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I am investigating on source code of a program called hostapd. And I am looking into Diffii-Hellman part of it.

/* RFC 3526, 3. Group 14 - 2048 Bit MODP
 * Generator: 2
 * Prime: 2^2048 - 2^1984 - 1 + 2^64 * { [2^1918 pi] + 124476 }
 */
static const u8 dh_group14_generator[1] = { 0x02 };
static const u8 dh_group14_prime[256] = {
  0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,
  ...
}; /* too long */
static const u8 dh_group14_order[256] = {
  0x7F, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF,
  ...
}; /* too long */

I understand that dh_group14_generator is a generator and dh_group14_prime is a prime for 2048 bit MODP from RFC 3526.

However, I can't find any clue of dh_group14_order. What does it stand for?

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Order here is not anything to do with how you might "order" a row of pencils in order of height (or some similar behaviour) -- but the number of elements in the group in question.

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    $\begingroup$ And, in this specific case, the order is the number of different values $g^i \bmod p$ can take on (for different values of $i$) $\endgroup$ – poncho Dec 6 '16 at 17:14
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$g$ is a generator of a multiplicative group ${Z_p}^*$ if ${Z_p}^*=\{g^i |1\leq i \leq p-1\}$ (every element of the group can be obtained as a power of $g$).
The order of the finite group ${Z_p}^*$is the number of its elements: in this case, $p-1$.

The order of an element $\alpha \in {Z_p}^*$ is the smallest positive integer $r$ that $\alpha^r\equiv 1(mod p) $.

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    $\begingroup$ Two problems: 1) the order of $\mathbb{Z}^*_p$ is $p-1$ (as 0 is not a member of that group), and 2) the question really was about the order of $g$ (which in this case is not a generator of the entire group) $\endgroup$ – poncho Dec 6 '16 at 19:23
  • $\begingroup$ 1) I correct immediately 2) I understood that the question was about the order of the multiplicative group, but I add the definition of the order of an element of the group. $\endgroup$ – M-elman Dec 6 '16 at 19:26
  • $\begingroup$ You might want to state that $p$ has to be a prime, otherwise all but the last statement are wrong. Since the order of the group is something different than the order of an element: You should state that the order of an element can only be a divisor of the group order. $\endgroup$ – tylo Dec 7 '16 at 12:19

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