The group $G$ has a normal Sylow $p$-subgroup if and only if the number $n_p$ of Sylow $p$-subgroup is $1$.

Proof.

We show that the group $G$ has a normal Sylow subgroup of order either $p,q$, or $r$.
Let $n_p$ be the number of the Sylow $p$-subgroups of $G$ and similarly define $n_q$ and $n_r$.

Seeking a contradiction, suppose that $G$ has no normal Sylow subgroups.
This is equivalent to saying that $n_p>1$, $n_q>1$, and $n_r>1$.

Sylow’s theorem yields that $n_r\equiv 1 \pmod r$ and $n_r|pq$. Since $n_r>1$ and $r>p,q$, we must have $n_r=pq$.

Also Sylow’s theorem implies that $n_q \equiv 1 \pmod q$ and $n_q|pr$. Since $n_q>1$ and $q>p$, we must have $n_q\geq r$.
We also have $n_p\equiv 1 \pmod p$ and $n_p|qr$ by Sylow’s theorem and $n_p>1$, we must have $n_p\geq q$.

Each Sylow $r$-subgroup contains $r-1$ elements of order $r$. Since distinct Sylow $r$-subgroups intersect trivially, there are $(r-1)n_r=(r-1)pq$ elements of order $r$ in $G$.

By the similar argument, the number of elements of order either $p, q$, or $r$ is greater than or equal to
\[(r-1)pq+(q-1)r+(p-1)q=pqr+qr-r-q.\]
Of course, this number must be less than or equal to $|G|=pqr$.

Hence $qr-r-q\leq 0$. This implies that
\[2 < q \leq \frac{r}{r-1}\leq 2\] and this is a contradiction.

Sylow Subgroups of a Group of Order 33 is Normal Subgroups
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We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]

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Let $G$ be a finite group of order $18$.
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Definition
Recall that a group $G$ is said to be solvable if $G$ has a subnormal series
\[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\]
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Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
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(b) The group $G$ is solvable.
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Determine whether a group $G$ of the following order is simple or not.
(a) $|G|=100$.
(b) $|G|=200$.
Hint.
Use Sylow's theorem and determine the number of $5$-Sylow subgroup of the group $G$.
Check out the post Sylow’s Theorem (summary) for a review of Sylow's […]

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Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.
Hint.
Use Sylow's theorem.
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Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow's theorem.
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Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
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Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
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I made the same observation. I believe he means the line

(r−1)pq + (q−1)r + (p−1)q = pqr + qr − r − q

should instead be

(r−1)pq + (q−1)r + (p−1)q + 1 = pqr + qr − r − q + 1.

Indeed, I agree you may omit the identity element when counting the elements of the Sylow p-subgroups, but in the end, I think you need to add it back in. This then gives us q=3, so p=2 and r=5. The result is straightforward from here.

I think you may forget to count identity element.

Dear Xiaoqi Wei,

I think I don’t have to count the identity element because we need to count different elements. Please let me know if I didn’t get your point.

I made the same observation. I believe he means the line

(r−1)pq + (q−1)r + (p−1)q = pqr + qr − r − q

should instead be

(r−1)pq + (q−1)r + (p−1)q + 1 = pqr + qr − r − q + 1.

Indeed, I agree you may omit the identity element when counting the elements of the Sylow p-subgroups, but in the end, I think you need to add it back in. This then gives us q=3, so p=2 and r=5. The result is straightforward from here.

Please correct me if I misunderstand.

Sincerely,

a lowly undergraduate