**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.2 Exercise 4.2.12**

Solution: If $\pi[G] \leq S_G$ contains an odd permutation, then $\pi[G] \not\leq A_G$. Now $A_G \vartriangleleft S_G$ has index 2, which is prime. By Exercise 3.3.3, we have $[ \pi[G] : A_G \cap \pi[G] ] = 2$. Since the action of $G$ on itself by left multiplication is faithful, $G \leq S_G$ can be identified with $\pi[G]$. Thus $G$ has a subgroup of index 2; namely the preimage of $A_G \cap \pi[G]$.