TITLE: $A\oplus B\cong A\oplus C$ implies $B\cong C$?  (No, it does not)
QUESTION [5 upvotes]: I am asked to prove that for $p\in (1, \infty)$, $$L_{p}[0,1]\cong L_{p}[0,1]\oplus \ell_{2}$$ on a homework assignment, and I think I can show using results from class that $\ell_2\oplus \ell_2\cong \ell_2$.
From this I could say that 
$L_{p}[0,1]\oplus \ell_{2}\cong L_{p}[0,1]\oplus \ell_{2}\oplus\ell_{2}$
From here I feel like I should be able to conclude that 
$$L_{p}[0,1]\cong L_{p}[0,1]\oplus \ell_{2}$$
But I know of no such result that allows me to do this.  Can anyone tell me if it's true or false?
EDIT:  Definitely false.  (See counter example below from Arturo Magidin).
That is, if $B\oplus A\cong C\oplus A$ can I conclude that $B\cong C$?
Proper Solution (based on hints below from t.b.):
1) Prove that $\ell_2\oplus \ell_2\cong \ell_2$
2) Use the fact that $\ell_2$ is complemented in $L_p[0,1]$ to write $L_p[0,1] = \ell_2\oplus (\ell_2)^{c}$.
3) Then I combine these to obtain:
$L_p[0,1]\cong \ell_2\oplus (\ell_2)^{c}\cong (\ell_2\oplus \ell_2) \oplus (\ell_2)^{c} \cong \ell_2 \oplus (\ell_2\oplus (\ell_2)^{c})\cong \ell_2\oplus L_p[0,1]$.
I skipped some pieces of your more general argument.  I was just wondering if I did anything illegal, so to speak.

REPLY [3 votes]: Since Arturo explained why your idea doesn't work, here's a 
Hint:
Let $X = L^p$ and let $Y = L^p \oplus \ell^2$.
Prove: 

$X^2 \cong X$ and $Y^2 \cong Y$.
$X$ is isomorphic to a complemented subspace of $Y$ and vice versa (you probably know that $L^p$ has a complemented subspace isomorphic to $\ell^2$, see e.g. Proposition 6.4.2 in Albiac-Kalton or follow David Mitra's suggestion, Corollary 9.2 on  page 90 in Carother's A Short Course on Banach Space Theory).
Now you are in position to apply Pełczyński's argument to conclude that $X \cong Y$ (see point 3. of my question here for the details).