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
The answer is positive and we’ll prove that it is the case for the symmetric group \(S_3\).

Consider the morphism \[
\Phi : & S_3 & \longrightarrow & \text{Aut}(S_3) \\
& a & \longmapsto & \varphi_a \end{array}\]
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?