There are a couple of applications of the Lefschetz Hyperplane Theorem I am struggling to wrap my head around. Hopefully someone knows how these facts are deduced directly from the theorem.
Suppose $X$ is a smooth projective variety which is a complete intersection of hypersurfaces in some projective space. If $\dim(X)=n$ and the degrees of the hypersurface are sufficiently large, then $X$ cannot have a subvariety isomorphic to $\mathbb{P}^{n-1}$.
My attempt. If I apply the hyperplane theorem with $\mathbb{C}$ coefficients, then I know that $H^{i}(X,\mathbb{C})\rightarrow H^i(\mathbb{P}^{n-1},\mathbb{C})$ is an isomorphism for $i<n-1$ and injective for $i=n-1$. Now if $i=n-1$ is odd, then the right hand side is $0$. Then I want to show that $H^{n-1}(X,\mathbb{C})\neq 0$. I think I can always try to use the Hodge decomposition to try and compute this (i.e. I need some $h^{p,q}(X)\neq 0$, but I do not see how to do it in general.
A variation of this question is to use the Lefschetz Hyperplane Theorem to rule out existence of certain subvarieties with various degrees. The above case is for a linear subvariety.
I have seen this used often enough and my general understanding is that if $X$ is given, then I can rule out small codimension small degree subvarieties from being inside $X$. For example I was told that if $\dim(X)=5$, then $X$ cannot have a quadric threefold.
To indicate I have done some research, I've found this post, but have been unable to determine if this implies everything I needed it to.
Assume $X$ is a complete intersection in $\mathbb{P}^N$. Consider the restriction morphism $$ H^2(\mathbb{P}^N, \mathbb{Z}) \to H^2(X, \mathbb{Z}). $$ If $\dim(X) > 2$, it is an isomorphism by LHT. This means that the cohomology class of any divisor in $X$ is a multiple of the restriction of the hyperplane class, in particular, its degree (with respect to the embedding into $\mathbb{P}^N$) is a multiple of the degree of $X$. Thus, if $\deg(X) > 1$ (i.e., $X$ is not a linear subspace), the degree of any divisor is greater than $1$, hence such a divisor cannot be equal to $\mathbb{P}^{n-1}$ linearly embedded into $\mathbb{P}^N$.