January 17th, 2018, 06:55 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Sylow's Theorem
I have gone through some videos & notes about Sylow's theorem but I am not able to understand it exactly. Kindly someone explain it in simple terms : The theorem statements & proof. Thank you so much ðŸ˜Š

January 17th, 2018, 01:21 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,510 Thanks: 584 
From Wikipedia "In mathematics, specifically in the field of finite group theory, the Sylow theorems are a collection of theorems named after the Norwegian mathematician Ludwig Sylow (1872) that give detailed information about the number of subgroups of fixed order that a given finite group contains. The Sylow theorems form a fundamental part of finite group theory and have very important applications in the classification of finite simple groups." I know nothing about the subject, but it seems you need to be more specific. 
January 17th, 2018, 05:54 PM  #3  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
I am looking for theorems & proof explanation  
January 17th, 2018, 07:14 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,951 Thanks: 1599  
January 18th, 2018, 02:51 AM  #5 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 186 Thanks: 55 Math Focus: Algebraic Number Theory, Arithmetic Geometry 
I'll state the theorems for you, and give some very brief/basic motivation for them. The proofs are relatively long, and certainly longer than I could bear typing up for this, but I expect you could find them in most basic algebra or group theory books. To study finite groups, it can be very helpful to understand how their subgroups behave. This itself can be quite hard to do: if I just told you the order of a group, it's hard for you to say much about its subgroups in general. However, Sylow's theorems allow us to deduce a fair amount of information at least about the subgroups of prime power order. For some basic examples (which I'll explain after stating the theorems), this is all the information we need. If you haven't seen Cayley's theorem, I'd recommend looking at that first. It says: if $G$ is a finite group whose order is divisible by a prime $p$, then $G$ has a subgroup of order $p$. We're interested in looking at $p$subgroups of a group $G$ (i.e. subgroups whose order is a power of $p$), and "Sylow $p$subgroups" (subgroups whose order is the largest power of p dividing $G$). To be concrete, if $G$ has order $7203 = 3^1 * 7^4$, then a Sylow $7$subgroup is a subgroup of order $7^4$, a Sylow $3$subgroup is a subgroup of order $3^1 = 3$, and a Sylow $p$subgroup for any other prime $p$ is a group of order $p^0 = 1$ (i.e. the trivial subgroup). Some notation: I'll write $S_p(G)$ to denote the number of Sylow $p$subgroups of $G$. Let $G$ be a finite group and $p$ be a prime. Then Sylow's 1st Theorem: $G$ has at least one Sylow $p$subgroup. In other words, $S_p(G) > 0$. (Note: for the primes $p$ not dividing $G$, this particular theorem doesn't say anything useful  it just says $G$ has a subgroup of order 1. If $p$ divides $G$ but $p^2$ doesn't divide $G$, then this tells us nothing more than Cayley's theorem. It is most interesting when a larger power of $p$ divides $G$  it then gives us new information.) Sylow's 2nd Theorem $S_p(G) \equiv 1 \bmod p$ (Notice that this implies the first theorem. However, it turns out to be easier to first prove the 1st theorem and then use it to prove the 2nd compared to trying to prove the 2nd theorem from scratch.) Sylow's 3rd Theorem Every $p$subgroup of $G$ is contained in a Sylow $p$subgroup. Sylow's 4th Theorem The Sylow $p$subgroups form a single conjugacy class of subgroups. That is, if $P$ and $Q$ are Sylow $p$subgroups, then there is a $g \in G$ such that $Q = gPg^{1}$. Corollary $S_p(G)$ divides $G$. ________ Let's look a basic application. What are the groups (up to isomorphism) of order $15$? More generally, what are the groups of order $pq$ for any primes $q > p$? Let $G$ be a group of order $15$. Let's consider the Sylow $3$subgroups and Sylow $5$subgroups (i.e. the subgroups of order $3$ and $5$, respectively). By Sylow's 2nd theorem, $S_5(G) \equiv 1 \bmod 5$ so $S_5(G) = 1, 6$ or $11$. By the corollary to Sylow's 4th theorem, $S_5(G)$ divides $15$ so we must have $S_5(G) = 1$. This means there are precisely four elements of order $5$ in $G$  there must be at least four (as the subgroup of order $5$ must have four elements of order $5$) and there can't be any more, else they'd generate another Sylow $5$subgroup. Similarly, $S_3(G) \equiv 1 \bmod 3$ and $S_3(G) \, \, 15$ implies $S_3(G) = 1$ and there are exactly two elements of order $3$ in $G$. Now note that there are $15$ elements of $G$, and each one's order must divide $15$, i.e. each one's order must be $1$, $3$, $5$ or $15$. But there's just one element of order $1$, and we've seen that there are two and four elements of order $3$ and $5$, respectively. This only accounts for seven elements of $G$  the remaining eight must then have order $15$. In particular, $G$ has an element of order $15$ so is cyclic. This shows the only group of order $15$ is the cyclic one. Turning to the more general case (groups of order $pq$, where $q > p$ are primes), pretty much the same argument works if $q \not \equiv 1 \bmod p$, and we find that the cyclic group of order $pq$ is the only group of order $pq$. However, if $q \equiv 1 \bmod p$, it turns out there are two groups of order $pq$. If you're familiar with semidirect products of groups, Sylow's theorems make this very easy to prove, and I'll leave it as an exercise for you! 
February 27th, 2018, 05:22 AM  #6  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
 
February 27th, 2018, 05:33 AM  #7  
Senior Member Joined: Oct 2009 Posts: 400 Thanks: 138  Quote:
Again I deeply apologize if I am being mean in any way, but for years I have always made sure that math students had the right prerequisite knowledge for the right things, and to detect and rectify a lack of knowledge quickly. Hence my reply. Last edited by Micrm@ss; February 27th, 2018 at 05:37 AM.  

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