Krdytkyu

Following the idea that a group is its structure, and reminding of Cayley theorem, I'm wondering whether we can build up virtually any finite group $G=lbrace a_0,dots,a_{n-1} rbrace$ by searching pairs of subgroups of $operatorname{Sym}(G)$ (the symmetric group over the set $G$), say $Theta=lbracetheta_i,i=0,dots,n-1rbrace$ and $Gamma=lbracegamma_j,j=0,dots,n-1rbrace$, such that:

i) $theta_i(a_j)=gamma_j(a_i)$ for all $i,j=0,dots,n-1$

ii) $theta_igamma_j=gamma_jtheta_i$ for all $i,j=0,dots,n-1$

Supposing to have found out such a pair, we could use their elements to define right and left multiplications, where i) would ensure the identity $a_ia_j=a_ia_j$ for all $i,j$, and ii) the associativity of the composition law "under construction". Moreover, the constraint i) entails that $Theta=Gamma Rightarrow theta_i=gamma_i$ for all $i$, so that $G$ is abelian if and only if $Theta=Gamma$ [Proof: $Theta=Gamma Rightarrow$ $exists sigma in operatorname{Sym}(n)$ such that $theta_i=gamma_{sigma(i)}$ for all $i Rightarrow$ $theta_i(a_j)=gamma_{sigma(i)}(a_j)$ for all $i,j Rightarrow$ (by virtue of i)) $gamma_j(a_i)=gamma_{sigma(i)}(a_j)$ for all $i,j Rightarrow$ $gamma_{sigma(i)}(a_i)=gamma_{sigma(i)}(a_{sigma(i)})$ for all $i Rightarrow$ ($gamma_{sigma(i)}$ is 1-1) $a_i=a_{sigma(i)}$ for all $i Rightarrow$ ($a_k$ are distinct by hypothesis) $sigma(i)=i$ for all $i Rightarrow$ $theta_i=gamma_i$ for all $i$. #]

As a first test for this approach, let's consider $rho in operatorname{Sym}(G)$ defined by $rho(a_k):=a_{k+1 mod n}$, $k=0,dots,n-1$. It is: $rho^i(a_j)=a_{j+i mod n}=a_{i+j mod n}=rho^j(a_i)$; therefore, if we set $gamma_i=theta_i:=rho^i$, we have that either i) and ii) are fulfilled. The subgroups of $operatorname{Sym}(G)$ (here coincident) $Theta=lbrace theta_i=rho^i rbrace$ and $Gamma=lbrace gamma_i=rho^i rbrace$ define the (abelian) composition law $a_ia_j=a_{j+i mod n}$, whence $a_i^k=a_{ki mod n}$ and then $a_1^k=a_{k mod n}=a_k$ for $k=0,dots,n-1$. Thus, we are finally led to $G=lbrace a_k, k=0,...,n-1 rbrace= lbrace a_1^k, k=0,...,n-1 rbrace= langle a_1 rangle$, and $G$ is cyclic. This result is irrespective of $n$, so cyclic groups exist for any order $n$ (not a surprising result, indeed, but here what I'm focused on is rather the approach to get it).

Yet another way to start appreciating this approach, by rediscovering basic facts by means of it, could be the following. Condition ii) implies that $ThetaGamma=GammaTheta$ and then $ThetaGamma le operatorname{Sym}(G)$. So, once set $l:=|Theta cap Gamma|$, $n:=|Theta|$ (=$|G|$) and noticing that $|ThetaGamma|=n^2/l$, we get: $l le n le n^2/l le n!$, with (Lagrange) $l|n wedge (n^2/l)|n!$. Now, $Theta ne Gamma Rightarrow l < n < n^2/l le n!$. Then, if $|G|=n=p$, with $p$ prime, we have $l=1$ and then $p^2|p!$: contradiction. Then we are left with $|G|=p$ ($p$ prime) $Rightarrow Theta=Gamma Rightarrow G$ abelian.

Could this approach be used to search for other, less trivial group structures?

You can find a lot of structure by looking the representation of a group in its symmetry group, as well the symmetry group of other groups. Examples, if you have a group $G$ with $|G|=p^km$ where $p$ is prime, and $m$ is relatively prime to $p$, then you can show that $G$ is not simple if $m!<p^km$ by looking at the symmetry group $text{Sym}_{m}leqtext{Sym}(G)$.

Or if you have an injective homomorphism $phi:Gtotext{Sym}_{G}$ that contains cycles $(1,2),(1,2,...,n)$, then $G$ contains an isomoprhic copy of $text{Sym}_{n}.$

I forget the exact relation, but I recall that you can deduce properties of symmetric groups by looking at representations of Galois Groups inside symmetric groups.

In general representing groups inside well known groups like the Symmetric groups or Matrix groups can be very helpful.