# A Group of Order $pqr$ Contains a Normal Subgroup of Order Either $p, q$, or $r$ ## 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 $r$. Add to solve later

## 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. Add to solve later

### 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

(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,

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 $r$ […]

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

###### More in Group Theory ##### 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