Skim over the chapters on groups from your 210A textbook. You can also take a look at some of the additional books and resources on the course webpage.
Try to work on some of the following problems. It’s better to understand one of them deeply (even if you don’t completely solve it) then to just have vague ideas for all. Hopefully as a team we can make it through all of them.
Orbit Equation. Let $G$ be a finite group acting on a finite set $X$. Let $\mathcal{O}_1, \ldots, \mathcal{O}_n$ be the orbits of $G$ that contain more than $1$ element, and let $x_1,\ldots x_n \in X$ be such that $x_i \in \mathcal{O}_i$. Then $|X| = |\operatorname{Fix}(G)|+\sum_{i=1}^n|G:G_{x_i}|.$
p-group Fixed-Point Lemma. Let $p$ be prime, and let $P$ be a finite $p$-group1. If $P$ acts on a finite set $X$, then $|\operatorname{Fix}(P)| \equiv |X|$ modulo $p$.
Lemma. Let $p$ be prime. If $P$ is a nontrivial finite $p$-group, then $Z(P)$ is nontrivial.
Sylow’s Theorem. Let $G$ be a finite group of order $mp^k$ with $p$ prime and $p$ not dividing $m$.
- $G$ has a subgroup of order $p^k$ (called a Sylow $p$-subgroup).
- All Sylow $p$-subgroups are conjugate.
- The number of Sylow $p$-subgroups is congruent to 1 mod $p$.
- Every $p$-subgroup of $G$ is contained in some Sylow $p$-subgroup.
Hint: To prove conjugacy and that the number of Sylows is congruent to 1 mod $p$, try letting $G$ act on $\operatorname{Syl}_p(G)$ (i.e. the set of Sylow $p$-subgroups) by conjugation. Let $\mathcal{O}$ be an orbit. The goal is to show $\mathcal{O}=\operatorname{Syl}_p(G)$ and $|\mathcal{O}| \equiv 1$ mod $p$. Choose $P \in \mathcal{O}$, and towards a contradiction, assume that $Q\in \operatorname{Syl}_p(G)$ but $Q\notin \mathcal{O}$. The key is to consider how $P$ and $Q$ act on $\mathcal{O}$ and use the p-group Fixed-Point Lemma above; to make this work, you’ll want to show that $R\in \operatorname{Fix}(P) \iff RP \text{ is a group of order } |R|\cdot|P|/|R\cap P| \iff R = P$ (and similarly with $Q$ in place of $P$.)
Let $p$ be prime. A $p$-group is a group in which every elements has order that is a power of $p$. By Lagrange, we have that $P$ is a finite $p$-group if and only if $P$ has order $p^k$. ↩