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I was reading our university slide on the Diffie-Hellman where it is mentioned that one of the disadvantages of D-H is that For large prime, $p-1$ is an even number so, $\mathbb{Z}^*_p$ will have a subgroup of order 2

What does this statement mean?

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Let $p$ be a prime number and denote by $\mathbb{Z}_p$ the ring of integers modulo $p$. The set $\mathbb{Z}_p^*$ typically denotes the units of $\mathbb{Z}_p$. This set is defined as $$\mathbb{Z}_p^* = \{n \in \mathbb{Z}_p : \exists m \in \mathbb{Z}_p, nm = 1\},$$ where arithmetic is done modulo $p$. An important fact about $\mathbb{Z}_p$ is that it is a field. This means that we can simplify the above to $\mathbb{Z}_p^* = \{1, 2, \ldots, p-1\}$. This set forms a group under multiplication, and it is a standard fact that this is a cyclic group, i.e., it is isomorphic to $\mathbb{Z}_{p-1}$ where the group operation is addition. In symbols, $(\mathbb{Z}_p^*,\times) \cong (\mathbb{Z}_{p-1},+)$.

Since $p-1$ is even, consider the two-element subset of $\mathbb{Z}_{p-1}$ given by $\left\{0,\frac{p-1}{2}\right\}$. As $\frac{p-1}{2} + \frac{p-1}{2} = p-1 = 0$ in $\mathbb{Z}_{p-1}$, this set forms a subgroup of $\mathbb{Z}_{p-1}$. The order of a subgroup is its cardinality as a set, so this is a subgroup of order two.

Via the isomorphism between $(\mathbb{Z}_{p-1},+)$ and $(\mathbb{Z}_p^*,\times)$, this corresponds to a subgroup of $\mathbb{Z}^*_p$ of order two.

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    $\begingroup$ On the other hand, if we restrict ourselves to even primes, there are not that many choices... $\endgroup$
    – gnasher729
    Oct 22, 2019 at 19:35
  • $\begingroup$ The condition is that $p-1$ is even, not $p$. $\endgroup$ Oct 22, 2019 at 20:21

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