ALG2_T4

tutorial-4.pdf

1

a

Let ${\varphi }_1,{\varphi }_2\in \text{Aut}(G)$. Then, ${\varphi }_1\circ {\varphi }_2$ is a bijection and a homomorphism. Thus, it is an isomorphism. An identity exists, and associativity follows from the associativity of function composition. If ${\varphi }_1\in \text{Aut}(G)$, ${\varphi }_1^{-1}$ is also an isomorphism, and hence is in $\text{Aut}(G)$.

b

An automorphism must map a generator to a generator. For ${\mathbb{Z}}_n$, the generators are the set of numbers less than $n$ and coprime to $n$. Thus, $|\text{Aut}({\mathbb{Z}}_n)|=\varphi (n)$.

c

Let $\text{Aut}({\mathbb{Z}}_8)=\{{\varphi }_1,{\varphi }_3,{\varphi }_5,{\varphi }_7\}$. Note that every element in $\text{Aut}({\mathbb{Z}}_8)$ has order 2.

d

$Q_8$ is generated by $i$ and $j$. Note that any automorphism will map $1$ to $1$ and $-1$ to $-1$ (since -1 is the only element with order 2). The rest of the elements are of order 4.

1	-1	$i$	$-i$	$j$	$-j$	$ij$	$ji$
fixed	fixed

Consider the pairs $\{\{i,-i\},\{j,-j\},\{ij,ji\}\}$. If $i$ maps to one of $\{i,-i\}$, $j$ cannot map to one of $\{i,-i\}$. Thus, the total number of automorphisms is $\left(\genfrac{}{}{0ex}{}{3}{2}\right)\cdot 2\cdot 2=24$. The leaves us with the choices $S_3\times {\mathbb{Z}}_4$, $S_4$, and ${\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_2\times {\mathbb{Z}}_3$. One can find at least three elements of order 3 in $\text{Aut}(Q_8)$, such as $(j,ij)$, $(j^3,ij)$, and $(j,ji)$. $S_4$ is the only group with 3 or more elements of order 3.