# Proofs of Sylow’s Theorems in Group Theory- Part 1

I will try to give a breakdown of the proof of Sylow’s theorems in group theory. These theorems can be tricky to understand, and especially retain even if you’ve understood the basic line of argument.

1. Sylow’s First Theorem- If for a prime number $p$, $o(G)=p^km$, and $p\nmid m$, then there is a subgroup $H\leq G$ such that $o(H)=p^k$.

First we will try to understand what an orbit of a group is. Take a set $S$, and a group $G$. $S$ does not necessarily have to be a subset of $G$. $Orbit(S)=\{gs|s\in S, g\in G\}$. Note that this is right multiplication.

$Stabilizer(S)=\{g\in G|gs=s\}$.

Now the Orbit-Stabilizer Theorem- $|Orbit||Stabilizer|=|Group|$. Something to remember from this is that there is nothing special about this theorem. It is just another way of saying $|Image||Kernel|=|Group|$, as derived from the First Isomorphism Theorem.

The stabilizer of a set $S$ is a subgroup. Proof: Let $as=s$ and $bs=s$. Then $abs=as=s$. Also, if $as=s$, then taking left inverse on both sides, $a^{-1}s=s$.

Now we move to a combinatorial argument: what power of $p$ divides ${p^km\choose p^k}$? On expanding, we get $\displaystyle{\frac{p^km(p^km-1)(p^km-2)\dots (p^km-p^k+1)}{1,2,3\dots p^k}}=m\displaystyle{\Pi}_{j=1}^{p^k-1}\frac{p^km-j}{p^k-j}$. The highest power of $p$ that can divide $j$ is $\leq k-1$. Hence, the power of $p$ that divides $p^k-j$ is the same that which divides $p^km-j$. In fact, $p$ does not have to be prime for this property to be true. More generally speaking, just to gain a feel for this kind of argument, provided $j\leq p^k-1$, the fraction of the form $\frac{a-j}{b-j}$ not be divided by any power of $p$ except for $1$, as long as both $a$ and $b$ are divisible by powers of $p$ greater than or equal to $k$. $j$ is clearly the constraint here. Remember that $p\nmid m$ is a necessary condition for this.

Now we prove ${p^km\choose p^k}\equiv m (mod p)$.

Now we will define the set $S$ to contain all subsets of $G$ with $p^k$ elements. Clearly, $|S|={p^km\choose p^k}$, which as shown above is not divisible by $p$. Take every $g\in G$, and multiply it with every element of $S$. Note that very element of $S$ is itself a set containing $p^k$ elements of $G$. If $S_i\in S$, then $gS_i\in S$. Moreover, equivalence classes are formed when all elements of $G$ are multiplied with all elements of $S$, and these equivalence classes partition $S$. The proof of this: let $GS_i$ be the orbit of $S_i$. Then if $S_k\in Orb(S_i), gS_i=S_k$. Moreover, if $S_l\in S=g'S_i$, then $g'g^{-1}S_k=S_l$. This gives a feel for why these objects of $S$ are all in the same class.

Let there be $r$ such partitions of $S$. Then $|S|=|S_1|+|S_2|+\dots+|S_r|$, where $S_1,S_2$,etc are partitions of $S$. We know $|S|$ is not divisible by $p$. Hence, there has to be at least one $|S_k|$ which is not divisible by $p$. Let $S_k\in [S_k]$, where $[S_k]$ is an equivalence class partitioning $S$. Consider a mapping $f:G\to S_k$ such that $f(g)=gS_k$. The image is $[S_k]$, and the kernel is the stabilizer. By the orbit-stabilizer theorem, $|G|=|Stabilizer||Orbit|$. We know that the orbit is not divisible by $p^k$. Hence, the stabilizer has to be divisible by $p^k$. Also, the stabilizer is a subgroup.

The rest of the proof will be continued next time.