The First Sylow Theorem Proof

Use induction on on the order of the group. If $|G|=1$, then what do you know? If $|G|=2$, then what do you know. Assume for some $n\in \mathbb{N}$ that the first Sylow theorem is true for all groups of order $k<n$. Then show that the theorem is true if $|G|=n$.

At some point in your work, you'll need to use two problems that we have already shown to be true.

1. Cauchy's Theorem for Abelian Groups - If $p$ divides $|G|$, then $G$ has an element of order $p$. This applies to the center of every group, so if $G$ is not Abelian you can still apply the theorem to the center $Z(G)$ of $G$.
2. If $G/N$ has a subgroup $B$, then $G$ has a subgroup $H$ with $H/N=B$. If $G$ is finite, then we know $|H|/|N|=|B|$.

What do you do? If $G$ has order $p^k$, then you're done. If $G$ has a proper subgroup of order $p^k$, then you're done. So suppose it does not have a proper subgroup of order $p^k$. What does this mean about $|C(a)|$? In particular, what does this mean about $|G:C(a)|$ which equals $|\text{cl}|$? Can you prove that $p$ must divide this index for each $a\notin Z(G)$? You may have to use the induction hypothesis your work here. If you then use the class equation, you should be able to show that the center of $G$ contains an element $x$ of order $p$. Once you do that, you can create the factor group $G/\left<x\right>$, which has order less than $G$. Here's where you get to use the induction hypothesis again. Problem 7 from Wednesday last week should help you to get to a subgroup in $G$ from knowing things about subgroups in $G/\left<x\right>$. I believe Nick presented this in class on Monday (at the very end).