monoalphabetic substitution perfect secrecy I have this question that I'm trying to answer:


  Consider a monoalphabetic substitution cipher applied to plaintexts
   consisting of just a single letter. That is, $$M = \{0, 1, . . . ,
 25\}$$ The keyspace $K$ is the set of all permutations of $M$. (As
   usual, we identify the letter $A, . . . , Z$ with this set of
   integers.) Show that this is perfectly secret.


I understand that to show perfect secrecy it suffices to show that $|K|\ge |M|$. I'm just not understanding how to answer this question. Would it make sense to say $K = \{0, 1, ..., 25\}\Rightarrow |K|\ge |M|$?
IF $K$ is randomly chosen (and well distributed over the key space with size 26! - the total number of possible permutations) then each $C$ (ciphertext letter) is also equally likely. An attacker may choose any $K$ but it is impossible to gain any advantage by that; each message $M$ is still as likely as before.

Basically this scheme is a one-time-pad using a per-character permutation rather than XOR as often used.



As indicated by the first IF, what's missing is the requirement that the choice of $K$ is perfectly random. Generally we try and proof ciphers correct by looking at the set $\operatorname{Gen}, \operatorname{E}, \operatorname{D}$. The $\operatorname{Gen}$ is missing from the question.