Maximal subgroup of group of prime power order

Definition
Suppose $$G$$ is is a group of prime power order (with underlying prime $$p$$) and $$H$$ is a subgroup of $$G$$. We say that $$H$$ is a maximal subgroup of group of prime power order if $$G$$ is a group of prime power order and $$H$$ satisfies any of the following equivalent conditions:


 * 1) $$H$$ is a maximal subgroup of $$G$$.
 * 2) $$H$$ is a maximal normal subgroup of $$G$$.
 * 3) $$H$$ is a normal maximal subgroup of $$G$$.
 * 4) $$H$$ is a subgroup of prime index in $$G$$.
 * 5) $$H$$ is a normal subgroup of $$G$$ and the quotient group $$G/H$$ is a group of prime power order.
 * 6) $$H$$ contains the Frattini subgroup $$\Phi(G)$$ of $$G$$ and the quotient $$H/\Phi(G)$$ is a codimension one subspace in the Frattini quotient $$G/\Phi(G)$$, viewed as a vector space over the field of $$p$$ elements.

Equivalence of definitions
See equivalence of definitions of maximal subgroup of group of prime power order. The key ingredients to the proof are prime power order implies nilpotent, nilpotent implies every maximal subgroup is normal, and equivalence of definitions of group of prime order (which shows that any simple abelian group must be cyclic of prime order).

Weaker properties

 * Stronger than::Maximal subgroup of finite nilpotent group
 * Stronger than::Subgroup of prime index
 * Stronger than::Normal subgroup of prime index
 * Stronger than::Maximal normal subgroup
 * Stronger than::Isomorph-normal subgroup
 * Stronger than::Order-normal subgroup