Prove there is no simple group of order $729$

Let G be a group of order $729$.

$729 = 3^6$ so by Sylow’s Theorem G has a Sylow $3$-subgroup of order $729$. And there are $x$ of them. $r \equiv 1 \pmod 3$ and $r\ |\ 1$. So $x=1$? Is this correct?

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Firstly, Egreg’s answer is solid; I will use slightly different notation here that you may find easier to understand:

Let G be a group, with $|G| = 729 = 3^{6}$.
Let G act on itself under the conjugation action. We want to try and show that the centre of G, $Z(G)$, is a non-trivial normal subgroup of G. We already know that it is normal, so we just need to show that it is non-trivial.

First, note that all of the conjugacy classes $Cl_{G}(g)$ of G partition G, since they are either equal or disjoint. So, if there are $m\geq 1$ distinct conjugacy classes, then
\begin{align*}
G = Cl_{G}(g_{1}) \cup Cl_{G}(g_{2}) \cup \dots \cup Cl_{G}(g_{m}).
\end{align*}

Now, $g \in Z(G) \iff Cl_{G}(g) = \left \{ g \right \}$.

So, $Z(G) = \bigcup_{\left | Cl_{G}(g) \right |=1}Cl_{G}(g)$, which is the union of all conjugacy classes of size 1.

So, we can now say that there are $n$ distinct conjugacy classes of size at least 2, where $1 \leq n \leq m$, and
\begin{align*}
& G = Z(G) \cup Cl_{G}(g_{1}) \cup Cl_{G}(g_{2}) \cup \dots \cup Cl_{G}(g_{n}) \\
\Rightarrow & \left | G \right | = \left | Z(G) \right | + \left | Cl_{G}(g_{1}) \right | + \dots + \left | Cl_{G}(g_{n}) \right |.
\end{align*}

By the Orbit-Stabiliser theorem,
\begin{align*}
\left | Cl_{G}(g) \right | = \frac{\left | G \right |}{\left | C_{G}(g) \right |} \\
\Rightarrow \left | Cl_{G}(g) \right | \big| \left | G \right|, \forall g \in G.
\end{align*}
Since we now have that $\left | Cl_{G}(g) \right | \geq 2$, then $\left | Cl_{G}(g) \right | \in \{ 3, 3^{2}, 3^{3}, 3^{4}, 3^{5} \}$.

( Note that $\left | Cl_{G}(g) \right | < 3^{6} = \left| G \right|, \forall g \in G$, since $\left | Z(G) \right | \geq 1$ ).

For each of the n distinct conjugacy classes, if we denote the size of the $i^{th}$ class as $3^{x_{i}}$, then
\begin{align*}
\left | G \right | = \left | Z(G) \right | + 3^{x_{1}} + \dots + 3^{x_{n}} = 3^{6} \\
\iff \left | Z(G) \right | = 3^{6} – \left( 3^{x_{1}} + \dots + 3^{x_{n}} \right)
\end{align*}

Now,
\begin{align*}
3 \big| 3^{6} – \left( 3^{x_{1}} + \dots + 3^{x_{n}} \right) & \iff 3 \big| \left | Z(G) \right| \\
& \iff \left | Z(G) \right| \geq 3
\end{align*}
$\Rightarrow Z(G)$ is non trivial.

Thus, we have shown that $Z(G)$ is non-trivial, and we know that $Z(G)$ is a normal subgroup of $G$. If $Z(G) \neq G$, then $Z(G)$ is a proper non-trivial normal subgroup of $G$, so $G$ is not simple.

Suppose that $Z(G) = G$. Then $G$ is abelian. There is a theorem (this can be easily proved) stating that a simple abelian group is cyclic of prime order. If $G$ is simple, then $G$ is a simple abelian group, and thus cyclic of prime order. But $\lvert G \rvert = 3^{6}$, which is not prime. Thus, G is not simple.

Therefore, we can conclude that G cannot be simple.

This is a special case of a general theorem that says a finite $p$-group has non trivial center. A $p$-group is a group with order $p^n$, for a prime $p$.

Consider the equivalence relation $x\sim y$ if and only if there exists $g\in G$ such that $y=gxg^{-1}$.

The elements $x$ which satisfy $[x]_{\sim}=\{x\}$ are those in the center $Z(G)$, because this means $gx=xg$ for all $g\in G$. Now show that, for any $x$, $|[x]_{\sim}|$ divides $|G|$.

Hint: set $C_G(x)=\{g\in G: gx=xg\}$, show it is a subgroup and that
$$
|[x]_{\sim}|=\frac{|G|}{|C_G(x)|}
$$

Now apply this to your case: take a representative for each equivalence class with more than one element; say they are $x_1,x_2,\dots,x_r$ and set $|[x_i]_{\sim}|=p^{k_i}$, with $k_i>0$. Then
$$
|G|=|Z(G)|+p^{n_1}+\dots+p^{n_r}=p^n
$$
so $p$ divides $|Z(G)|$. Since $Z(G)$ is normal in $G$, in particular $G$ is not simple.

Let $G$ be a group of order $p^n$ where $p>2$ is a prime and $n\ge 2$. Then by Sylow’s First Theorem, $G$ has a normal subgroup of order $p^{n−1}$ (for a proof on MSE see here). Hence $G$ is not simple. We have $729=p^6$ with $p=3$.