity map]] is clearly an [[automorphism]] of $C_{2} \times C_{2}$. - $\phi_{(123)}$: we have $\phi_{(123)}(ab)=\phi_{(123)}(c)=a=bc=\phi_{(123)}(a)\phi_{(123)}(b)$ and $\phi_{(123)}(bc)=\phi_{(123)}(a)=b=ca=\phi_{(123)}(b)\phi_{(123)}(c)$ and the only case left to check is $\phi_{(123)}(ac)=\phi_{(123)}(b)=c=ba=\phi_{(123)}(a)\phi_{(123)}(b).$ Since [[classification of small groups|the Klein Four Group is]] [[abelian group|abelian]] these are all of the cases and we conclude that $\phi_{(123)}$ is an [[automorphism]] of $C_{2} \times C_{2}$. - $\phi_{(132)}$: This is an [[automorphism]] too; for essentially identical reasoning to that for $\phi_{(123)}$ above. - $\phi_{(12)}:$ We have $\phi_{(12)}(bc)=\phi_{(12)}(a)=b =\phi_{(12)}(b)\phi_{(12)}(c)$ and $\phi_{(12)}(ab)=\phi_{(12)}(c) =c = ba =\phi_{(12)}(a)\phi_{(12)}(b)$ and $\phi_{(12)}(ac)=\phi_{(12)}(b) =a = bc =\phi_{(12)}(a)\phi_{(12)}(c)$, so $\phi_{(12)}$ is an [[automorphism]] of $C_{2} \times C_{2}$. - $\phi_{(23)}$ : This is an [[automorphism]] too; for essentially identical reasoning to that for $\phi_{(12)}$ above. - $\phi_{(13)}$ : This is an [[automorphism]] too; for essentially identical reasoning to that for $\phi_{(23)}$ above. Therefore all $6$ maps are [[automorphism]]s and $|\text{Aut } C_{2} \times C_{2}|=6$, and $\text{Aut } C_{2} \times C_{2}$ is [[group isomorphism|isomorphic]] to $S_{3}$, where $\phi_{(123)}$ is the element of order $3$ $(\tau)$ and $\phi_{(12)},\phi_{(23)},\phi_{(13)}$ are the elements of order $2$. # $\text{Aut }\mathbb{Z}$ $\mathbb{Z}$ is generated by $1$ and $-1$, thus a prospective [[group isomorphism|isomorphism]] $\phi$ is determined by where it sends these elements. If we map $1 \mapsto 1$ then $\phi$ is the [[identity map]] $\phi_{e}$ and hence an [[automorphism]]. If we map $1 \mapsto -1$ then the resulting map looks like $\phi(k)=-k$ for all $k \in \mathbb{Z}$. This map is [[linear operator|linear]] over $\mathbb{Z}$ and [[bijection|bijective]]; hence an [[automorphism]] of $\mathbb{Z}$. So $\text{Aut }\mathbb{Z}$ has [[order of a group|order]] $2$. It is thus [[group isomorphism|isomorphic]] to $C_{2}$. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```