A Group of Order $pqr$ Contains a Normal Subgroup of Order Either $p, q$, or $r$

Group Theory Problems and Solutions in Mathematics

Problem 81

Let $G$ be a group of order $|G|=pqr$, where $p,q,r$ are prime numbers such that $p<q<r$.

Show that $G$ has a normal subgroup of order either $p,q$ or $r$.
LoadingAdd to solve later

Sponsored Links

Hint.

Show that using Sylow’s theorem that $G$ has a normal Sylow subgroup of order either $p,q$, or $r$.

Review Sylow’s theorem (Especially (3) and (4) in the theorem).

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.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Sylow Subgroups of a Group of Order 33 is Normal SubgroupsSylow Subgroups of a Group of Order 33 is Normal Subgroups Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.   Hint. 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 $P$ […]
  • Group of Order 18 is SolvableGroup of Order 18 is Solvable Let $G$ be a finite group of order $18$. Show that the group $G$ is solvable.   Definition Recall that a group $G$ is said to be solvable if $G$ has a subnormal series \[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\] such […]
  • Group of Order $pq$ Has a Normal Sylow Subgroup and SolvableGroup of Order $pq$ Has a Normal Sylow Subgroup and Solvable Let $p, q$ be prime numbers such that $p>q$. If a group $G$ has order $pq$, then show the followings. (a) The group $G$ has a normal Sylow $p$-subgroup. (b) The group $G$ is solvable.   Definition/Hint For (a), apply Sylow's theorem. To review Sylow's theorem, […]
  • Are Groups of Order 100, 200 Simple?Are Groups of Order 100, 200 Simple? 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 […]
  • Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4 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. (See Sylow’s Theorem (Summary) for a review of Sylow's theorem.) Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is […]
  • A Group of Order $20$ is SolvableA Group of Order $20$ is Solvable Prove that a group of order $20$ is solvable.   Hint. Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow's theorem. See the post summary of Sylow’s Theorem to review Sylow's theorem. Proof. Let $G$ be a group of order $20$. The […]
  • If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal SubgroupIf a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup 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. It follows from […]
  • Non-Abelian Group of Order $pq$ and its Sylow SubgroupsNon-Abelian Group of Order $pq$ and its Sylow Subgroups Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$. Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.   Hint. Use Sylow's theorem. To review Sylow's theorem, check […]

You may also like...

6 Responses

  1. Xiaoqi Wei says:

    I think you may forget to count identity element.

    • Yu says:

      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.

      • Lowly Undergraduate says:

        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

  1. 12/19/2016

    […] A group of order pqr contains a normal subgroup […]

  2. 01/06/2017

    […] A group of order $pqr$ contains a normal subgroup of order either $p, q$, or $r$ […]

  3. 01/06/2017

    […] A group of order $pqr$ contains a normal subgroup of order either $p, q$, or $r$ […]

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Group Theory
Group Theory Problems and Solutions in Mathematics
If the Order is an Even Perfect Number, then a Group is not Simple

(a) Show that if a group $G$ has the following order, then it is not simple. $28$ $496$ $8128$ (b)...

Close