Given that $G$ is a group of order 52, we can determine the number of subgroups of order 13 using the Sylow theorems.

Step 1: Apply the First Sylow Theorem

The First Sylow Theorem states that for a prime $p$, there exists at least one Sylow $p$-subgroup of $G$, where $p^n$ divides the order of the group $|G|$.

Here, $|G| = 52 = 4 \times 13$, so the primes to consider are $p = 2$ and $p = 13$.

Step 2: Focus on Sylow $13$-Subgroups

We are interested in the subgroups of order 13. According to the Sylow theorems, the number of Sylow $p$-subgroups, denoted by $n_p$, must satisfy: $n_{13} \equiv 1 \pmod{13}$ $n_{13} \mid \frac{|G|}{13} = \frac{52}{13} = 4$

So, $n_{13}$ must be a divisor of 4 and satisfy $n_{13} \equiv 1 \pmod{13}$.

Step 3: Determine the Possible Values of $n_{13}$

The divisors of 4 are 1, 2, and 4. Now, we check which of these satisfy $n_{13} \equiv 1 \pmod{13}$:

• $1 \equiv 1 \pmod{13}$ — This is true.
• $2 \equiv 2 \pmod{13}$ — This is not true.
• $4 \equiv 4 \pmod{13}$ — This is not true.

Thus, $n_{13} = 1$.

Conclusion

There is exactly one Sylow 13-subgroup of $G$.

Do you want details or have any questions?

Here are some related questions to expand your understanding:

1. How can you use Sylow theorems to find the number of subgroups of a different prime order?
2. What is the significance of the Sylow theorems in group theory?
3. Can you determine how many Sylow 2-subgroups exist in $G$?
4. What is the order of the automorphism group of a cyclic group of order 13?
5. If a group has exactly one Sylow $p$-subgroup, what can you conclude about this subgroup?
6. How do Sylow subgroups relate to the structure of a group?
7. Could the group $G$ be simple? Why or why not?
8. What is the role of the center of a group in determining its subgroups?

Tip: When dealing with Sylow theorems, always check the divisibility condition and the modularity condition carefully, as they guide you to the correct number of subgroups.