Simple group of order $168$ embeds into $A_8$

Question

Show that a simple group of order $168$ must be isomorphic to a subgroup of $A_8$.

I read the link here and I didn't find direct information about this particular problem. But here is my attempt at this.

Proof.

Now, $|G|=2^3 \cdot 3 \cdot 7$. By Sylow considerations, $n_7 \in \{1,8\}$ but $n_7 \neq 1$ by simplicity. So $n_7 = 8$. This induces a homomorphism $\varphi:G \to S_8$. Now, we know

$$G/ \ker \varphi \cong \varphi (G) \leq S_8$$

By simplicity, $\ker \varphi = 1$ or $G$ and clearly $\ker \varphi \neq G$ by Sylow. So $\ker \varphi = 1$. Toward a contradiction, suppose $\varphi (G) \not\subseteq A_8$. Then $G$ contains an odd permutation. Then $|\varphi(G): \varphi(G) \cap A_8|=2$. But now,

$$|\varphi(G):\varphi(G) \cap A_8| = \frac{|\varphi(G)|}{|\varphi(G) \cap A_8|} = 2 \implies |\varphi(G) \cap A_8| = 168/2 = 84$$

Because $A_8 \trianglelefteq S_8$, then $\varphi(G) \cap A_8 \trianglelefteq S_8$ which is impossible because $A_n$ is the only non-trivial proper normal subgroup of $S_n$ for $n \geq 5$.

$$\hspace{10cm} \blacksquare$$

Answer 1

I follow everything till you have embedded $G$ in $S_8$ by the injective homomorphism $\varphi\colon\,G\to S_8$. Now to show that this is an embedding in $A_n$, you have to show that there are no odd permutations in $\varphi(G)$.

Since $G$ is simple, so is its isomorph $\varphi(G)$. If there was an odd permutation in $\varphi(G)$ then half the elements in $\varphi(G)$ would be even permutations and the other half odd permutations. The even permutations would form a subgroup of index $2$ contradicting simplicity of $\varphi(G)$.