The number of Sylow \( 5 \)-subgroups is \( 17 \).
Show Solution
The Correct Option isB
Solution and Explanation
Step 1: Understanding the Concept.
Here is a different way to see that $G$ must have a proper normal subgroup: instead of pinning down $n_5$ directly by congruence, count elements and show that the Sylow $5$-subgroup and the Sylow $7$-subgroup cannot both fail to be normal at the same time.
Step 2: Key Formula or Approach.
If a Sylow $p$-subgroup is not normal, then $n_p>1$, and since every Sylow $p$-subgroup here has prime order $p$, any two distinct ones intersect only in the identity. So all $n_p$ of them together contribute $n_p(p-1)$ distinct non-identity elements of order $p$ to $G$. If two different primes both give $n_p>1$, we can add up these counts and compare with $|G|$.
Step 3: Detailed Explanation.
As before, $595=5\times7\times17$. For the Sylow $5$-subgroups, $n_5$ divides $119$ and is $\equiv1\pmod5$; among the divisors $1,7,17,119$ of $119$, only $1$ and $119$ can possibly satisfy this (checking mod $5$ gives $1,2,2,4$, so really only $1$ works, but suppose for the sake of this alternate check that $n_5$ could be as large as $119$). For the Sylow $7$-subgroups, $n_7$ divides $85$ and is $\equiv1\pmod7$; the divisors of $85$ are $1,5,17,85$, and mod $7$ these are $1,5,3,1$, so $n_7\in\{1,85\}$.
Suppose, for contradiction, that both $n_5>1$ and $n_7>1$, so $n_5=119$ and $n_7=85$. Then the number of elements of order $5$ is $119\times(5-1)=476$, and the number of elements of order $7$ is $85\times(7-1)=510$. These elements are all distinct from each other and from the identity, so just these two counts already add up to $476+510=986$, which is more than $|G|=595$. That is impossible.
So $n_5$ and $n_7$ cannot both be greater than $1$ at the same time; at least one of the Sylow $5$-subgroup or the Sylow $7$-subgroup must be the unique one of its kind, and a unique Sylow subgroup is always normal. Either way, $G$ has a normal subgroup of order $5$ or $7$, and both are proper subgroups of a group of order $595$.
Step 4: Final Answer.
Counting elements rules out having both the Sylow $5$-subgroup and the Sylow $7$-subgroup non-normal at once, so $G$ is forced to have a proper normal subgroup.
\[ \boxed{\text{Option B: } G \text{ must have a proper normal subgroup}} \]
