Sylow p-subgroup
Sylow p-subgroup
Sylow p-subgroup represents a pivotal concept in Group Theory. Introduced through the groundbreaking work of Ludwig Sylow, these subgroups underpin the study of Mathematics and Abstract Algebra. Their historical importance lies in their ability to classify finite groups efficiently. For more insight, refer to related Mathematics problems or Abstract Algebra problems.
What You Will Learn?
In this post, you will explore:
- Definition: Sylow p-subgroup
- Theorem-1: Let \((G,\circ)\) be a finite group. Then for each prime \(p\), \(G\) has a Sylow p-subgroup.
- Theorem-2: Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) be a prime and \(r,m\) be positive integers and \(p,m\) be relatively prime. If \(H\) is a subgroup of \(G\) of order \(p^{i}\) where \( 1\leq i \lt r \) then there exists a subgroup \(K\) of \(G\) of order \(p^{i+1}\) and \(H\) is normal in \(K\).
- Theorem-3: Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime. Let \(H\) be a subgroup of \(G\). Then \(H\) is a Sylow p-subgroup of \(G\) if and only if \(\big|H\big|=p^{r}\).
- Theorem-4: Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime. If \(H\) is a p-group then any conjugate of \(H\) is a p-group.
- Theorem-5: Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime. If \(H\) is a Sylow p-subgroup of \(G\) then any conjugate of \(H\) is a Sylow p-subgroup of \(G\).
- Theorem-6: Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime. If \(H\) is the only Sylow p-subgroup of \(G\) then \(H\) is normal in \(G\).
- Theorem-7: Let \((G,\circ)\) be a group and \(H\) be a subgroup of \(G\). If \(H\) and \(G/H\) are p-groups then \(G\) is p-group.
Things to Remember
- Set Theory
- Relations
- Mappings
- Group Theory
Introduction
The Sylow p-subgroup is a cornerstone in Mathematics and particularly in Abstract Algebra. These subgroups are characterized as maximal p-subgroups of a finite group and are crucial for solving problems in Group Theory. Explore related Mathematics problems for deeper understanding.
Sylow p-subgroup
Definition:
Let \((G,\circ)\) be a finite group and \(p\) be a prime. Then a subgroup of \(H\) of \(G\) is said to be a Sylow p-subgroup of \(G\) if \(H\) is a p-subgroup and is not properly contained in any other p-subgroup of \(G\) i.e., \(H\) is a maximal p-subgroup of \(G\).
Theorem-1
Statement:
Let \((G,\circ)\) be a finite group. Then for each prime \(p\), \(G\) has a Sylow p-subgroup.
Proof:
Given that \((G,\circ)\) is a finite group.
Let \(p\) be a prime.
- Case-1: Let \(\big|G\big|=1\)
Let \(e\) be the identity element of \(G\). Then the trivial subgroup \(\set{e}\) is the required Sylow p-subgroup of \(G\). - Case-2: Let \(p\cancel{\big|}~\big|G\big|\)
Then the trivial subgroup \(\set{e}\) is the required Sylow p-subgroup of \(G\). - Case-3: Let \(p\big|~\big|G\big|\)
Let \(\big|G\big|=p^{r}m\) where \(r,m\) are positive integers and \(p,m\) are relatively prime.
Then by Sylow’s theorem, \(G\) has subgroups \(H_{i}~i=0,1,2,..,r \) of order \(p^{i},~i=0,1,2,..,r \).
Implies \(H_{i}~i=0,1,2,..,r\) are p-subgroups of \(G\).
Therefore \(H_{r}\) is the required Sylow p-subgroup of \(G\).
Hence the theorem is proved.
Theorem-2
Statement:
Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) be a prime and \(r,m\) be positive integers and \(p,m\) be relatively prime. If \(H\) is a subgroup of \(G\) of order \(p^{i}\) where \( 1\leq i \lt r \) then there exists a subgroup \(K\) of \(G\) of order \(p^{i+1}\) and \(H\) is normal in \(K\).
Proof:
Given that \((G,\circ)\) is a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) be relatively prime.
Let \(H\) be a subgroup of \(G\) of order \(p^{i}\) where \( 1\leq i \lt r \).
To prove there exists a subgroup \(K\) of \(G\) of order \(p^{i+1}\) and \(H\) is normal in \(K\)
Now
\begin{align*}
& \big[G:H \big]=\frac{\big|G\big|}{\big|H\big|}\\
\implies & \big[G:H \big]=\frac{p^{r}m}{p^{i}}\\
\implies & \big[G:H \big]=p^{r-i}m\\
\implies & p\big|~\big[G:H \big]\\
\end{align*}
We have
\begin{align*}
&\big[N(H):H \big]\equiv\big[G:H \big]\big( mod~p\big) \\
\implies & p\big|~\big[N(H):H \big]-\big[G:H \big]\\
\implies & p\big|~\big[N(H):H \big]~\because p\big|~\big[G:H \big]\\
\implies & p\big|~\big[N(H)/H \big]~\because H \text{ is normal in }N(H)
\end{align*}
Then by Cauchy’s theorem, \(\exists\) a subgroup \(K/H\) of order \(p\) where \(H\) is normal in \(K\). Now
\begin{align*}
& \big|K/H\big|=p\\
\implies & \frac{\big|K\big|}{\big|H\big|}=p\\
\implies & \big|K\big| =p.\big|H\big|\\
\implies & \big|K\big| =p^{i+1}
\end{align*}
Hence \(K\) is the required subgroup of \(G\).
Theorem-3
Statement:
Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime. Let \(H\) be a subgroup of \(G\). Then \(H\) is a Sylow p-subgroup of \(G\) if and only if \(\big|H\big|=p^{r}\).
Proof:
Given that \((G,\circ)\) is a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) be relatively prime. And \(H\) is a subgroup of \(G\).
- Let \(H\) be a Sylow p-subgroup of \(G\).
To prove \(\big|H\big|=p^{r}\)
If possible let, \(\big|H\big|=p^{i}\) where \(1\leq i \lt r\).
Then \(\exists\) a subgroup \(K\) of \(G\) of order \(p^{i+1}\) and \(H\) is normal in \(G\).
A contradiction since \(H\) be a Sylow p-subgroup of \(G\).
Therefore our assumption is wrong.
Hence \(\big|H\big|=p^{r}\).
- Conversely, let \(\big|H\big|=p^{r}\).
To prove \(H\) is a Sylow p-subgroup of \(G\)
Since \(\big|G\big|=p^{r}m\) where \(p,m\) be relatively prime.
Then \(H\) is a maximal p-subgroup of \(G\).
Hence \(H\) is a Sylow p-subgroup of \(G\).
Theorem-4
Statement:
Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime. If \(H\) is a p-group then any conjugate of \(H\) is a p-group.
Proof:
Given that \((G,\circ)\) is a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) be relatively prime. And \(H\) is a subgroup of \(G\).
Let \(H\) be a p-group.
To prove any conjugate of \(H\) is a p-group.
Let \(a\in G\) then \(aHa^{-1}\) is a conjugate of \(H\). Let
\begin{align*}
& \alpha\in aHa^{-1} \\
\implies & \alpha=aha^{-1} \text{ for some }h\in H\\
\implies & \big|\alpha \big|=\big|h \big| \\
\end{align*}
Therefore \(\big|\alpha \big|\) is power of \(p\) since \(\big|h \big|\) is power of \(p\) since \(H\) is a p-group.
Since \(\alpha\) is an arbitrary element of \(aHa^{-1} \), then \(aHa^{-1} \) is a p-group.
Hence any conjugate of \(H\) is a p-group since \(aHa^{-1} \) is an arbitrary conjugate of \(H\).
Theorem-5
Statement:
Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime. If \(H\) is a Sylow p-subgroup of \(G\) then any conjugate of \(H\) is a Sylow p-subgroup of \(G\).
Proof:
Given that \((G,\circ)\) is a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) be relatively prime. And \(H\) is a subgroup of \(G\).
Let \(H\) be a Sylow p-subgroup of \(G\).
To prove any conjugate of \(H\) is a Sylow p-subgroup of \(G\).
Let \(a\in G\) then \(aHa^{-1}\) is a conjugate of \(H\). Then we have
\begin{align*}
& \big|aHa^{-1} \big|=\big|H \big|\\
\implies & \big|aHa^{-1} \big|=p^{r} ~\because \big|H \big|=p^{r}\\
\end{align*}
Therfore \(aHa^{-1}\) is a Sylow p-subgroup of \(G\).
Hence any conjugate of \(H\) is a Sylow p-subgroup of \(G\) since \(aHa^{-1} \) is an arbitrary conjugate of \(H\).
Theorem-6
Statement:
Let \((G,\circ)\) be a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime. If \(H\) is the only Sylow p-subgroup of \(G\) then \(H\) is normal in \(G\).
Proof:
Given that \((G,\circ)\) is a finite group of order \(p^{r}m\) where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) be relatively prime. And \(H\) is a subgroup of \(G\).
Let \(H\) be the only Sylow p-subgroup of \(G\).
To prove \(H\) is normal in \(G\).
Let \(a\in G\) then \(aHa^{-1}\) is a conjugate of \(H\). Then we have
\begin{align*}
& \big|aHa^{-1} \big|=\big|H \big|\\
\implies & \big|aHa^{-1} \big|=p^{r} ~\because \big|H \big|=p^{r}\\
\end{align*}
\(\therefore\) \(aHa^{-1}\) is a Sylow p-subgroup of \(G\).
\(\therefore\) any conjugate of \(H\) is a Sylow p-subgroup of \(G\) since \(aHa^{-1} \) is an arbitrary conjugate of \(H\).
Since \(H\) is the only Sylow p-subgroup of \(G\) then \(H=aHa^{-1}~\forall~a\in G\).
Hence \(H\) is normal in \(G\).
Theorem-7
Statement:
Let \((G,\circ)\) be a group and \(H\) be a subgroup of \(G\). If \(H\) and \(G/H\) are p-groups then \(G\) is p-group.
Proof:
Given that \((G,\circ)\) is a group and \(H\) is a subgroup of \(G\).
Let \(H\) and \(G/H\) be p-groups.
To prove \(G\) is p-group.
Let
\begin{align*}
& a\in G \\
\implies & aH\in G/H \\
\implies & \big|aH\big|=p^{k}~\text{ for some positive integer }k~\because G/H \text{ is a p-group} \\
\implies & \big(aH\big)^{p^{k}}=H \\
\implies & a^{p^{k}}H=H \\
\implies & a^{p^{k}}\in H \\
\implies & \big|a^{p^{k}}\big|=p^{m}~\text{ for some positive integer }m~\because H \text{ is a p-group} \\
\implies & \big(a^{p^{k}}\big)^{p^{m}}=e~\text{ where }e~\text{ is the identity element of }G \\
\implies & a^{p^{m+k}}=e\\
\end{align*}
Since \(p\) is a prime then \(\big|a\big| \) is power of \(p\).
Since \(a\) is an arbitrary element of \(G\) then the order of every element of \(G\) is a power of \(p\).
Hence \(G\) is p-group.
Applications
Group Actions are crucial in a wide range of applications across mathematics and science. In geometry, group actions help classify shapes and structures based on their symmetries. In physics, they are used to study conservation laws and quantum mechanics. Group actions also play a role in coding theory, providing solutions to problems in communication systems. For further study, explore Relations and Ring Theory.
Conclusion
In summary, Sylow p-subgroup serves as a fundamental tool in understanding finite groups within Abstract Algebra. Their applications in Mathematics and Group Theory ensure their continued relevance. Delve into Mathematics problems and Abstract Algebra problems for practical examples.
References
- Introduction to Group Theory by Benjamin Steinberg
- Topics in Group Theory by Geoffrey Smith
- Abstract Algebra by David S. Dummit and Richard M. Foote
- Algebra by Michael Artin
- Symmetry and Group Theory by Mark A. Armstrong
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FAQs
-
What is a Sylow p-subgroup?
Sylow p-subgroup is a maximal subgroup of a finite group where the order is a power of p, a prime number. Refer to Group Theory for details. -
Why is the Sylow p-subgroup significant?
The Sylow p-subgroup aids in understanding the structure of finite groups, crucial to Abstract Algebra. -
Who introduced Sylow p-subgroups?
Ludwig Sylow introduced Sylow p-subgroups, advancing the study of Mathematics. -
Where are Sylow p-subgroups applied?
Sylow p-subgroups are applied extensively in Group Theory and Mathematics problems. -
How do Sylow p-subgroups relate to finite groups?
Sylow p-subgroups help classify finite groups and their structures in Mathematics. -
What is the Sylow theorem?
The Sylow theorem establishes the existence and conjugacy of Sylow p-subgroups in finite groups. -
How are Sylow p-subgroups used in problem-solving?
Sylow p-subgroups simplify complex Mathematics problems and Abstract Algebra problems. -
What are the key properties of Sylow p-subgroups?
The Sylow p-subgroup is unique for a given p in some groups, enhancing studies in Group Theory. -
Are Sylow p-subgroups relevant today?
Yes, Sylow p-subgroups remain critical in contemporary Mathematics research. -
Can Sylow p-subgroups be linked to other mathematical concepts?
Yes, Sylow p-subgroups are foundational to exploring finite groups and other areas of Abstract Algebra.
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