Prove $Q_8$ is isomorphic to a subgroup of $S_8$ [duplicate]

Solution 1:

Following mark Brandenburg's comment, I suggest to first note the following:

If $A,B$ are two sets and $f\colon A\to B$ is a bijection (so $f^{-1}$ exists), then we obtain a group isomorphism $\phi_f\colon \operatorname{Sym}(A)\to \operatorname{Sym}(B)$ by letting $\phi_f(\sigma)=f\circ \sigma\circ f^{-1}$ for each $\sigma\in \operatorname{Sym}(A)$.

To see that $\phi_f$ is a homomorphism note that $$\phi_f(\sigma)\phi_f(\tau)=f\circ \sigma\circ f^{-1}\circ f\circ \tau\circ f^{-1}=f\circ \sigma\circ \tau\circ f^{-1}=\phi_f(\sigma\tau);$$ to see that $\phi_f$ is an isomorphism note that we can explicitly name the inverse homomoprhism $\operatorname{Sym}(B)\to\operatorname{Sym}(A)$, namely we have $(\phi_f)^{-1}=\phi_{f^{-1}}$: $$\phi_{f^{-1}}(\phi_f(\sigma))= f^{-1}\circ \phi_f(\sigma)\circ f = f^{-1}\circ f\circ \sigma\circ f^{-1}\circ f=\sigma,$$ so $\phi_{f^{-1}}\circ\phi_f=\operatorname{Id}_{\operatorname{Sym}(A)}$ and similarly $\phi_{f}\circ\phi_{f^{-1}}=\operatorname{Id}_{\operatorname{Sym}(B)}$.

Now if $|G|=n$, this just states that there exists a bijection $G\to \{1,\ldots,n\}$ of the underlying set of $G$ with the set of the first $n$ naturals. By waht we just saw, $\operatorname{Sym}(G)$ is isomorphic to $\operatorname{Sym}(\{1,\ldots,n\})=S_n$. Therefore, to show that $G$ is isomorphic to a subgroup of $S_n$, it is sufficient to show that $G$ is isomorphic to a subgroup of $\operatorname{Sym}(G)$: Any injective group homomorphism $G\hookrightarrow\operatorname{Sym}(G)$ produces an injective group homomorphism $G\hookrightarrow S_n$ via the isomorphism.

That being said, you essentially did the right thing: you considered left multiplication, i.e. for $a\in G$ the left multiplication by $a$, i.e. the map $$\begin{align}\lambda_a\colon G&\to G\\ x&\mapsto ax\end{align} $$ We immediately verify that $\lambda_a(\lambda_b(x))=abx=\lambda_{ab}(x)$ (the law of associativity hides in here) so that $$\tag1\lambda_a\circ\lambda_b=\lambda_{ab}.$$ Moreover $\lambda_1$ is the identity map: $\lambda_1(x)=1x=x$). We conclude that $\lambda_a\circ\lambda_{a^{-1}}=\lambda_{a^{-1}}\circ \lambda_a=\lambda_1=\operatorname{Id}_G$, i.e. $\lambda_a$ is bijective with inverse $\lambda_{a^{-1}}$. Thus for each $a\in G$ we have an element $\lambda_a\in\operatorname{Sym}(G)$. Moreover, the map $$\begin{align}\lambda\colon G&\to \operatorname{Sym}G\\ a&\mapsto \lambda_a\end{align} $$ is a homomorphism because of $(1)$. This homomorphism has trivial kernel because $\lambda_a(1)=a\ne 1$ whenever $a\ne 1$. Thus we have our desired injective gorup homomorphism $\lambda\colon G\hookrightarrow \operatorname{Sym}(G)$.