Read: Read Section 6.1.
- If you haven’t already, watch the following two Socratica videos:
- If you have extra time over the break, you can get ahead by starting Section 6.2 and watching the following video:
Turn in: 5.4, 5.21, 5.25, 5.33, 6.9, 6.13, 6.20
- To help you check your work, here’s part of the computation for 5.4. The group is $Q_8$ one of the subgroups is \(H = \langle i \rangle = \{1,-1,i,-i\}\). To compute the left cosets of $H$ in $Q_8$, we take individual elements of \(Q_8\) and multiply $K$ on the left by them. (Remember: $ij = k$, $ji = -k$, $ik = -j$, $ki = j$, $jk=i$, $kj = -i$.)
- \(1H = \{1\cdot1,1\cdot(-1), 1\cdot i, i\cdot (-i)\} = \{1,-1, i, -i\} = H\)
- \(-1H =\) (you do it)
- \(iH = \{i\cdot1,i\cdot(-1), i\cdot i, i\cdot (-i)\} = \{i,-i, -1, 1\} = H\)
- \(-iH =\) (you do it)
- \(jH = \{j\cdot1,j\cdot(-1), j\cdot i, j\cdot (-i)\} = \{j,-j, -k, k\}\)
- \(-jH =\)(you do it)
- \(kH =\)(you do it)
- \(-kH = \{(-k)\cdot1,(-k)\cdot(-1), (-k)\cdot i, (-k)\cdot (-i)\} = \{-k,k,-j, j\} = jH\)
In the end, for this part, you should only have two different cosets of $H$ in $Q_8$ (the rest are repeats).
- Let me clarify 5.33. Let’s name the subgroups: $A = \langle s \rangle$, $B = \langle r^2, sr^2\rangle$. To show $A\trianglelefteq B$, you need to show $bA = Ab$ for every $b\in B$; however, if you see that 5.32 applies (wink, wink), you can use that. Next, you should explain why $B \trianglelefteq D_4$. Finally, to show $B\not\trianglelefteq D_4$, you only need to find one element $g\in D_4$ such that $gB \neq Bg$.
Extra practice: 6.2, 6.10, 6.14, 6.18