Read: Continue with 5.1.
Turn in: 5.4(c), 5.6, 5.7, 5.9 (only \(\sim_L\)), 5.12
- In 5.6 and 5.7, you will need to adapt the definition of cosets to additive notation: left cosets now look like \(a+H := \{a+h\mid h\in H\}\) and right cosets like \(H+a := \{h+a\mid h\in H\}\).
- For 5.7, there will be only finitely many cosets, but each coset will be infinite.
- Note that you will not need to use any of the previous results to prove 5.12; just work through the definitions of one-to-one and onto.
Remark about working with cosets: Please carefully read over 5.9–5.11. One thing to take away is that 5.9 and 5.10 together yield an important criteria for determining when two left (respectively right) cosets are equal:
\[aH = bH \iff b\in aH \iff a^{-1}b\in H.\] \[Ha = Hb \iff b\in Ha \iff ba^{-1}\in H.\]You are welcome to use this!
Extra practice: 5.4(a),5.5, 5.8, 5.10–5.11
- To help you check your work, here’s part of the computation for 5.4(a). 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).
- To carefully prove 5.8, you want to work with $aH$ and $Ha$ as (potentially infinite) sets. You need to show that $g\in aH$ if and only if $g\in Ha$. Also, remember that a statement like $g\in aH$ means that $g$ has a specific form: $g=ah$ for some $h\in H$.