Some result on realizability of groups as automorphism groups

The problem of realizability is one of the problems i find most enterteining, maybe not the deepest, but i approach it just for the game.

I do not know how long this will be and how much times i will update it. This will be the content table, which will be updated as the document grows.

1. Theorem and Lemma i will use

1.1 Cyclic inner automorphism

Proof. Let's apply $\varphi$ to the product of two elements $x,y \in G$. We have $\varphi(xy) = (xy)^{-1} = y^{-1}x^{-1}$. On the other hand, since $\varphi$ is an automorphism, we have $\varphi(xy) = \varphi(x)\varphi(y) = x^{-1}y^{-1}$. Therefore, we have $y^{-1}x^{-1} = x^{-1}y^{-1}$, which implies that $xy = yx$. Since this holds for any $x,y \in G$, we conclude that $G$ is abelian. $\blacksquare$

Proof. Remind that $Inn(G) \cong G/Z(G)$, where $Z(G)$ is the center of $G$. If $Inn(G)$ is cyclic, then $G/Z(G)$ is cyclic. Let $gZ(G)$ be a generator of $G/Z(G)$. Then for any $x \in G$, we have $xZ(G) = (gZ(G))^n$ for some integer $n$. This implies that $x = g^nz$ for some $z \in Z(G)$. Then for any $x,y \in G$, we have $x = g^nz_1$ and $y = g^mz_2$ for some integers $n,m$ and $z_1,z_2 \in Z(G)$. Which clearly commute. Therefore $G$ is abelian. $\blacksquare$

1.2 Locally cyclic groups

Proof. Let $H$ be a finitely generated subgroup of $\Q$. Then there exist rational numbers $q_1, q_2, \ldots, q_n \in \Q$ such that $H = \langle q_1, q_2, \ldots, q_n \rangle$. We can write each $q_i$ in the form $q_i = \frac{a_i}{b_i}$, where $a_i \in \Z$ and $b_i \in \Z^+$ are coprime. Let $B = \text{lcm}(b_1, b_2, \ldots, b_n)$ be the least common multiple of the denominators and consider $q = \gcd(a_1B/b_1, a_2B/b_2, \ldots, a_n B/b_n)$ the greatest common divisor. We claim that $H = \langle \frac{q}{B} \rangle$. By definition of greatest common divisor, there exist integers $k_1, k_2, \ldots, k_n$ such that $a_i \cdot (B/b_i) = k_i q$ for each $i$. Therefore, we have $q_i = \frac{a_i}{b_i} = \frac{k_i q}{B}$, which implies that $q_i \in \langle \frac{q}{B} \rangle$ for each $i$. Hence, $H \subseteq \langle \frac{q}{B} \rangle$. Conversely, by Bezout's theorem we have that $q = \sum_{i=1}^{n} \lambda_i a_i \cdot (B/b_i)$ for some integers $\lambda_1, \lambda_2, \ldots, \lambda_n$. Therefore, we have $\frac{q}{B} = \sum_{i=1}^{n} \lambda_i \frac{a_i}{b_i} = \sum_{i=1}^{n} \lambda_i q_i$, which implies that $\frac{q}{B} \in H$. Hence, $\langle \frac{q}{B} \rangle \subseteq H$. Combining both inclusions, we conclude that $H = \langle \frac{q}{B} \rangle$, which shows that $H$ is cyclic. Actually we could conclude earlier that $H$ is cyclic by noting that $H$ is a subgroup of cyclic group $\langle \frac{q}{B}\rangle$, therefore $H$ is cyclic. $\blacksquare$

Proof. Let $x, y$ two element of $G$. Consider the subgroup $X = \langle x, y \rangle$ generated by $x$ and $y$. Let $\overline{X} = X * Z(G)/Z(G)$ be the image of $X$ in the quotient group. By an homomorphism theorem, we have that $\overline{X} \simeq X/(X \cap Z(G))$. Though it is a finitely generated subgroup of $G/Z(G) \simeq Inn(G) \leq Aut(G) \simeq H$. Since $H$ is locally cyclic, we have that $\overline{X}$ is cyclic. Finally $X/Z(X) \simeq (X/(X \cap Z(G))) / (Z(X) / x \cap Z(G))$ is cyclic being a quotient of a cyclic group. Therefore by Theorem 1.2, we have that $X$ is abelian. In particular, $x$ and $y$ commute. Since $x$ and $y$ are arbitrary elements of $G$, we conclude that $G$ is abelian. $\blacksquare$

2. Unrealizability of $\Z$ and $\Z/n\Z$ with $n$ odd

Let $G$ be a group such that $Aut(G) \simeq \Z$. Then $Inn(G) \leq Aut(G)$ is cyclic. Therefore by Theorem 1.2, we have that $G$ is abelian. Then $\varphi: G \to G$ defined by $\varphi(x) = x^{-1}$ is an automorphism of $G$. Therefore we have that $Aut(G)$ contains an element of order that divides $2$. But $\Z$ does not contain any element of order $2$. Therefore $\varphi$ has to be the identity automorphism. Therefore $G = \mathbb{F}_2^n$ for some natural number $n$. But $Aut(\mathbb{F}_2^n)$ contains an element of order $2$, which is the one that swap the first two coordinates. Therefore $Aut(G)$ contains an element of order $2$, which is a contradiction. Therefore there is no group $G$ such that $Aut(G) \simeq \Z$.

The same argument works on $\Z/n\Z$ with $n$ odd, since the point is to find an element of order 2.

2. Unrealizability of $\Q$

By the Theorems of 1.2 section, we easily conclude that if $\Q$ were an automorphism group, it would have had an element of order 2, which is not the case. Though $\Q$ is not such a group.