Read
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.
- Review definitions and basic properties for groups.
- Highlights include: (1) definitions of index of a subgroup, conjugates of elements and subgroups, centralizers of elements and subgroups, normalizers of subgroups, the center of a group, normal subgroup, and simple groups; (2) Lagrange’s Theorem; (3) Cauchy’s Theorem; (4) Sylow’s Theorem(s); (5) the classification of the finite abelian groups.
- Review definitions and basic properties for group actions.
- Highlights include: (1) if $G$ acts on a set $X$, then the orbits of $G$ form a partition of $X$; (2) the Orbit-Stabilizer Theorem: if $G$ acts on a set $X$ and $x\in X$, then the orbit containing $x$ has cardinality $|G:G_x|$. (In words: the size of the orbit containing $x$ equals the index of the stabilizer of $x$.)
- Notation: for $G$ a group acting on a set $X$ and $x\in X$, we’ll tend to write $G_x$ for the stabilizer of $x$ in $G$ and write $\operatorname{Fix}(G)$ for the common fixed points of $G$; that is, $x\in \operatorname{Fix}(G)$ if and only if $g\cdot x = x$ for all $g\in G$ (i.e. if the orbit containing $x$ has size $1$).
To Work On
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.
- Prove the following lemma, which should look a lot like the class equation. You’ll want to combine the fact that the orbits of $G$ partition $X$ together with the Orbit-Stabilizer Theorem, which tells you how big each orbit is.
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 $|X| = |\operatorname{Fix}(G)|+\sum_{i=1}^n|G:G_{x_i}|.$
- Use the Orbit Equation to prove the following lemma. Hint: Lagrange tells you something about $|G:G_{x_i}|$.
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 $p$.
- Use the $p$-group Fixed-Point Lemma to prove that finite $p$-groups have a nontrivial center. Hint: let $P$ act on itself by conjugation. What is $\operatorname{Fix}(G)$ with respect to this action?
Lemma. Let $p$ be prime. If $P$ is a nontrivial finite $p$-group, then $Z(P)$ is nontrivial.
- Put together a clean proof (that you can explain to the group) of one of the parts of Sylow’s Theorem.
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 $1\le n \le 75$. Using the results above, determine as many values of $n$ as possible for which you can prove that there is no simple group of order $n$.
Footnotes
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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 $p^k$. ↩