Prove the proposition that every finite group is isomorphic to a subgroup of $S_n$ for some $n$. I think I understand this proof, but I want to be sure I'm using terminology correctly.
I'm trying to prove the proposition that every finite group is isomorphic to a subgroup of $S_n$ for some $n$.
I defined the map $\phi: G \to \mathrm{Perm}(G)$ by $g \mapsto m_g$, where $m_g$ is the map sending $x \mapsto gx$. I proved that $m_g$ is in fact a bijection and that $\phi$ is a homomorphism. Further, I proved that $\phi$ is injective, so $G$ is isomorphic to its image. Since $|G| = n$, I can then say that $\phi(G) \cong S_{|G|}$, so I have
$$G \cong \mathrm{Im}(\phi) \subset \mathrm{Perm}(G) \cong S_{|G|}.$$
The question of terminology comes in here. I want to say that I have "identified $G$ isomorphically with a subgroup of $S_{|G|}$. Really what I'm doing is asserting the existence of an isomorphism from $\mathrm{Perm}(G)$ to $S_n$, and if I consider $\phi(G)$ under tht isomorphism, I will have a subgroup of $S_{|G|}$. But I worry I'm overloading notation, because "identify" is often used, I believe, to denote a bijection of sets.
 A: I understand that you have already proved that $G$ is isomorphic to a subgroup of $\operatorname{Perm}(G)$, and now you want to prove that $\operatorname{Perm}(G)\cong S_{|G|}$, so as to give the sentence "to identify $G$ isomorphically with a subgroup of $S_{|G|}$" a justification. If so, take any bijection $f\colon G\longrightarrow \{1,\dots,|G|\}$ and define $\psi^{(f)}\colon\operatorname{Perm}(G)\longrightarrow S_{|G|}$ by $\sigma\longmapsto f\sigma f^{-1}$. You can prove that $\psi^{(f)}$ is indeed an isomorphism. Therefore:
$$G\cong (\psi^{(f)}\phi)(G)\le S_{|G|} \tag 1$$
So, yes, you can say that you have "identified $G$ isomorphically with a subgroup of $S_{|G|}$". (Side note: such an identification is noncanonical, depending on the choice of the bijection $f$.)
A: Question: "I think I understand this proof, but I want to be sure I'm using terminology correctly. I'm trying to prove the proposition that every group is isomorphic to a subgroup of Sn for some n."
Answer: If $G:=\{g_1,..,g_n\}$ are the elements in $G$ and $S:=G$ you get a map
$$\sigma: G \rightarrow Aut(S)$$
defined by
$$\sigma_g(g_i):=gg_i.$$
It follows $\sigma_{gh}(g_i)=(gh)g_i=g(hg_i)=\sigma_g(\sigma_h(g_i))$ hence $\sigma_{gh}=\sigma_g \sigma_h$ and $\sigma$ is a map of groups. The inverse of $\sigma_g$ is $\sigma_{g^{-1}}$, hence $\sigma_g$ is an automorphism of $S$ and the map is "well defined".
If $\sigma_g=\sigma_h$ it follows $\sigma_g(e)=ge=g=\sigma_h(e)=h$ ($e$ is the identity element) hence $\sigma$ is injective.
Hence $\sigma$ realize $G$ as a subgroup of $Aut(S)\cong S_n$, which is the symmetric group on $n$ elements.
There are many different proofs of this fact:
https://en.wikipedia.org/wiki/Cayley%27s_theorem
Are you able to give another proof that is "essentially different" from this one?
