I have a question from Stinson: 7.14. The question states:

Suppose that $p > 3$ is an odd prime, and $a,b$ is an element of $\mathbb Z_p$. Further, suppose that the equation $x^3 + ax + b$ is congruent to $0$ ($\mod p$) has 3 distinct roots in $\mathbb Z_p$. Prove that the corresponding elliptic curve group $(E, +)$ is not cyclic. Hint: Show that the points of order two generate a subgroup of $(E, +)$ that is isomorphic to $\mathbb Z_2 \times\mathbb Z_2$.

I've read through the elliptic curve material in the chapter preceding this question but I'm still a little hopeless. Can anyone explain or push me in the right direction on how to show/answer this?

I understand that points of order two (from the hint) follow the formula: $[2]P = \mathcal{O}$ but I'm not sure how this relates.


Can anyone explain or push me in the right direction on how to show/answer this?

The first obvious thing to ponder is "what does it mean that a 'group is cyclic'? What properties does a cyclic group have, that noncyclic groups do not?"

(Broader hint: a cyclic group has a generator; noncyclic (finite) groups do not)

I understand that points of order two (from the hint) follow the formula: 2P = 0 but I'm not sure how this relates.

It relates quite a bit, actually. Can a cyclic group have two different elements of order 2?

Update: since Dexter has solved the problem, I'll lay out the proof I was originally trying to suggest.

One fundamental property of all finite groups is that, if $n$ is the number of group elements, we have $xP = (x \bmod n)P$ (for any integer $x$ and any element $P$).

In addition, a finite group is cyclic if it has a generator, that is, an element $G$ for which all elements can be expressed as $xG$ (for some $x$). In addition, for any $P$, we always have $xG = P$ for some $0 \le x < n$. In particular, $xG \ne 0$ for any $0 < x < n$.

Now, for the group in question, we have two distinct elements $P_1$ and $P_2$, both of order two (actually, we have three, we only need two). That is, $P_1 \ne P_2 \ne 0$, and $2P_1 = 2P_2 = 0$.

Now, suppose the group is cyclic; if so, we have a generator $G$. We know $P_1$ can be expressed as $P_1 = x_1G$ for some $0 \le x_1 < n$. We also have $2P_1 = 0$, that is, $2 (x_1 G) = (2 x_1)G = 0$. We know that $(2x_1)G = (2x_1 \bmod n)G$, and if that is the neutral element, we must have $2x_1 \bmod n = 0$. The only value $0 < x_1 < n$ that can satisfy that is $x_1 = n/2$ (which implies that $n$ must be even).

However, by the same reasoning, if we have $P_2 = x_2G$, we have $x_2 = n/2$, but that would mean that $P_1 = (n/2)G = P_2$, and so the two points of order two were the same, but we know they aren't.

Hence, there cannot be a generator, and hence the group cannot be cyclic.

This proof may sound a bit lengthy, however, that's mostly because I spelled out the logic quite explicitly, listing the various assumed properties of groups. Once you get more acquainted with group theory, this logic would be more natural.

  • $\begingroup$ So a group is cyclic if it contains an element that, when raised to powers, calculates all of the other elements of that group... So I have to show that the group (E, +) is not cyclic by showing it has no generators? $\endgroup$ – Dexter Mar 24 '20 at 21:25
  • $\begingroup$ @Dexter: that would work - how could you use the existence of two elements of order two to show that there cannot be a generator? $\endgroup$ – poncho Mar 24 '20 at 21:26
  • $\begingroup$ What is the group (E, +)? $\endgroup$ – Dexter Mar 24 '20 at 21:26
  • $\begingroup$ Well I think that if two elements have both order two, would they have to equal because they both equal 0? Suppose 2P = 0 and 2Q = 0, that means 2P = 2Q => P = Q, but this can't be true correct? $\endgroup$ – Dexter Mar 24 '20 at 21:33
  • $\begingroup$ The question says that the elliptic curve is mod p, which is an odd prime. If this is true, then the possible cycle length of generators is either 1 or of length p... Wouldn't this prove that the group is cyclic instead of not cyclic? $\endgroup$ – Dexter Mar 24 '20 at 21:35

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