3
$\begingroup$

Prove that every group of order $567$ has a normal subgroup of order $27$.

Let $G$ be such a group. Then $|G| = 3^4\cdot7.$ Let $H\in\text{Syl}_3(G).$ From the Sylow theorems, we have that $n_3 | 7, $ $n_3\equiv 1\pmod{3}.$ This leaves us with either $n_3=1$ or $n_3=7$. If it's the former, then we're done since $H\triangleleft G$ and by Sylow, there exists $N_1<H$ of order $3^3=27$. And since $H\triangleleft G$ and $N_1\subset H$, we have $N_1\triangleleft G$.

Now if $n_3=7$, this seems trickier. For some $g\in G$, let $$N = \bigcap_{P\in\text{Syl}_3(G)} gPg^{-1}.$$ I know that $N\triangleleft G$, and for any normal $3$-subgroup $K$ of $G$, $K\subset N.$ So it must be that $|N|=27$ since $|N|=81$ is the case we already considered. But how would you show that the intersection has order $27$?

If there are $27$ elements in $N$, then there are $54$ elements in $P\setminus N$ for each $p\in \text{Syl}_3(G)$. And the $7$-Sylow subgroup intersects trivially with any of the $3$-Sylow subgroups. So that would mean there are

$$6+7(3^4-27)+27 = 411$$ elements in the group, which doesn't add up.

$\endgroup$
1
  • 3
    $\begingroup$ In your first paragraph you say "since $H\triangleleft G$ and $N_1\subset H$, we have $N_1\triangleleft G$". I don't see this. It's possible that if you conjugate something from $N_1$ by something from $G$, you get an element in $H$ that is outside of $N_1$. $\endgroup$ – alex.jordan Jul 26 '16 at 17:55
1
$\begingroup$

A different way (from Alex Jordan's +1 answer) of looking at this problem in the case $n_3=7$.

I also begin by observing that the group has a unique, hence normal, Sylow $7$-subgroup $P$ because the residue class of $3$ is of order six modulo 7.

Let us fix a generator $x$ of $P$, and set as our goal to figure out the size of the conjugacy class $[x]\subset G$ of $x$. All the conjugates of $x$ in $G$ have order seven, so they must be elements of $P$. Therefore $x$ has at most six conjugates in $G$. But the number of conjugates is a factor of $|G|$; it is equal to the index $[G:C]$ of the centralizer $C=C_G(x)$. Hence we can conclude that $|[x]|$ must be either $1$ or $3$.

Let $P'$ be any of the Sylow $3$-subgroups. Because $[G:N_G(P')]=n_3=7$ we see that $P'$ is its own normalizer in $G$. In particular it cannot be normalized by $x$. Hence the element $x$ cannot be in the center of $G$, and we can deduce that $x$ has exactly three conjugates. Therefore $|C|=567/3=3^3\cdot7$.

Let $Q$ be a Sylow $3$-subgroup of $C$. It has $27$ elements, so it is a subgroup of some Sylow $3$-subgroup $P''$ of $G$. Because $[P'':Q]=3$ is the smallest prime factor of $|P''|$ we deduce that $Q\unlhd P''$ (this is the standard fact from the theory of $p$-groups that maximal subgroups are normal). But, $Q$ is also normalized by $x$. Therefore $N_G(Q)$ has order that is divisible by both $3^4$ and by $7$. Hence $Q\unlhd G$ and we are done.

From the above facts it also follows easily that $Q$ is the intersection of all the Sylow $3$-subgroups of $G$.

$\endgroup$
5
  • $\begingroup$ For extra credit: Show that $[x]=\{x,x^2,x^4\}$. $\endgroup$ – Jyrki Lahtonen Jul 31 '16 at 14:15
  • $\begingroup$ From what fact does $x\in N_G(Q)$ follow? $\endgroup$ – user346096 Aug 1 '16 at 16:01
  • $\begingroup$ @user346096 $x$ centralizes $Q$, so it also normalizes $Q$. $\endgroup$ – Jyrki Lahtonen Aug 1 '16 at 19:10
  • $\begingroup$ And that is because $Q<C$? $\endgroup$ – user346096 Aug 1 '16 at 19:19
  • $\begingroup$ Correct, @user346096! $\endgroup$ – Jyrki Lahtonen Aug 1 '16 at 19:27
3
$\begingroup$

Note that there is exactly one Sylow-7 subgroup when you consider the powers of $3$ mod $7$.

If there is only one Sylow-3 subgroup, then $G$ is the direct product of a group of order $3^4$ and a group of order $7$. In this case the question boils down to whether the group of order $3^4$ has a normal subgroup of order $3^3$. Any $p$-group has a normal subgroup of order $p^k$ for each $p^k$ dividing the order of the group. (See here for example.)

So assume that there are seven Sylow-3 subgroups. Then $G$ permutes these by conjugation, so there is a homomorphism from $G$ to $\mathrm{Sym}_7$. The image of this homomorphism has an order divisible by $7$ since that is the size of the orbit. So the kernel has order some power of $3$. It's not $3^0$ or $3^1$, since then $3^4$ or $3^3$ (respectively) would have to divide $7!$.

If the kernel has order $3^4$, then it's a normal Sylow-3 subgroup, so it's the only Sylow-3 subgroup, a case already covered.

If the kernel has order $3^2$, then the image of this homomorphism is a subgroup $K$ of $\text{Sym}_7$ with order $3^27=63$. There is no such subgroup of $\mathrm{Sym}_7$. You can catalog all four groups of order $63$ using Sylow theorems. Three of them have elements of order $9$ or $21$, which is not possible within $\mathrm{Sym}_7$ because the order of an element is the lcm of the orders of disjoint cycles. That leaves the group $C_3\times(C_3\ltimes C_7)$. This cannot be a subgroup of $\mathrm{Sym}_7$ either because the elements of order $7$ from $(C_3\ltimes C_7)$ are $7$-cycles, and don't commute with other elements from $\mathrm{Sym}_7$ such as the ones from the lone $C_3$.

So the kernel must have order $3^3$. And any kernel of a homomorphism is normal.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.