I am working on a series of cryptography problems regarding the Pohlig-Hellman algorithm when I came across a notation I was not familiar with in regards to finite fields. In particular, I cannot discern the meaning of $F^*_p$ in regards to a finite field. The term is used elsewhere in the textbook, but not defined. The question in particular is listed below.

Let $F_p$ be a finite field and let $N | p − 1$. Prove that $F^∗_p$ has an element of order $N$. This is true in particular for any prime power that divides $p − 1$. (Hint. Use the fact that $F^*_p$ has a primitive root.)

To be clear, I am not looking for the solution to the problem, only clarification as to the likely meaning of $F^*_p$, and how it would relate to this particular question.

  • $\begingroup$ If you have a query about something in a textbook or paper, it's really helpful to say what the book or paper is: it helps the community to answer your question and it helps other readers of the book or paper who subsequently encounter the same problem as you and refer to MSE for assistance. $\endgroup$
    – Rob Arthan
    Oct 11, 2016 at 20:15

1 Answer 1


$\mathbb{F}_p^*$ denotes definitely the multiplicative group of $\mathbb{F}_p$. It has $p-1$ elements and it is cyclic, so it has a primitive root. Recall that a field $K$ has an additive group $(K,+)$ and a multiplicative group $(K^*,\cdot)$, where $K^*=K\setminus\{0\}$.

  • $\begingroup$ Excellent! So this notation would be equivalent to $(Z/pZ^*)$, as I am more familiar with. Makes sense. $\endgroup$ Oct 11, 2016 at 20:05
  • $\begingroup$ Almost correct - with $(\mathbb{Z}/p\mathbb{Z})^*$ :) $\endgroup$ Oct 11, 2016 at 20:06
  • $\begingroup$ That is what I meant, just a typo. $\endgroup$ Oct 11, 2016 at 20:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .