Why does there exist a non-split sequence with the condition that $\mathrm{pd} M = \infty$?

# Remarks.

- I am reading

wherein on p. 476 there is the

**Theorem.** Let $\Lambda$ be a left artinian ring and assume that each indecomposable finitely generated left $\Lambda$-module has finite projective dimension or finite injective dimension. Then $\Lambda$ has finite left global dimension.

*Proof of the Theorem*. Assume that $\Lambda$ is a left artinian ring of infinite global dimension where each indecomposable module has finite projective dimension or finite injective dimension. We shall show that this leads to a contradiction.
Clearly we may assume that no simple module has both infinite projective dimension and infinite injective dimension. Let $S$ be a simple module of infinite projective dimension. Recall that if
$$\cdots \longrightarrow P_{n} \longrightarrow P_{n-1} \longrightarrow \cdots
\longrightarrow P_{1}\longrightarrow P_{0} \longrightarrow S\longrightarrow 0$$
is a minimal projective resolution of S and T is a simple $\Lambda$-module, then $\mathrm{Ext}_{\Lambda}^{n}(S,T)\neq0$ if and only if the projective cover of $T$ is a direct summand of $P_{n}$. Since $S$ has infinite projective dimension and sup $\{{\rm id}\ Y$ $|$ $Y$ is simple
and of finite injective dimension$\}$ is finite, say equal to $n$, each simple module $T$ with $\mathrm{Ext}_{\Lambda}^{n+1}(S,T)\neq0$
is of infinite injective dimension. Now since $S$ has infinite projective dimension, there exists a direct summand M of $\Omega^{n}(S)$ and a nonsplit exact sequence $0 \longrightarrow T \longrightarrow E\longrightarrow M \longrightarrow 0$
with $\mathrm{pd} M=\infty$. Where $\mathrm{pd} M$ denotes the projective dimension of $M$.
${\rm id} Y$ denotes the injective dimension of $M$.
$ \Omega^{n}(S)$ is the $n$th syzygy of $S$.

I cannot understand that why there exists a direct summand $M$ of $ \Omega^{n}(S)$ and a nonsplit exact sequence $0 \longrightarrow T \longrightarrow E\longrightarrow M \longrightarrow 0$ with $\mathrm{pd} M=\infty$.

In fact, since $\mathrm{Ext}_{\Lambda}^{1}(\Omega^{n}(S),T)=\mathrm{Ext}_{\Lambda}^{n+1}(S,T)\neq0$, we know that there exists a direct summand $M$ of $ \Omega^{n}(S)$ such that $\mathrm{Ext}_{\Lambda}^{1}(M,T)\neq0$.But why $\mathrm{pd} M=\infty?$