To analyze these statements, they all involve properties related to the order of groups and Lagrange’s Theorem. Let's go through them carefully:

1. "If $d \mid o(G)$ then $G$ has a subgroup of order $d$."

   • True, by Cauchy’s theorem and Lagrange’s theorem, if a divisor $d$ of the order of the group $o(G)$ exists, $G$ will have a subgroup of that order.

2. "If a prime $p \mid o(G)$, then $G$ has a subgroup of order $p$."

   • True, by Cauchy’s theorem, if $p$ is a prime divisor of the order of $G$, there is a subgroup of order $p$.

3. "If $p^r \mid o(G)$, $p$ is prime, $r$ is a positive integer, $G$ has a subgroup of order $p^r$."

   • True, by Sylow’s Theorem, there is a subgroup of order $p^r$ if $p^r$ divides $o(G)$.

4. "If $d \mid o(G)$ and $G$ is cyclic, then $G$ has a subgroup of order $d$."

   • True, if $G$ is cyclic, then every divisor of the group’s order corresponds to a unique subgroup of that order.

Conclusion:

The wrong statement here is none of the above. All four statements are correct based on group theory theorems like Cauchy’s theorem, Lagrange’s theorem, and Sylow's theorems.

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Let me know if you need further details on any of these theorems. Here are some relative questions:

1. What is Cauchy's theorem, and how does it apply to finite groups?
2. How does Sylow’s theorem provide insights into subgroups of prime power orders?
3. Why does the divisibility condition guarantee the existence of subgroups in cyclic groups?
4. Can there be exceptions to these rules in infinite groups?
5. How does Lagrange’s theorem limit the possible orders of subgroups?

Tip: Always remember that in cyclic groups, divisors of the group order directly correspond to the orders of subgroups.