$\mathbb{Z}_8^*$ is not cyclic because no element has order 4 (= the group order) — every element squares to 1. $\mathbb{Z}_9^*$ is cyclic because element 2 has order 6 and generates the entire group.

$\mathbb{Z}_8^* = \{1, 3, 5, 7\}$, $|\mathbb{Z}_8^*| = \varphi(8) = 4$:

No element has order 4 → no generator exists → not cyclic.

$\mathbb{Z}_9^* = \{1, 2, 4, 5, 7, 8\}$, $|\mathbb{Z}_9^*| = \varphi(9) = 6$:

Element 2 generates the entire group → cyclic.

When is $\mathbb{Z}_n^*$ cyclic?

• $n = 1, 2, 4$ → always cyclic
• $n = p^k$ or $n = 2p^k$ (for odd prime $p$) → cyclic
• Otherwise → not cyclic (like $n = 8 = 2^3$, which fails because the exponent is > 1 for $p = 2$)

Cryptographic relevance: DH and ECC require cyclic groups. If your group isn't cyclic, the discrete logarithm problem doesn't work properly. This is why parameter selection matters.

Go deeper:

• Primitive root modulo n (Wikipedia) — exactly which ℤ_n* have a generator, and which don't.