Is the following a valid approach to determine group types of small orders?
Let $G=\{a_1,\dots,a_n\}$ be a set. Suppose that $\Theta=\{\theta_1,\dots,\theta_n\}$ and $\Gamma=\{\gamma_1,\dots,\gamma_n\}$ are subgroups of $S_n$ such that:
- $\theta_i(j)=\gamma_j(i), \forall i,j=1,\dots,n$;
- $\theta_i\gamma_j=\gamma_j\theta_i, \forall i,j=1,\dots,n$.
Then the composition law:
$$a_ia_j:=a_{\theta_i(j)}=a_{\gamma_j(i)}$$
gives $G$ a group structure. In fact:
- closure: by definition;
- associativity: by 1 and 2 above, $$\begin{align} (a_ia_j)a_k&=a_{\theta_i(j)}a_k\\ &=a_{\gamma_k(\theta_i(j))}\\ &=a_{(\gamma_k\theta_i)(j)}\\ &=a_{(\theta_i\gamma_k)(j)}\\ &=a_{\theta_i(\gamma_k(j))}\\ &=a_ia_{\gamma_k(j)}\\ &=a_i(a_ja_k). \end{align}$$
- unit: since $\Theta$ and $\Gamma$ are groups, $\exists \bar i, \bar j$ such that $\theta_{\bar i}=\gamma_{\bar j}=\iota$; then, $$a_{\bar i}a_j=a_{\theta_{\bar i}(j)}=a_{\iota(j)}=a_j, \forall j$$ and $a_{\bar i}$ is left unit; likewise, $a_{\bar j}$ is right unit; now, $a_{\bar j}=a_{\bar i}a_{\bar j}$, because $a_{\bar i}$ is left unit; but also $a_{\bar i}a_{\bar j}=a_{\bar i}$, because $a_{\bar j}$ is right unit; then, $a_{\bar j}=a_{\bar i}$;
- inverse: $\forall j, \exists k$ such that $\theta_j(k)=\bar i$; then, $a_ja_k=a_{\theta_j(k)}=a_{\bar i}$, and $a_k$ is right inverse of $a_j$; likewise, $\forall j, \exists m$ such that $\gamma_j(m)=\bar i$; then, $a_ma_j=a_{\gamma_j(m)}=a_{\bar i}$, and $a_m$ is left inverse of $a_j$; but $$\begin{align}a_k&=a_{\bar i}a_k\\ &=(a_ma_j)a_k\\ &=a_m(a_ja_k)\\ &=a_ma_{\bar i}\\ &=a_m. \end{align}$$
As a first use of this formulation, I notice that, for any $n$-cycle $\sigma$ and $\bar k \in \{1,\dots,n\}$, the (abelian) option
$$\theta_{\sigma^i(\bar k)}(\sigma^j(\bar k))=\gamma_{\sigma^i(\bar k)}(\sigma^j(\bar k)):=\sigma^{i+j}(\bar k), i,j=1,\dots,n \tag 1$$
is a valid one, i.e. fulfills the conditions 1 and 2, for every $n$, and gives rise to the cyclic group type. In fact, from $a_{\sigma^i(\bar k)}a_{\sigma^j(\bar k)}=a_{\sigma^{i+j}(\bar k)}$, we get $a_{\sigma^m(\bar k)}=a_{\sigma(\bar k)}^m$ (induction on $m$) and then $G=\langle a_{\sigma(\bar k)} \rangle$. Then, cyclic groups exist of every order: this is trivial, but here my focus is to look at ciclicity just as an entry test for the "allowed structures" approach (1 and 2).
I'm wondering whether we can pursue this way (to explicit $\Theta$ and $\Gamma$ fulfilling 1 and 2) to show that:
a) if $n$ is prime, then $(1)$ is the only valid option (for some suitable choice of $\sigma$ and $\bar k$);
b) if $n=4$, then the only other option is the one establishing Klein's Group;
c) if $n=6$, then the only other option is the one establishing $S_3$.
REMARK. I'm aware that finding the number of groups of a given order $n$ (up to isomorphism) has purely group-theoretic answers (based on Cauchy and Sylow theorems), but I'd like to retrieve the known results a), b) and c) from "structure constraints" 1 and 2.
Edit (Aug 20)
Note that, by conditions 1 and 2:
$$\begin{align} (\theta_i\theta_j)(k)&=\theta_i(\theta_j(k))\\ &=\theta_i(\gamma_k(j))\\ &=(\theta_i\gamma_k)(j)\\ &=(\gamma_k\theta_i)(j)\\ &=\gamma_k(\theta_i(j))\\ &=\theta_{\theta_i(j)}(k) \end{align}$$
so that: $$\theta_i\theta_j=\theta_{\theta_i(j)}$$
Likewise,
$$\begin{align} (\gamma_i\gamma_j)(k)&=\gamma_i(\gamma_j(k))\\ &=\gamma_i(\theta_k(j))\\ &=(\gamma_i\theta_k)(j)\\ &=(\theta_k\gamma_i)(j)\\ &=\theta_k(\gamma_i(j))\\ &=\gamma_{\gamma_i(j)}(k) \end{align}$$
so that: $$\gamma_i\gamma_j=\gamma_{\gamma_i(j)}$$
Therefore, if we define $\varphi(\theta_i):=\gamma_i$, we get:
$$\begin{align} \varphi(\theta_i\theta_j)&=\varphi(\theta_{\theta_i(j)})\\ &=\gamma_{\theta_i(j)}\\ &=\gamma_{\gamma_j(i)}\\ &=\gamma_j\gamma_i\\ &=\varphi(\theta_j)\varphi(\theta_i) \end{align}$$
and finally
$$\Theta \cong \Gamma^{\operatorname{op}}$$