# Group of Order 18 is Solvable ## Problem 118

Let $G$ be a finite group of order $18$.

Show that the group $G$ is solvable. Add to solve later

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## Definition

Recall that a group $G$ is said to be solvable if $G$ has a subnormal series
$\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G$ such that the factor groups $G_i/G_{i-1}$ are all abelian groups for $i=1,2,\dots, n$.

## Proof.

Since $18=2\cdot 3^2$, the number $n_3$ of Sylow $3$-subgroups is $1$ by the Sylow theorem.
(Sylow’s theorem implies that $n_3 \equiv 1 \pmod{3}$ and $n_3$ divides $2$.)
Hence the unique Sylow $3$-subgroup $P$ is a normal subgroup of $G$.

The order of $P$ is $9$, a square of a prime number, thus $P$ is abelian.
(See A group of order the square of a prime is abelian.)
Also, the order of the quotient group $G/P$ is $2$, thus $G/P$ is an abelian (cyclic) group.

Thus we have a filtration
$G \triangleright P \triangleright \{e\}$ whose factors $G/P, P/\{e\}$ are abelian groups, hence $G$ is solvable.

## Related Question.

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