What Are The Necessary And Sufficient Conditions For A Finite Group G To Have A Normal Subgroup Isomorphic To The Quaternion Group Q8, Such That The Corresponding Quotient Group G/Q8 Is Isomorphic To The Cyclic Group Of Order 5, And The Conjugation Action Of G On Q8 Induces A Faithful Representation Of G/Q8 In The Automorphism Group Of Q8?

To determine the necessary and sufficient conditions for a finite group ${ G }$ to have a normal subgroup isomorphic to the quaternion group ${ Q_8 }$, such that the corresponding quotient group ${ G/Q_8 }$ is isomorphic to the cyclic group of order 5, and the conjugation action of ${ G }$ on ${ Q_8 }$ induces a faithful representation of ${ G/Q_8 }$ in the automorphism group of ${ Q_8 }$:

1. Understanding the Groups:

   • The quaternion group ${ Q_8 }$ is a non-abelian group of order 8.
   • The quotient group ${ G/Q_8 }$ is cyclic of order 5, implying ${ G }$ has order 40.

2. Automorphism Group of ${ Q_8 }$:

   • The automorphism group ${ \text{Aut}(Q_8) }$ is isomorphic to the symmetric group ${ S_3 }$, which has order 6.

3. Faithful Representation:

   • The conjugation action of ${ G }$ on ${ Q_8 }$ induces a homomorphism from ${ G }$ to ${ \text{Aut}(Q_8) }$.
   • Since ${ G/Q_8 }$ is cyclic of order 5, the image of this homomorphism must be a cyclic subgroup of order 5 in ${ \text{Aut}(Q_8) }$.

4. Contradiction:

   • ${ \text{Aut}(Q_8) }$ (isomorphic to ${ S_3 }$) does not have a cyclic subgroup of order 5 because 5 does not divide 6.
   • Therefore, the homomorphism from ${ G }$ to ${ \text{Aut}(Q_8) }$ cannot be injective, making a faithful representation impossible.

5. Conclusion:

   • Since the required faithful representation is impossible, no such group ${ G }$ exists.

Thus, the necessary and sufficient condition is that no such group ${ G }$ exists.

${ \boxed{\text{No such group } G \text{ exists.}} }$