Hypothetical DES weakness: dual encryption identical to single encryption with specific keys If there always existed a $k_3$ such that $\operatorname{DES}(k_2, \operatorname{DES}(k_1, M)) = \operatorname{DES}(k_3, M)$,  how would that affect the security of the DES block cipher?
I will take this to mean that it's possible to efficiently find a $k_3$ so that

$$E(k_2,E(k_1,M))=E(k_3,M)$$

which would mean the cipher $E$ has too much structure and given a keylength $n,$ a double encryption like above is as weak as a single encryption and brute force key search would take $O(2^n)$ encryptions at worst.

Mind you, a meet in the middle attack on double encryption for an arbitrary cipher $E$ can break it in time $O(2^n)$ and a few plaintext/ciphertext pairs $(M,C)$ if
memory of size $O(2^{(n+\log n)})$ is also available. First form a hash-sorted list of $$\{(k_2,E(k_1,M)):k_1 \in \{0,1\}^n\}$$ sorted on the second coordinate
and then keep computing $E^{-1}(k_2,C)$ as $k_2$ ranges over $\{0,1\}^n,$ and looking up to see if it is in the table.

If you find a hit, then try with another P/C pair to confirm the guess is not spurious. This is because the composite permutation may have fixed points. 

Edit:. I didn't originally want to make this too abstract, but as pointed out in the comments, what you're asking is whether $E$ forms a group, which it most certainly does not when $E$ is DES.