Given a simple group $G$ of order 60, we want to find the number of Sylow 3-subgroups and Sylow 5-subgroups, and also understand why these numbers must take on specific values.

Step 1: Apply Sylow's Theorems

1. Sylow 3-Subgroups

The order of a Sylow 3-subgroup is $3^1 = 3$. The number of Sylow 3-subgroups, $n_3$, must satisfy:

1. $n_3$ divides $\frac{60}{3} = 20$.
2. $n_3 \equiv 1 \pmod{3}$.

The divisors of 20 are $1, 2, 4, 5, 10, 20$.

Checking the congruences modulo 3:

• $1 \equiv 1 \pmod{3}$
• $2 \equiv 2 \pmod{3}$
• $4 \equiv 1 \pmod{3}$
• $5 \equiv 2 \pmod{3}$
• $10 \equiv 1 \pmod{3}$
• $20 \equiv 2 \pmod{3}$

Thus, $n_3$ could be 1, 4, or 10.

However, if $n_3 = 1$, then the Sylow 3-subgroup would be normal in $G$ because a unique Sylow subgroup is always normal. This would contradict the simplicity of $G$ (a simple group has no nontrivial normal subgroups). Therefore, $n_3$ cannot be 1, and $n_3$ must be either 4 or 10.

2. Sylow 5-Subgroups

The order of a Sylow 5-subgroup is $5^1 = 5$. The number of Sylow 5-subgroups, $n_5$, must satisfy:

1. $n_5$ divides $\frac{60}{5} = 12$.
2. $n_5 \equiv 1 \pmod{5}$.

The divisors of 12 are $1, 2, 3, 4, 6, 12$.

Checking the congruences modulo 5:

• $1 \equiv 1 \pmod{5}$
• $2 \equiv 2 \pmod{5}$
• $3 \equiv 3 \pmod{5}$
• $4 \equiv 4 \pmod{5}$
• $6 \equiv 1 \pmod{5}$
• $12 \equiv 2 \pmod{5}$

Thus, $n_5$ could be 1 or 6.

Again, if $n_5 = 1$, the Sylow 5-subgroup would be normal in $G$, contradicting the simplicity of $G$. Therefore, $n_5$ cannot be 1, and $n_5$ must be 6.

Conclusion

• The number of Sylow 3-subgroups $n_3$ must be either 4 or 10.
• The number of Sylow 5-subgroups $n_5$ must be 6.

Why These Values?

• The restriction that $G$ is simple forces $n_3$ and $n_5$ to be greater than 1 because otherwise, the corresponding Sylow subgroup would be normal, violating the simplicity of $G$.

• The numbers $n_3 = 4$ or $10$ and $n_5 = 6$ are the smallest possible values that satisfy the divisibility and congruence conditions imposed by Sylow's theorems, while also respecting the simplicity of $G$.

Would you like more details or have any questions? Here are some related questions to consider:

1. How can we decide between $n_3 = 4$ and $n_3 = 10$?
2. What implications do these Sylow subgroup numbers have for the structure of $G$?
3. Could $n_3$ ever be equal to $n_5$ in a simple group?
4. How does the existence of Sylow subgroups help in classifying finite groups?
5. What are the characteristics of normal Sylow subgroups?
6. What can be inferred about the simplicity of $G$ if $n_3$ or $n_5$ equals 1?
7. How can the simplicity of $G$ be used to deduce more about the group's structure?
8. Are there examples of simple groups of order 60 in classical group theory?

Tip: Understanding why certain numbers of Sylow subgroups exist in a group gives deep insight into the group's overall structure, especially in distinguishing simple groups.