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.

  • 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

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 $G$.

  • $\begingroup$ 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

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.


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