This is the first set of homework problems Link to main web page for the course

You do not need to hand these problems in, but be prepared to present the solutions in class.

First problem:

Look up and or prove the Orbit-Stabilizer Theorem. Use it to prove that the number of permutations with $m_i$ cycles of length $i$ (for $1 \leq i \leq n$) is equal to $\frac{n!}{\prod_{i=1}^n i^{m_i} m_i!}.$ I asked you to verify this in the case that the permutations have cycle structure $(a_1 a_2 a_3) (a_4 a_5) (a_6 a_7)$ by using the formula and listing them but I just realized that this conjugacy class has 210 elements in it. While I think it is a good idea to understand the inner workings of how this enumeration works, it is not necessarily productive to list 210 elements.

 Sirous agreed to do this in class. He put up a proof of the orbit stabilizer theorem that showed the map $\phi : G/G_s \rightarrow O_s$ defined by $\phi( g G_s ) = g s$ was well defined, 1-1 and onto.

Second problem:

Fill in the details of the two statements at the end of class. Namely,
(1) Let $G$ be a finite group and $V$ a $G$-module. If $V$ has a non-trivial submodule $W$ and a $G$-invariant scalar product (one that satisfies $\left< u, v \right> = \left< gu, gv \right>$), then the subspace $W' = {\mathcal L}\{ v \in V : \left< v, w \right> = 0 \hbox{ for all } w \in W\}$ is also a $G$-module and $V = W \bigoplus W'$.
(2) If $V$ has a basis $\{ v_1, v_2, \ldots, v_n \}$, then set $\left< v_i, v_j \right> = \begin{cases} 1&\hbox{ if }i=j\\ 0&\hbox{ else } \end{cases}.$ It may or may not be that $\left< , \right>$ is $G$ invariant. But if it isn't $G$ invariant, then $$\left< u, v \right>' = \sum_{g \in G} \left< gu, gv \right>$$ is a $G$ invariant scalar product.

 Farid agreed to do this in class. I didn't copy exactly what he wrote, but I will try to recall the details that he added. He actually went further and proved: Machke's Theorem: If $G$ is a finite group and a $V$ is a $G$-module over ${\mathbb C}$, then $$V = W_1 \oplus W_2 \oplus \cdots \oplus W_k$$ where each of the $W_i$ are irreducible $G$-submodules. Lemma 1: Let $\{ v_1, v_2, \ldots, v_n \}$ be a basis for $V$, then set $\left< v_i, v_j \right> = \begin{cases} 1&\hbox{ if }i=j\\ 0&\hbox{ else } \end{cases}.$ $\left< u,v \right>$ is a inner product. (Proof of Lemma 1): We need to show that (a) $\left< u, v \right> = {\overline {\left< v, u \right>}}$, (b) $\left< a v + b w, u \right> = a \left< v, u \right> + b \left< w, u\right>$ and (c) $\left< v, v \right> > 0$ if $v \neq {\bf 0}$. We can ensure that that (a) and (b) hold by definition since we have defined this scalar product on a basis. (c) holds because if $v = \sum_{i=1}^n c_i v_i$, then $\left< v, v\right> = \sum_{i=1}^n c_i {\overline c_i}$ and this is a positive number as long as it is not the case that $v = {\bf 0}$. Lemma 2: Let $\left< u, v \right>$ be the inner product defined in Lemma 1, then either this inner product is $G$ invariant or $\left< u, v \right>' = \sum_{g \in G} \left< g u, g v \right>$ is a $G$ invariant inner product. (Outline of proof of Lemma 2): Assume that $\left< u, v \right>$ as defined in Lemma 1 is not $G$-invariant, then $$\left< hu, hv \right>' = \sum_{g \in G} \left< gh u, gh v \right> = \sum_{k in G} \left< k u, k v \right> = \left< u, v \right>'~.$$ Next show that $\left< u, v \right>'$ satisfies the conditions of a scalar product (a), (b), (c) as in the proof of Lemma 1. The details are left to the reader. Lemma 3: If $W$ is a submodule of $V$, and $\left< u, v \right>$ is a $G$ invariant inner product, then $W^\perp = \{ v \in V: \left< w, v \right> = 0 \}$ is also a submodule and $V = W \oplus W^\perp$. (Proof of Lemma 3): Since $W$ is a subspace (as a vector space) then it is always the case that $V = W \oplus W^\perp$. This is a result that is known from linear algebra. Roughly the proof goes, let $\{ e_1, e_2, \ldots, e_k \}$ be an orthonormal basis of $W$ (and you need to know that we can construct this), then for $v \in V$, let $$w = \left< v, e_1 \right> e_1+ \left< v, e_2 \right> e_2+ \cdots + \left< v, e_k \right> e_k$$ then $v = w + (v-w)$ and $w \in W$ and $v-w \in W^\perp$. $W^\perp$ is a submodule because if $v \in W^\perp$ then for all $w \in W$, it will be the case that $g^{-1} w \in W$ and so $$\left< g v, w \right> = \left< g^{-1} g v, g^{-1} w \right> = \left< v, g^{-1} w \right> = 0$$ and hence $gv \in W^\perp$. (Proof of Machke's Theorem) If $V$ is irreducible, then the conclusion of Machke's theorem holds since $V$ itself is a direct sum of irreducible modules. So assume that $V$ is not irreducible, then there exists a proper submodule $W$ (that is, $W$ is not the $0$ module and it is not $V$ itself). By Lemma 1 and 2 there is a $G$ invariant scalar product, and by Lemma 3 $W^\perp$ is a submodule of $V$ as well and $V = W \oplus W^\perp$. Now since both $W$ and $W^\perp$ are of dimension smaller than $V$, we can assume that $W = W_1 \oplus \cdots \oplus W_\ell$ and $W^\perp = W_{\ell + 1} \oplus \cdots \oplus W_k$ for some $\ell$ and $k$ and irreducible submodules $W_i$ and hence $V = W_1 \oplus W_2 \oplus \cdots \oplus W_k$.