Classification of finite groups such that every subgroup is a quotient group of the group by one of its subgroup.

Question

Let $G$ be a finite group such that $\forall H\le G$ $\exists K\le G$ such that $H\cong G/K$

What can be said about the group?

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Suppose $G$ is a finite group.

1. We know that any finite abelian group has the said property.

2. Let $G$ is finite non abelian group.

$S_3$ doesn't satisfy the above property as for $H=\langle (123)\rangle$ $\not\exists K\le G$ such that $H\cong S_3/K$

What I have observed that to satisfy the above property,$\forall d\mid o(G) $ there must exists $K\lhd G $ such that $o(K) =d$

But this condition is not sufficient as $H=\langle i\rangle \le Q_8$ there doesn't exists $K\le Q_8$ such that $H\cong Q_8/K$ as $Q_8$ has a unique (normal) subgroup of order $2$ , namely $Z(Q_8) $ which yields a non cyclic factor group (which is in fact isomorphic to $K_4$)

Give an example of a non abelian group having the above property.

What can be said about the finite group having the above property?