Automorphism Groups

Proposition 1. If $G$ is an infinite cyclic group, then $\operatorname{Aut}(G)$ is a cyclic group of order $2$.

Proof : Let $G=\langle g\rangle$ be an infinite cyclic group generated by $g\in G$. Then the only generators of $G$ are $g$ and $g^{-1}$. Since every automorphism sends a generator to a generator, and an automorphism of a cyclic group is uniquely determined by its action on a generator, it follows that for every $\sigma\in \operatorname{Aut}(G)$, we must have
$$
\sigma(g)=g \quad \text{or} \quad \sigma(g)=g^{-1}.
$$
If $\sigma(g)=g$, then $\sigma$ is the identity automorphism. If $\sigma(g)=g^{-1}$, then $\sigma$ is also an automorphism. Therefore $\operatorname{Aut}(G)$ has exactly two elements, and hence it is a cyclic group of order $2$.

Proposition 2. Let $U_n$ be the group of all invertible elements of $\mathbb{Z}_n$ under multiplication. If $G$ is a cyclic group of order $n$, then
$$
\operatorname{Aut}(G)\cong U_n.
$$

Proof : Let $a$ be a generator of $G$. Then $a^k$ is a generator of $G$ if and only if $(k,n)=1$. There are exactly $\varphi(n)$ such integers $k$ with $1\le k<n$, where $\varphi(n)$ is Euler’s totient function. Denote these integers by
$$
k_1,k_2,\cdots,k_{\varphi(n)},
$$
where $k_1=1$. Then
$$
a^{k_1},a^{k_2},\cdots,a^{k_{\varphi(n)}}
$$
are precisely all generators of $G$.

Since every automorphism sends a generator to a generator, and is uniquely determined by its value on a generator, for every $\tau\in \operatorname{Aut}(G)$ there exists $i\in \{1,\cdots,\varphi(n)\}$ such that
$$
\tau(a)=a^{k_i}.
$$
Conversely, for each $k_i$, define a map $\tau_i:G\to G$ by
$$
\tau_i(a^l)=a^{k_i l},\quad \forall l\in \mathbb{Z}.
$$
It is easy to verify that $\tau_i$ is an automorphism of $G$. Hence
$$
\operatorname{Aut}(G)=\{\tau_1,\cdots,\tau_{\varphi(n)}\}.
$$

Now define a map
$$
f:\operatorname{Aut}(G)\longrightarrow U_n
$$
by
$$
f(\tau_i)=\overline{k_i}.
$$
We show that $f$ is well-defined and injective. Indeed,
$$
\tau_i=\tau_j
\iff \tau_i(a)=\tau_j(a)
\iff a^{k_i}=a^{k_j}
\iff n\mid (k_i-k_j)
\iff \overline{k_i}=\overline{k_j}.
$$
Thus $f$ is well-defined and injective. Since
$$
|\operatorname{Aut}(G)|=\varphi(n)=|U_n|,
$$
it follows that $f$ is also surjective, hence bijective.

Finally, for any $\tau_i,\tau_j\in \operatorname{Aut}(G)$, we have
$$
\tau_i\tau_j(a)=\tau_i(a^{k_j})=a^{k_i k_j}.
$$
Since $\overline{k_i},\overline{k_j}\in U_n$, and $U_n$ is a group under multiplication, there exists $s\in \{1,\cdots,\varphi(n)\}$ such that
$$
\overline{k_i k_j}=\overline{k_s}.
$$
Hence
$$
\tau_i\tau_j(a)=a^{k_s},
$$
so
$$
f(\tau_i\tau_j)=\overline{k_s}=\overline{k_i k_j}=\overline{k_i},\overline{k_j}=f(\tau_i)f(\tau_j).
$$
Therefore $f$ is a group homomorphism. Thus $f$ is an isomorphism, and so
$$
\operatorname{Aut}(G)\cong U_n.
$$

Proposition 3. Determine $\operatorname{Aut}(\mathbb{Z}_6)$.

Solution : By Proposition 2, we have
$$
\operatorname{Aut}(\mathbb{Z}_6)\cong U_6=\{\bar 1,\bar 5\}.
$$
Hence
$$
|\operatorname{Aut}(\mathbb{Z}_6)|=2.
$$
Therefore
$$
\operatorname{Aut}(\mathbb{Z}_6)=\{\sigma,\tau\},
$$
where $\sigma$ is the identity automorphism, and $\tau$ is given by
$$
\tau(\bar a)=\overline{5a}.
$$

Proposition 4. Let $p$ be a prime number. Determine the order of $\operatorname{Aut}(\mathbb{Z}_{p^n})$, where $n\in \mathbb{N}^*$.

Solution : By Proposition 2, we have

$$
\operatorname{Aut}(\mathbb Z_{p^n})\cong U_{p^n}.
$$

Moreover,

$$
\bar a\in U_{p^n}\iff (a,p^n)=1.
$$

The number of such integers $a$ is
$$
\varphi(p^n)=p^n-p^{n-1}.
$$
Therefore,

$$
|\operatorname{Aut}(\mathbb Z_{p^n})|=|U_{p^n}|=\varphi(p^n)=p^n-p^{n-1}.
$$

Proposition 5. Let $G$ be a group, let $\operatorname{Inn}(G)$ be the group of inner automorphisms of $G$, and let $C$ be the center of $G$. Then:

$(1)$ $\operatorname{Inn}(G)\lhd \operatorname{Aut}(G)$;

$(2)$ $\operatorname{Inn}(G)\cong G/C$.

Proof : For every $a,b\in G$, define
$$
\varphi_a:x\mapsto axa^{-1},\qquad \varphi_b:x\mapsto bxb^{-1}.
$$
These are inner automorphisms induced by $a$ and $b$, respectively. Clearly,
$$
\varphi_a\varphi_b=\varphi_{ab},\qquad \varphi_a^{-1}=\varphi_{a^{-1}}.
$$
Hence $\operatorname{Inn}(G)$ is a subgroup of $\operatorname{Aut}(G)$.

Now let $f\in \operatorname{Aut}(G)$. Then for every $x\in G$,
$$
f\varphi_a f^{-1}(x)
=f(\varphi_a(f^{-1}(x)))
=f(af^{-1}(x)a^{-1})
=f(a)xf(a)^{-1}
=\varphi_{f(a)}(x).
$$
Thus
$$
f\varphi_a f^{-1}=\varphi_{f(a)}\in \operatorname{Inn}(G).
$$
Therefore
$$
\operatorname{Inn}(G)\lhd \operatorname{Aut}(G).
$$

For $(2)$, define a map
$$
\phi:G\longrightarrow \operatorname{Inn}(G)
$$
by
$$
\phi(a)=\varphi_a.
$$
Clearly $\phi$ is a surjective homomorphism. Moreover,
$$
\operatorname{Ker}\phi
=\{a\in G:\varphi_a(x)=x,\forall x\in G\}
=\{a\in G:ax=xa,\forall x\in G\}
=C.
$$
By the First Isomorphism Theorem,
$$
G/C\cong \operatorname{Inn}(G).
$$

Proposition 6. Let $G$ be a finite group, and suppose that there exists $a\in G$ such that $a^2\neq e$. Prove that
$$
|\operatorname{Aut}(G)|>1.
$$

Proof : We give two methods.

Method 1. Suppose, to the contrary, that
$$
|\operatorname{Aut}(G)|\le 1.
$$
Since the identity automorphism $id\in \operatorname{Aut}(G)$, it follows that
$$
|\operatorname{Aut}(G)|=1.
$$
Since $\operatorname{Inn}(G)\lhd \operatorname{Aut}(G)$, we have
$$
|\operatorname{Inn}(G)|=1.
$$
By Proposition 5,
$$
\operatorname{Inn}(G)\cong G/C,
$$
so
$$
|G/C|=1.
$$
Hence
$$
G=C,
$$
that is, $G$ is abelian.

Therefore the map
$$
\varphi:x\mapsto x^{-1}
$$
is an automorphism of $G$. Since $|\operatorname{Aut}(G)|=1$, we must have
$$
\varphi=id.
$$
Thus for every $a\in G$,
$$
a^{-1}=a,
$$
so
$$
a^2=e,
$$
contrary to the assumption. Therefore
$$
|\operatorname{Aut}(G)|>1.
$$

Method 2. If $G$ is abelian, then the map
$$
\varphi:x\mapsto x^{-1}
$$
is an automorphism of $G$. Since there exists $a\in G$ with $a^2\neq e$, we have $\varphi\neq id$, and hence
$$
|\operatorname{Aut}(G)|>1.
$$

If $G$ is nonabelian, then there exists $b\in G\setminus C$. The map
$$
\varphi_b:x\mapsto bxb^{-1}
$$
is an inner automorphism of $G$, and since $b\notin C$, we have $\varphi_b\neq id$. Thus
$$
|\operatorname{Aut}(G)|>1.
$$

In either case,
$$
|\operatorname{Aut}(G)|>1.
$$

Proposition 7. Let $G$ be a finite group such that
$$
|\operatorname{Aut}(G)|=1.
$$
Prove that
$$
|G|\le 2.
$$

Proof : Since
$$
|\operatorname{Aut}(G)|=1,
$$
and $\operatorname{Inn}(G)\lhd \operatorname{Aut}(G)$, we have
$$
|\operatorname{Inn}(G)|=1.
$$
By Proposition 5,
$$
\operatorname{Inn}(G)\cong G/C,
$$
so
$$
|G/C|=1.
$$
Hence
$$
G=C,
$$
that is, $G$ is abelian.

Therefore the map
$$
\varphi:x\mapsto x^{-1}
$$
is an automorphism of $G$. Since $|\operatorname{Aut}(G)|=1$, we must have
$$
\varphi=id.
$$
Thus for every $a\in G$,
$$
a^{-1}=a,
$$
so
$$
a^2=e.
$$

Now suppose that $|G|>2$. Then $G$ cannot be cyclic, for otherwise $G$ would contain an element of order $|G|>2$, contradicting the fact that every element satisfies $a^2=e$.

Let
$$
a_1,a_2,\cdots,a_n,\qquad n\ge 2,
$$
be a minimal generating set of $G$. Since $G$ is abelian and every element satisfies $a^2=e$, each element of $G$ can be written uniquely in the form
$$
a_1^{\epsilon_1}a_2^{\epsilon_2}\cdots a_n^{\epsilon_n},
$$
where each $\epsilon_i$ is either $0$ or $1$.

Define a map $\sigma:G\to G$ by
$$
\sigma\bigl(a_1^{\epsilon_1}a_2^{\epsilon_2}\cdots a_n^{\epsilon_n}\bigr)
=
a_2^{\epsilon_1}a_1^{\epsilon_2}\cdots a_n^{\epsilon_n}.
$$
It is easy to verify that $\sigma$ is an automorphism of $G$. Moreover,
$$
\sigma(a_1)=a_2\neq a_1,
$$
so
$$
\sigma\neq id.
$$
Hence
$$
|\operatorname{Aut}(G)|>1,
$$
a contradiction. Therefore we must have
$$
|G|\le 2.
$$

Proposition 8. Let $G$ be a finite abelian group which is not cyclic. Prove that $\operatorname{Aut}(G)$ is nonabelian.

Proof : Let
$$
a_1,a_2,\cdots,a_t
$$
be a minimal generating set of $G$. Since $G$ is not cyclic, we have
$$
t\ge 2.
$$
Because $G$ is abelian, every element of $G$ has the form
$$
a_1^{m_1}a_2^{m_2}\cdots a_t^{m_t}.
$$

Let the order of $a_1$ be $m$, and the order of $a_2$ be $n$. We claim that
$$
(m,n)\neq 1.
$$
Indeed, if $(m,n)=1$, then $a_1a_2$ has order $mn$. Since there exist $u,v\in \mathbb{Z}$ such that
$$
um+vn=1,
$$
we have
$$
(a_1a_2)^{vn}=a_1^{1-um}a_2^{vn}=a_1,
$$
so
$$
a_1\in \langle a_1a_2\rangle.
$$
Similarly,
$$
a_2\in \langle a_1a_2\rangle.
$$
This contradicts the minimality of the generating set $a_1,a_2,\cdots,a_t$. Therefore,
$$
(m,n)=d\neq 1.
$$

Now define an endomorphism $\varphi_1$ by
$$
\varphi_1(a_1)=a_1a_2^k,\qquad \varphi_1(a_i)=a_i \quad (i\ge 2),
$$
where
$$
k=\frac{n}{d}.
$$
Then $\varphi_1$ is invertible, with inverse given by
$$
\varphi_1^{-1}(a_1)=a_1a_2^{-k},\qquad \varphi_1^{-1}(a_i)=a_i \quad (i\ge 2).
$$
Hence
$$
\varphi_1\in \operatorname{Aut}(G).
$$

Similarly, define $\varphi_2$ by
$$
\varphi_2(a_2)=a_1^s a_2,\qquad \varphi_2(a_i)=a_i \quad (i\neq 2),
$$
where
$$
s=\frac{m}{d}.
$$
By the same reasoning,
$$
\varphi_2\in \operatorname{Aut}(G).
$$

Now compute:
$$
\varphi_1\varphi_2(a_1)=\varphi_1(a_1)=a_1a_2^k,
$$
while
$$
\varphi_2\varphi_1(a_1)
=\varphi_2(a_1a_2^k)
=\varphi_2(a_1)\varphi_2(a_2)^k
=a_1(a_1^s a_2)^k
=a_1^{1+sk}a_2^k.
$$
Hence
$$
\varphi_1\varphi_2(a_1)=\varphi_2\varphi_1(a_1)
\iff a_1^{sk}=e
\iff m\mid sk.
$$
Similarly,
$$
\varphi_1\varphi_2(a_2)=\varphi_2\varphi_1(a_2)
\iff n\mid sk.
$$
Therefore,
$$
\varphi_1\varphi_2=\varphi_2\varphi_1
\iff sk
$$
is a common multiple of $m$ and $n$.

However, the least common multiple of $m$ and $n$ is
$$
[m,n]=\frac{mn}{d},
$$
whereas
$$
sk=\frac{m}{d}\cdot \frac{n}{d}=\frac{mn}{d^2}.
$$
Since $d\neq 1$, we have
$$
\frac{mn}{d}>\frac{mn}{d^2},
$$
that is,
$$
[m,n]>sk.
$$
Thus $sk$ cannot be a common multiple of $m$ and $n$. Therefore
$$
\varphi_1\varphi_2\neq \varphi_2\varphi_1.
$$
Hence $\operatorname{Aut}(G)$ is nonabelian.

The cover image of this article was taken at Nanyang Technological University in Singapore.