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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 $\Theta=\lbrace\theta_i,i=0,\dots,n-1\rbrace$ and $\Gamma=\lbrace\gamma_j,j=0,\dots,n-1\rbrace$, such that:

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

ii) $\theta_i\gamma_j=\gamma_j\theta_i$ 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 $\Theta\Gamma=\Gamma\Theta$ and then $\Theta\Gamma \le \operatorname{Sym}(G)$. So, once set $l:=|\Theta \cap \Gamma|$, $n:=|\Theta|$ (=$|G|$) and noticing that $|\Theta\Gamma|=n^2/l$, 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?

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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 $\text{Sym}_{m}\leq\text{Sym}(G)$.

Or if you have an injective homomorphism $\phi:G\to\text{Sym}_{G}$ 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.

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  • $\begingroup$ That's encouraging - thanks for the feedback. I've edited my post with some more basic stuff in that direction, hoping it fits! $\endgroup$ – Luca Dec 10 '18 at 14:11
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    $\begingroup$ " $G$ is simple if $m!<p^k m$" doesn't seem right to me as there is always a cyclic group of any given order.(And did you mean "$p$ is prime and $m$ is coprime to $p$"?) $\endgroup$ – ancientmathematician Dec 10 '18 at 15:13
  • $\begingroup$ @ancientmathematician You're right, type-o. It's not simple, and we can show that by a non-trivial non-injective homomorphism $\phi:G\to\text{Sym}_{\text{Syl}_p(G)}.$ $\endgroup$ – Melody Dec 10 '18 at 18:39
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    $\begingroup$ You've still got to tidy up the hypotheses on $p$ (it is to be "prime") and $m$ (it's for to be "relatively prime to $p$" - not to $m$). $\endgroup$ – ancientmathematician Dec 11 '18 at 7:35
  • $\begingroup$ My gosh, I can't believe I wrote that. Thank you again. $\endgroup$ – Melody Dec 11 '18 at 16:44

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