Problem Set 02

Read

Continue to skim over the chapters on groups from your 210A textbook or from the additional books and resources on the course webpage.

• Make sure to review group actions some more; focus on how an action of $G$ on a set $X$ corresponds to a homomorphism from $G$ to $\operatorname{Sym}(X)$.
  • Notation: when $G$ is a group acting on a set $X$, we have a homomorphism $\varphi : G \rightarrow \operatorname{Sym}(X)$ defined by $\varphi(g)$ is the permutation that takes each $x\in X$ to $g\cdot x$. We’ll refer to this homomorphism as the associated permutation representation. Also, if $|X| = m$, then $\operatorname{Sym}(X) \cong \operatorname{Sym}(m)$, so we can view $\varphi$ as a map from $G$ to $\operatorname{Sym}(m)$.

To Work On

Work on some of the following problems. Again, 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.

1. Our discussion of Problem Set 01 led us to the following results. Carefully tie together our work to give proofs of them.

   Lemma. Let $G$ be an abelian group. Prove that $G$ is simple if and only if $G$ has prime order.

   Lemma. Let $p$ be prime, and let $P$ be a finite $p$-group. Prove that $P$ is simple if and only if $P$ has prime order.

2. If $H$ is a subgroup of $G$, then $G$ acts by left multiplication on the left cosets of $H$ in $G$, and then we can look at the associated permutation representation $\varphi : G \rightarrow \operatorname{Sym}(m)$ where $m = |G:H|$. Use this idea to prove the next lemma. In doing so, here are a couple other things to have in mind: (1) $\operatorname{Alt}(m)\trianglelefteq\operatorname{Sym}(m)$; (2) if $N\trianglelefteq G$ and $H$ is any subgroup, then $H\cap N\trianglelefteq H$.

   Lemma. Let $G$ be a nonabelian simple group. If $G$ has a proper subgroup of index $m$, then $|G|$ is a divisor of $m!/2$.

   Remark: the above lemma typically gets used to show that simple groups cannot have subgroups of “small” index or, in the contrapositive, that groups with a subgroup of “small” index are not simple. Also, this lemma is a special case of a slightly more general lemma that you could try to prove instead: If $G$ has a proper subgroup of index $m$, then $G$ has a proper normal subgroup of index dividing $m!$.

3. For the next lemma, we consider the action of $G$ on itself by left multiplication. In this case, the associated permutation representation is called the left regular representation.

   Lemma. Let $G$ be a group of order $n$, and let $\lambda : G \rightarrow \operatorname{Sym}(n)$ be the left regular representation. Prove that if $g\in G$ and $|g| = k$, then the cycle decomposition for $\lambda(g)$ consists of $n/k$ many cycles each of length $k$.

   Hint: let $H = \langle g\rangle$. The cycles in the decomposition for $\lambda(g)$ are exactly the orbits of $H$; do you see this? So the question becomes: can you show that every orbit of $H$ has size $|H|$? Think orbit-stabilizer, and remember that you’re working with the action by left multiplication.

4. Here is a fun application of the previous lemma.

   Lemma. Let $G$ act be a group of order $2m$. Prove that if $m$ is an odd number, then $G$ has a subgroup of index $2$, and, consequently, $G$ is not simple unless $m=1$.

   Hint: Consider the left regular representation $\lambda : G \rightarrow \operatorname{Sym}(2m)$. Use what you learned above to argue that $\lambda$ is injective, so $G\cong \lambda(G)$. Thus, you can instead show that $\lambda(G)$ has a subgroup of index $2$. There are a couple of things to leverage: (1) think about how $\lambda(G)\cap \operatorname{Alt}(2m)$ relates to $\lambda(G)$; (2) by Cauchy, $G$ has an element $i$ of order $2$, and the previous lemma allows you to determine if $\lambda(i)$ is an even or odd permutation.

5. Let’s explicitly bring Sylow theory to bear on our questions of simplicity with a couple of classics. If you’ve done these before, try to redo them without notes.

   Lemma. Let $p,q,r$ be distinct primes. If $G$ has order $pq$ or $pqr$, then $G$ is not simple.

6. A continuation/reformulation from last week: using the results from Problems Sets 01 and 02, try to determine all simple groups of order $n$ (if any) for as many values of $n$ between $1$ and $70$ as possible.