user81940: One thing which I am wondering for a long time and to which I did not find an answer after doing a web search and hope to find an answer here.
When we construct the elliptic curve over a prime field why do we actually select a cyclic subgroup instead of taking the entire group of the elliptic curve?
On a side note the thing that confuses me most about this choice: We know that the cyclic subgroup of prime order p is isomorphic to Z/pZ and finding the isomorphism would mean solving the discrete log. 
Switching to a cyclic group seems actually rather like making the problem easier in comparison to staying with the full elliptic curve.

user208: The security of the discrete logarithm problem in a group is only as hard as that of the largest prime subgroup.
Because of this, there is no security gain from working in the larger group. However; it's worse. If you're working in the larger group you must make sure to not accidentally ending up in a (much smaller) subgroup when multiplying.

81940: <p>One thing which I am wondering for a long time and to which I did not find an answer after doing a web search and hope to find an answer here.</p>

<p>When we construct the elliptic curve over a prime field why do we actually select a cyclic subgroup instead of taking the entire group of the elliptic curve?</p>

<p>On a side note the thing that confuses me most about this choice: We know that the cyclic subgroup of prime order <code>p</code> is isomorphic to <code>Z/pZ</code> and finding the isomorphism would mean solving the discrete log. </p>

<p>Switching to a cyclic group seems actually rather like making the problem easier in comparison to staying with the full elliptic curve.</p>
 208: <p>The security of the discrete logarithm problem in a group is only as hard as that of the largest prime subgroup.</p>

<p>Because of this, there is no security gain from working in the larger group. However; it's worse. If you're working in the larger group you must make sure to not accidentally ending up in a (much smaller) subgroup when multiplying.</p>
