Consider a prime number \(p\) and a finite p-group \(G\), i.e. a group of order \(p^n\) with \(n \ge 1\).

If \(n=1\) the group \(G\) is cyclic hence abelian.

For \(n=2\), \(G\) is also abelian. This is a consequence of the fact that the center \(Z(G)\) of a \(p\)-group is non-trivial. Indeed if \(\vert Z(G) \vert =p^2\) then \(G=Z(G)\) is abelian. We can’t have \(\vert Z(G) \vert =p\). If that would be the case, the order of \(H=G / Z(G) \) would be equal to \(p\) and \(H\) would be cyclic, generated by an element \(h\). For any two elements \(g_1,g_2 \in G\), we would be able to write \(g_1=h^{n_1} z_1\) and \(g_2=h^{n_1} z_1\) with \(z_1,z_2 \in Z(G)\). Hence \[
g_1 g_2 = h^{n_1} z_1 h^{n_2} z_2=h^{n_1 + n_2} z_1 z_2= h^{n_2} z_2 h^{n_1} z_1=g_2 g_1,\] proving that \(g_1,g_2\) commutes in contradiction with \(\vert Z(G) \vert < \vert G \vert\). However, all \(p\)-groups are not abelian. For example the unitriangular matrix group \[
U(3,\mathbb Z_p) = \left\{
\begin{pmatrix}
1 & a & b\\
0 & 1 & c\\
0 & 0 & 1\end{pmatrix} \ | \ a,b ,c \in \mathbb Z_p \right\}\] is a \(p\)-group of order \(p^3\). Its center \(Z(U(3,\mathbb Z_p))\) is \[
Z(U(3,\mathbb Z_p)) = \left\{
\begin{pmatrix}
1 & 0 & b\\
0 & 1 & 0\\
0 & 0 & 1\end{pmatrix} \ | \ b \in \mathbb Z_p \right\},\] which is of order \(p\). Therefore \(U(3,\mathbb Z_p)\) is not abelian.

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 \(S_3 \simeq \text{Aut}(S_3)\).

Can the symmetric group \(\mathcal{S}_n\) be generated by any transposition and any \(n\)-cycle for \(n \ge 2\) integer? is the question we deal with.

We first recall some terminology:

Symmetric group

The symmetric group \(\mathcal{S}_n\) on a finite set of \(n\) symbols is the group whose elements are all the permutations of the \(n\) symbols. We’ll denote by \(\{1,\dots,n\}\) those \(n\) symbols.

Cycle

A cycle of length \(k\) (with \(k \ge 2\)) is a cyclic permutation \(\sigma\) for which there exists an element \(i \in \{1,\dots,n\}\) such that \(i, \sigma(i), \sigma^2(i), \dots, \sigma^k(i)=i\) are the only elements moved by \(\sigma\). We’ll denote the cycle \(\sigma\) by \((s_0 \ s_1 \dots \ s_{k-1})\) where \(s_0=i, s_1=\sigma(i),\dots,s_{k-1}=\sigma^{k-1}(i)\).

Transposition

A transposition is a cycle of length \(2\). We denote below the transposition of elements \(a \neq b\) by \((a \ b)\) or \(\tau_{a,b}\).

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