When I was reading a paper, I saw something like:
If $F$ and $E$ are Banach spaces with symmetric bases (precisely, they are symmetric sequence spaces), and $F$ is isomorphic to a complemented subspace of $l\_2\oplus E$, then $F=l\_2$ or $F$ is isomorphic to a complemented subspace of $E$. 


The author claimed that the result follows by standard elementary arguments. We omit the details. I don't know what the argument is. Any clue?



complemented subspace of the direct sum of two Banach spaces
This is well known but not trivial. It follows primarily from Theorem 2.c.13 of [LT]. The Theorem says (applied to this situation) if every operator $T:E\to \ell\_2$ is strictly singular, then every complemented subspace of $\ell\_2\oplus E$ is of the form (up to isomorphism) $X'\oplus E'$ for some complemented subspace $E'$ of $E$, and $X'$ is either finite dimensional or isomorphic to $\ell\_2$. So if $E'$ is finite dimensional then of course $F$ is isomorphic to $\ell\_2$. In other cases, since $F$ has symmetric basis, $F$ isomorphic to $\ell\_2\oplus E'$ implies, $F$ is either isomorphic to complemented subspace of $E'$ (and hence of $E$) or $\ell\_2$. For the last statement see Prop 3.b.8 of [LT].


If there is an operator $T:E\to \ell\_2$ which is not strictly singular, then, using the definition of not strictly singular and the fact that every subspace of $\ell\_2$ is complemented, you get a complemented subspace $E'$ of $E$ isomorphic to $\ell\_2$. But then $E\oplus \ell\_2$ is isomorphic to $E$ by decomposition method (using $E$ has symmetric basis). 


[LT] Lindenstaruss-Tzafriri, Classical Banach spaces vol 1.


