1. If $H_1\leq G$, $H_2 \leq G$, $G$ a group, and $a \in G$. Show that the double coset $H_1aH_2$ is such that

    \begin{displaymath} H_1aH_2 = \cap_{h_2} H_1ah_2 \end{displaymath}

    where $h_2$ ranges over certain elements of $H_2$.

  2. Let $H_1\leq G$, $H_2 \leq G$, $G$ a group, and $a \in G$. Let $\phi : H_1aH_2 \rightarrow a^{-1}H_1aH_2$ be defined by $\phi(h_1ah_2) = a^{-1}h_1ah_2$, where $h_1 \in H_1$ and $h_2 \in H_2$. Show $\phi$ is well-defined, 1-1, and onto.

  3. Following the proof of part (b) of Theorem 8.2.2, prove part (a), i.e., the number of right cosets of $H_1$ in $H_1aH_2$ is $[H_2 : H_2 \cap a^{-1}H_1a]$.

  4. Find the double coset decomposition of $S_3$ with respect to $H_1 = H_2 = \{(1), (1,2)\}$.

  5. Let $G$ be a finite group and $H \leq G$ such that $N_G(H) = N(H) = H$, and any two distinct conjugate subgroups of $H$ have only the identity element in common. Let $N$ be the set of elements of $G$ not contained in $H$ nor in any of its conjugates, together with the identity. Show that $\vert H\vert\, \vert\, (\vert N\vert-1)$.

    HINT: First use the given together with Lagrange's Theorem (in particular equation (4.8)) and Theorem 6.1.1 to show that $\vert N\vert = [G : H]$. Next decompose $G$ into double cosets with respect to $N(H)$ and $H$ and use equation (8.6). Now the identity $e \in G$ belongs to some double coset, so we may assume that $a_1 = e$, in the line before equation (8.6). Finally this implies that in (8.6) $d_1 = \vert H\vert$, but all the other $d_j = 1$. (Why?) Use the resulting relation to get the desired result.

David Joyner 2007-08-06