Doubt regarding strong component in a graph
Your question is a bit too broad. There can be a lot of properties $P$ such that if $H$ is a subgraph of $G$ and $P(H)$ holds then all vertices of $H$ are in the same strong component.

For example:


$P(H) = $ true iff H is a (directed) clique
$P(H) = $ true iff H contains only one vertex
$P(H) = $ true iff for every edge $(u,v)$ of $H$, $(v,u)$ is also in $H$.
$P(H) = $ true iff $H$ has $n&gt;1$ vertices and more than $(n-1)^2$ edges.
...


You can prove that the last condition implies that the vertices of $H$ are in a strong component by induction on $n$.

If $n=2$ then the graph has more than $1^2 = 1$ edge, i.e., it has at least two edges. The claim trivially follows.

If $n&gt;2$, then let $H'$ be any (not necessarily proper) subgraph of $H$ with $(n-1)^2+1$ edges. Pick a vertex $v$ of minimum degree in $H'$. The degree $\delta$ of $v$ is at most $\frac{(n-1)^2+1}{n}$. Therefore $H-v$ has $n-1$ vertices and at least $(n-1)^2 + 1 - \delta = \frac{n(n-1)^2+n - (n-1)^2 - 1}{n} &gt; (n-2)^2$ edges. By inductive hypothesis $H-v$ is strongly connected. The claim follows since $\delta$ is at least $(n-1)^2 + 1 - (n-1)(n-2) = (n-1)(n-1 -n +2) = (n-1) + 1 = n$, showing that $v$ has at least one incoming and one outgoing edge.