More concretely, if $R$ is a ring, and $M$ is a simple $R$-module, I want to show that any short exact sequence $0 \rightarrow L \rightarrow M \rightarrow N \rightarrow0$ splits. To this end, I have already proved the fact that for any simple $R$-module $M$, there exists a maximal left ideal $I \triangleleft R$ such that $M \cong_R R/I$. Now my question is this: could I in some way identify $L \oplus N$ with $R/I$? In that way, I could compose the isomorphism $\phi : M \rightarrow R/I$ with some isomorphism $h : L \oplus N \rightarrow R/I$ to split the sequence. I'm a bit worried that the short-exactness doesn't seem te get into play here, though.
Every short exact sequence with a simple module splits
abstract-algebraexact-sequencemodulesring-theory
Related Solutions
As I said in my comment above:
From the exactness at $F$, you have only used $\ker p\subseteq i(E)$. Presumably, the [next] step would be to use $i(E)\subseteq \ker p$.
So that's what I'll do. We want to show that $p|_S$ is surjective, so the most natural thing would be to take a $g\in G$ and then go from there.
Since $p$ is surjective, there is an $f\in F$ such that $p(f) = g$.
Let $f = f_i+f_S$ be the decomposition of $f$ in $i(E)\oplus S$, with $f_i \in i(E)$ and $f_S\in S$. Then by $i(E)\subseteq \ker p$, we get that $$g = p(f) = p(f_i+f_S) = p(f_i) + p(f_S) = p(f_S)$$ This shows that $p|_S$ is surjective.
To address your first comment: you can be less imprecise there by being more explicit about the isomorphism. In particular, if the sequence $$ 0 \to \ker \varphi \xrightarrow{i} M \xrightarrow{\varphi} N \to 0 $$ splits you have an isomorphism $f : M \to \ker \varphi \oplus N$ such that $f \circ i = \iota_1$ and $\pi_2 \circ f = \varphi$, where $\iota_1 : \ker \varphi \to \ker \varphi \oplus N$ and $\pi_2 : \ker \varphi \oplus N \to N$ are the corresponding inclusion and projection.
Then you have for $\iota_2 : N \to \ker \varphi \oplus N$ that $\pi_2 \circ i_2 = \mathrm{Id}_N$ and you define $r : N \to M$ by $r = f^{-1} \circ \iota_2$. Then $\varphi \circ r = \varphi \circ f^{-1} \circ \iota_2 = \pi_2 \circ \iota_2 = \mathrm{Id}_N$ so you have a right inverse.
For the converse, assume you have a right inverse $r : N \to M$. Note that a right inverse is injective, since $r(x) = r(y)$ implies $x = \varphi(r(x)) = \varphi(r(y)) = y$. Thus, we can identify $N$ with the submodule $\operatorname{im} r$ of $M$.
Our goal is to show that $M = \ker \varphi \oplus N$.
Now, to see how this can be done, let $m \in M$ . Then $n = \varphi(m)$ is an element of $N$ and since $r$ is a right inverse, for $m' = r(n)$ you have that $$\varphi(m - m') = \varphi(m) - (\varphi \circ r \circ \varphi)(m) = \varphi(m) - \varphi(m) = 0.$$
Thus $m - m' = i(k)$ for some unique $k \in \ker \varphi$ and we can write $m = i(k) + r(n)$.
Define $f : M \to \ker \varphi \oplus N$ by this decomposition, i.e. $$ f(m) = (k, \varphi(m)). $$
You can verify this is an isomorphism with inverse $g : \ker \varphi \oplus N$ given by $g(k, n) = i(k) + r(n)$
Now, since $\varphi \circ i = 0$ you have $(f \circ i)(k) = (k, 0)$, i.e. is the inclusion $\iota_1 : \ker \varphi \to \ker \varphi \oplus N$. Similarly for $\pi_2 : \ker \varphi \oplus N \to N$ you have $(\pi_2 \circ f)(m) = \varphi(m)$, so $\pi_2 \circ f = \varphi$.
Thus the exact sequence splits, with $f$ giving the required isomorphism between exact sequences.
Best Answer
Let $0 \rightarrow L \rightarrow M \rightarrow N \rightarrow0$ be any short exact sequence. Note that im$f$ is a submodule of $M$. As $f$ is injective, we have $L \cong \text{im}f \in \{\{0_M\}, M\}$ (since these are the only submodules of $M$). Consider the map $r_1 : M \rightarrow L$, $m \mapsto 0_L$. If $L \cong \{0_M\}$ (i.e. $L = \{0_L\}$), the for all $l \in L$ we have $r_1 \circ f(l) = r_1(f(l)) = 0_L = l$, so $r_1 \circ f$ = id$_L$ and $r_1$ is a retraction of $f$. Hence the sequence splits. If, on the other hand, im $f = M$, $f$ is bijective, so $r_2 := f^{-1}$ exists. Then $r_2 \circ f =$ id$_L$, so the sequence splits again.