# $G$ of prime order can't embed into symmetric groups of degree smaller than $|G|$. Proof verification.

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.)

• 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. – Mees de Vries Nov 21 '19 at 12:22
• No problem, Mees. If you mean $|\mathcal{O}||G|=$ etc., that's Burnside's Lemma. – 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.