Encryption and decryption example using the Pohlig-Hellman Exponentiation Cipher Let $n=11, d=3, e=7$ and $M=3$.

Encryption:     

  M = BSK
  B -&gt; 1 , S -&gt; 18 -&gt; k -&gt; 10
  M1 =&gt; C1 = 1^7 mod 11 = 1
  M2 =&gt; C2 = 18^7 mod 11 = 6
  M3 =&gt; C3 = 10^7 mod 11 = 10

  C = 1 6 10 
  &lt;=&gt; 
  1=&gt;B, 6=&gt;G 10=&gt;K


Decryption:

  C1 =&gt; M1 = 1^3 mod 11 = 1
  C2 =&gt; M2 = 6^3 mod 11 = 18
  C3 =&gt; M3 = 10^3 mod 11 = 10

  M = BSK


I would like to know that I am correctly doing encryption and decryption?
Using the Pohlig-Hellman cipher, you encrypt your message $M = 3$ using an encryption key $(e,n) = (7,11)$. The encryption is $C = M^e \mod n = 3^7 \mod 11=9$. 

You decrypt with a decryption key $(d,n) = (3,11)$, such that $e\cdot d = 1 \mod (n-1)$. Since $3 \cdot 7 \mod (11-1) \equiv 21 \mod 10 \equiv 1$, this is satisfied.

So, to decrypt, compute $9^3 \mod 11 \equiv 3$, which was your $M$. You're done.

It looks like you are confusing the Pohlig-Hellman exponentiation cipher with the Pohlig-Hellman theorem used to solve a discrete logarithm problem. These are related problems.

You use the Pohlig-Hellman algorithm when you want to solve $g^x = h \mod p$, where $g,h$ are in a finite field $\mathbb{F}^*_p$, $g$ is a primitive root and $p$ is a prime. When $p-1$ factors into a bunch of small primes, you can solve several smaller discrete logarithm problems and reassemble $x$ from the solutions to those smaller problems via the Chinese remainder theorem.

Also, make sure you know why you're using this cipher in the first place.