As a test for my understanding of actions, I thought the following. Could you verify it down to the resulting claim, please?

Let $G$ be a finite group. A homomorphism $\varphi: G \rightarrow S_m$ (symmetric group of some degree $m$) is equivalent to an action of $G$ on a set $X$ of $m$ elements. Its kernel is:

$$\operatorname{ker}\varphi=\lbrace g \in G \mid \operatorname{Fix}(g)=X\rbrace$$

where $\operatorname{Fix}(g):=\{x \in X \mid g\cdot x=x\}$. Therefore:

\begin{alignat}{1} \operatorname{ker}\varphi=\lbrace e \rbrace &\Leftrightarrow |\operatorname{Fix}(g)|<|X|, \forall g \in G \setminus\lbrace e\rbrace \\ &\Rightarrow |\mathcal{O}||G|=\sum_{g \in G}|\operatorname{Fix}(g)|<|G||X| \\ &\Rightarrow |\mathcal{O}|<|X|\\ \tag 1 \end{alignat}

where $\mathcal{O}$ is the set of action's orbits. Therefore, $\varphi$ injective $\Rightarrow \exists x \in X$ s.t. $|O(x)|>1 \Rightarrow$ $\exists x \in X$ s.t. $|\operatorname{Stab}(x)|<|G|$ (Orbit-Stabilizer Theorem). If, in addition, $G$ has prime order, then $\exists x \in X$ s.t. $|\operatorname{Stab}(x)|=1$, and finally $\exists x \in X$ s.t. $|O(x)|=|G|$; but then $|X| \ge |G|$.

So, provided that so far so good:

Claim: If $G$ has prime order $p$ and $G \hookrightarrow S_m$, then $m \ge p$.

In words, $G$ of prime order can't embed into symmetric groups of degree smaller than $|G|$. (Said differently, Cayley's one is the "sharpest" embedding we can have in this case.)

  • $\begingroup$ I realize now that this was a proof-verification question. My bad. I don't see where you got the equality right after the first implication arrow in the display. $\endgroup$ – Mees de Vries Nov 21 '19 at 12:22
  • $\begingroup$ No problem, Mees. If you mean $|\mathcal{O}||G|=$ etc., that's Burnside's Lemma. $\endgroup$ – Luca Nov 21 '19 at 12:31

Note that if $|G| = p > m$ is prime, then $p \not\mid |S_m| = m!$. Therefore there cannot be an injective homomorphism $G \to S_m$, because its image would be a subgroup of order $p$, contradicting Lagrange's theorem.

Edit: I realized only after answering that this is a proof verification question: so to add, yes, your proof is correct, although it is much more complex than necessary (as demonstrated by the above, much shorter proof).

  • $\begingroup$ Much smarter, thanks. Though I was mainly interested in consolidating the few notions on actions I have, by retrieving simple results by means of them. So, I'll assume this is the case, and accept the answer. $\endgroup$ – Luca Nov 21 '19 at 11:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.