# A Group of Order $20$ is Solvable ## Problem 286

Prove that a group of order $20$ is solvable. Add to solve later

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## Hint.

Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow’s theorem.

See the post summary of Sylow’s Theorem to review Sylow’s theorem.

## Proof.

Let $G$ be a group of order $20$. The prime factorization of $20$ is $20=2^2\cdot 5$.
Let $n_5$ be the number of $5$-Sylow subgroups of $G$.

By Sylow’s theorem, we have
$n_5\equiv 1 \pmod{5} \text{ and } n_5|4.$ It follows from these constraints that we have $n_5=1$.

Let $P$ be the unique $5$-Sylow subgroup of $G$.
The subgroup $P$ is normal in $G$ as it is the unique $5$-Sylow subgroup.

Then consider the subnormal series
$G\triangleright P \triangleright \{e\},$ where $e$ is the identity element of $G$.
Then the factor groups $G/P$, $P/\{e\}$ have order $4$ and $5$ respectively, and hence these are cyclic groups and in particular abelian.

Therefore the group $G$ of order $20$ has a subnormal series whose factor groups are abelian groups, and thus $G$ is a solvable group. Add to solve later

### 1 Response

1. 02/12/2017

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