Question

Let \( G \) be a group of order \( 595 \). Which one of the following is TRUE?

• A. \( G \) cannot have a proper normal subgroup.
• B. \( G \) must have a proper normal subgroup.
• C. \( G \) cannot have an element of order \( 17 \).
• D. The number of Sylow \( 5 \)-subgroups is \( 17 \).

Hint:

Factor 595 = 5 x 7 x 17 and check which Sylow subgroup count is forced to equal 1 by the congruence condition.

Answer 1

The Correct Option is B

Step 1: Understanding the Question.
We are given a group \( G \) of order \( 595 \) and need to decide which statement about its subgroups must be true. The first thing to do is factor \( 595 \) into primes.

Step 2: Key Formula or Approach.
We use Sylow's theorems. If \( n_p \) is the number of Sylow \( p \)-subgroups of a group of order \( n \), then \( n_p \) divides \( n/p^{a} \) (where \( p^{a} \) is the highest power of \( p \) dividing \( n \)), and \( n_p \equiv 1 \pmod{p} \). If \( n_p = 1 \), that unique Sylow \( p \)-subgroup is normal in \( G \). We also use Cauchy's theorem: if a prime \( p \) divides \( |G| \), then \( G \) has an element of order \( p \).

Step 3: Detailed Explanation.
Factor \( 595 = 5 \times 7 \times 17 \), three distinct primes, each to the first power.
Look at the Sylow \( 5 \)-subgroups. Here \( n_5 \) divides \( 7\times17=119 \) and \( n_5 \equiv 1 \pmod 5 \). The divisors of \( 119 \) are \( 1, 7, 17, 119 \). Checking each mod \( 5 \): \( 1\equiv1 \), \( 7\equiv2 \), \( 17\equiv2 \), \( 119\equiv4 \). Only \( 1 \) satisfies the congruence, so \( n_5 = 1 \) is forced.
A unique Sylow \( 5 \)-subgroup is automatically normal, and it has order \( 5 \), a proper subgroup of \( G \) since \( 5<595 \). So \( G \) always has a proper normal subgroup, no matter which group of order \( 595 \) we picked. This makes option (B) TRUE and option (A) false.
Since \( 17 \) divides \( |G|=595 \), Cauchy's theorem guarantees \( G \) has an element of order \( 17 \), so option (C) is false.
For option (D), check \( n_{17} \): it divides \( 5\times7=35 \) and satisfies \( n_{17}\equiv1\pmod{17} \). The divisors of \( 35 \) are \( 1,5,7,35 \); mod \( 17 \) these are \( 1,5,7,1 \), so \( n_{17}\in\{1,35\} \). Also \( 17 \) itself does not even divide \( 35 \), so \( n_{17} \) can never equal \( 17 \). And separately, we already showed \( n_5=1 \), not \( 17 \). So option (D) is false either way it is read.

Step 4: Final Answer.
The unique, hence normal, Sylow \( 5 \)-subgroup of order \( 5 \) is always a proper normal subgroup of \( G \).
\[ \boxed{\text{Option B: } G \text{ must have a proper normal subgroup}} \]