Sylow’s Theorem (Summary)

In this post we review Sylow’s theorem and as an example we solve the following problem.

Problem 64

Show that a group of order $200$ has a normal Sylow $5$-subgroup.
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Review of Sylow’s Theorem

One of the important theorems in group theory is Sylow’s theorem.

Sylow’s theorem is a very powerful tool to solve the classification problem of finite groups of a given order.

In this article, we review several terminologies, the contents of Sylow’s theorem, and its corollary.
We also give an example that can be solved using Sylow’s theorem.

At the end of this post, the links to various Sylow’s theorem problems are given.

We first introduce several definitions.

Definition 1.

Let $G$ be a group and $p$ be a prime number.

1. A group of order $p^{\alpha}$ for some non-negative integer $\alpha$ is called a $p$-group.
2. A subgroup of $G$ which is a $p$-subgroup is called $p$-subgroup.

Definition 2.

Let $G$ be a finite group of order $n$. Let $p$ be a prime number dividing $n$.
Write $n=p^{\alpha}m$, where $\alpha, m \in \Z$ and $p$ does not divide $m$.
Then any subgroup $H$ of $G$ is called a Sylow $p$-group of $G$ if the order of $H$ is $p^{\alpha}$.

 

Sylow’s theorem

Let $G$ be a finite group of order $p^{\alpha}m$, where the prime number $p$ does not divide $m$.

1. There exists at least one Sylow $p$-subgroup of $G$.
2. If $P$ is a Sylow $p$-subgroup of $G$ and $Q$ is any $p$-subgroup of $G$, then there exists $g \in G$ such that $Q$ is a subgroup of $gPg^{-1}$.
   In particular, any two Sylow $p$-subgroups of $G$ are conjugate in $G$.
3. The number $n_p$ of Sylow $p$-subgroups of $G$ is
   \[n_p \equiv 1 \pmod p.\] That is, $n_p=pk+1$ for some $k\in \Z$.
4. The number $n_p$ of Sylow $p$-subgroup of $G$ is the index of the normalizer $N_G(P)$ in $G$ for any Sylow $p$-subgroup $P$, hence $n_p$ divides $m$.

 

Corollary

In the notation of the previous theorem, if the number $n_p$ of Sylow $p$-subgroup of $G$ is $n_p=1$, then the Sylow $p$-subgroup is a normal subgroup of $G$.

 

Example/Problem.

Now as an example we solve the problem.

Problem.

Show that a group of order $200$ has a normal Sylow $5$-subgroup.

Solution.

We have the factorization $200=2^3\cdot 5^2$.

By Sylow’s theorem the number of Sylow $5$-subgroup satisfies $n_5 \equiv 1 \pmod 5$ and $n_5$ divides $8$.

The numbers satisfies $n_5 \equiv 1 \pmod 5$ are $n_5=1, 6, 11, \cdots$.
Among these numbers, only $1$ divides $8$.

Thus the only number satisfies both conditions is $1$.
Hence $n_5=1$ and there is only one Sylow $5$-subgroup.

Then by the corollary, the Sylow $5$-subgroup is normal.

More Problems on Sylow’s theorem

Sylow’s theorem is a handy tool to determine the group structure of a finite group.
We list here several problems/examples that can be solved using Sylow’s theorem.

All solutions are given in the links below.

• Sylow subgroups of a group of order $33$ is normal subgroups
• Group of order pq has a normal Sylow subgroup and solvable
• If the order is an even perfect number, then a group is not simple
• A group of order pqr contains a normal subgroup
• Groups of order 100, 200. Is it simple?
• If a Sylow subgroup is normal in a normal subgroup, it is a normal subgroup
• subgroup containing all p-Sylow subgroups of a group
• A group of order $20$ is solvable
• Non-abelian group of order $pq$ and its Sylow subgroups
• Prove that a Group of Order 217 is Cyclic and Find the Number of Generators
• Every Group of Order 20449 is an Abelian Group
• Every Sylow 11-Subgroup of a Group of Order 231 is Contained in the Center $Z(G)$
• Every Group of Order 72 is Not a Simple Group

Click here if solved 72
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