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 $r$.
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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.


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  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 $r$ […]

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