# A group isomorphic to its automorphism group

We consider a group $$G$$ and we look at its automorphism group $$\text{Aut}(G)$$. Can $$G$$ be isomorphic to
$$\text{Aut}(G)$$?
The answer is positive and we’ll prove that it is the case for the symmetric group $$S_3$$.

Consider the morphism $\begin{array}{l|rcl} \Phi : & S_3 & \longrightarrow & \text{Aut}(S_3) \\ & a & \longmapsto & \varphi_a \end{array}$
where $$\varphi_a$$ is the inner automorphism $$\varphi_a : x \mapsto a^{-1}xa$$. It is easy to verify that $$\Phi$$ is indeed a group morphism. The kernel of $$\Phi$$ is the center of $$S_3$$ which is having the identity for only element. Hence $$\Phi$$ is one-to-one and $$S_3 \simeq \Phi(S_3)$$. Therefore it is sufficient to prove that $$\Phi$$ is onto. As $$|S_3|=6$$, we’ll be finished if we prove that $$|\text{Aut}(S_3)|=6$$.

Generally, for $$G_1,G_2$$ groups and $$f : G_1 \to G_2$$ a one-to-one group morphism, the image of an element $$x$$ of order $$k$$ is an element $$f(x)$$ having the same order $$k$$. So for $$\varphi \in \text{Aut}(S_3)$$ the image of a transposition is a transposition. As the transpositions $$\{(1 \ 2), (1 \ 3), (2 \ 3)\}$$ generate $$(S_3)$$, $$\varphi$$ is completely defined by $$\{\varphi((1 \ 2)), \varphi((1 \ 3)), \varphi((2 \ 3))\}$$. We have 3 choices to define the image of $$(1 \ 2)$$ under $$\varphi$$ and then 2 choices for the image of $$(1 \ 3)$$ under $$\varphi$$. The image of $$(2 \ 3)$$ under $$\varphi$$ is the remaining transposition.

Finally, we have proven that $$|\text{Aut}(S_3)|=6$$ as desired and $$S_3 \simeq \text{Aut}(S_3)$$.

# Playing with interior and closure

Let’s play with the closure and the interior of sets.

To start the play, we consider a topological space $$E$$ and denote for any subspace $$A \subset E$$: $$\overline{A}$$ the closure of $$A$$ and $$\overset{\circ}{A}$$ the interior of $$A$$.

### Warm up with the closure operator

For $$A,B$$ subsets of $$E$$, the following results hold: $$\overline{\overline{A}}=\overline{A}$$, $$A \subset B \Rightarrow \overline{A} \subset \overline{B}$$, $$\overline{A \cup B} = \overline{A} \cup \overline{B}$$ and $$\overline{A \cap B} \subset \overline{A} \cap \overline{B}$$.

Let’s prove it.
$$\overline{A}$$ being closed, it is equal to its closure and $$\overline{\overline{A}}=\overline{A}$$.

Suppose that $$A \subset B$$. As $$B \subset \overline{B}$$, we have $$A \subset \overline{B}$$. Also, $$\overline{B}$$ is closed so it contains $$\overline{A}$$, which proves $$\overline{A} \subset \overline{B}$$.

Let’s consider $$A,B \in E$$ two subsets. As $$A \subset A \cup B$$, we have $$\overline{A} \subset \overline{A \cup B}$$ and similarly $$\overline{B} \subset \overline{A \cup B}$$. Hence $$\overline{A} \cup \overline{B} \subset \overline{A \cup B}$$. Conversely, $$A \cup B \subset \overline{A} \cup \overline{B}$$ and $$\overline{A} \cup \overline{B}$$ is closed. So $$\overline{A \cup B} \subset \overline{A} \cup \overline{B}$$ and finally $$\overline{A \cup B} = \overline{A} \cup \overline{B}$$.

Regarding the inclusion $$\overline{A \cap B} \subset \overline{A} \cap \overline{B}$$, we notice that $$A \cap B \subset \overline{A} \cap \overline{B}$$ and that $$\overline{A} \cap \overline{B}$$ is closed to get the conclusion.

However, the implication $$\overline{A} \subset \overline{B} \Rightarrow A \subset B$$ doesn’t hold. For a counterexample, consider the space $$E=\mathbb R$$ equipped with the topology induced by the absolute value distance and take $$A=[0,1)$$, $$B=(0,1]$$. We have $$\overline{A}=\overline{B}=[0,1]$$.

The equality $$\overline{A} \cap \overline{B} = \overline{A \cap B}$$ doesn’t hold as well. For the proof, just consider $$A=[0,1)$$ and $$B=(1,2]$$. Continue reading Playing with interior and closure

# A discontinuous real convex function

Consider a function $$f$$ defined on a real interval $$I \subset \mathbb R$$. $$f$$ is called convex if: $\forall x, y \in I \ \forall \lambda \in [0,1]: \ f((1-\lambda)x+\lambda y) \le (1-\lambda) f(x) + \lambda f(y)$

Suppose that $$I$$ is a closed interval: $$I=[a,b]$$ with $$a < b$$. For $$a < s < t < u < b$$ one can prove that: $\frac{f(t)-f(s)}{t-s}\le \frac{f(u)-f(s)}{u-s}\le\frac{f(u)-f(t)}{u-t}.$ It follows from those relations that $$f$$ has left-hand and right-hand derivatives at each point of the interior of $$I$$. And therefore that $$f$$ is continuous at each point of the interior of $$I$$.
Is a convex function defined on an interval $$I$$ continuous at all points of the interval? That might not be the case and a simple example is the function: $\begin{array}{l|rcl} f : & [0,1] & \longrightarrow & \mathbb R \\ & x & \longmapsto & 0 \text{ for } x \in (0,1) \\ & x & \longmapsto & 1 \text{ else}\end{array}$

It can be easily verified that $$f$$ is convex. However, $$f$$ is not continuous at $$0$$ and $$1$$.

# Is the quotient group of a finite group always isomorphic to a subgroup?

Given a normal subgroup $$H$$ of a finite group $$G$$, is $$G/H$$ always isomorphic to a subgroup $$K \le G$$?

### The case of an abelian group

According to the fundamental theorem of finite abelian groups, every finite abelian group $$G$$ can be expressed as the direct sum of cyclic subgroups of prime-power order: $G \simeq \bigoplus_{i=1}^u \mathbb{Z}_{p_i^{\alpha_i}}$ where $$p_1, \dots , p_u$$ are primes and $$\alpha_1, \dots , \alpha_u$$ non zero integers.

If $$H \le G$$ we have $H \simeq \bigoplus_{i=1}^u \mathbb{Z}_{p_i^{\beta_i}}$ with $$0 \le \beta_1 \le \alpha_1, \dots, 0 \le \beta_u \le \alpha_u$$. Then $G/H \simeq \bigoplus_{i=1}^u \mathbb{Z}_{p_i^{\alpha_i-\beta_i}}$ which is a subgroup of $$G$$.

If $$G$$ is not abelian, then $$G/H$$ might not be isomorphic to a subgroup of $$G$$. Continue reading Is the quotient group of a finite group always isomorphic to a subgroup?