Read: Section 6.2.
Turn in: 6.15 (just “backwards” direction—see below), 6.18, 6.35(d,e), 6.36(a,b), 6.37
- For 6.15, just prove: “if $m$ and $n$ are relatively prime, then \(\mathbb{Z}_{m} \times \mathbb{Z}_{n}\) is cyclic.” Try to explicitly find a generator $(a,b)$ for \(\mathbb{Z}_{m} \times \mathbb{Z}_{n}\) by using 6.12 to find an element $(a,b)$ such that \(\vert(a,b)\vert = \vert\mathbb{Z}_{m} \times \mathbb{Z}_{n}\vert\); a key point is that $m$ and $n$ are relatively prime if and only if $\operatorname{lcm}(m,n) = mn$.
- On 6.18, for each group that is cyclic, please also write down what familiar group it is isomorphic to using 4.40.
- For 6.36(a), you could first use 6.33 (which follows from Lagrange’s Theorem) to determine the order of the group \(Q_8/ \langle -1 \rangle\). Knowing the order of the group will help you focus in on what familiar group it might be isomorphic to. You will need a little more information to decide, and computing orders of elements of \(Q_8/ \langle -1 \rangle\) (like in 6.35(b)) should help.
Extra practice: 6.20
Remark about aritmetic in quotient groups (from last time): Let $H\le G$. Remember the $G/H$ denotes the set of all left cosets of $H$ in $G$, i.e. $G/H = \{aH \mid a\in G\}$ (so $G/H$ is a set of sets).
Now, if $H\trianglelefteq G$ (i.e. if $H$ is a normal subgroup of $G$), then there is a natural binary operation that makes $G/H$ a group. This is a super important idea! So, what is the binary operation on $G/H$? It is simply derived from the binary operation for $G$ as follows. For all $aH,bH\in G/H$, define
- $(aH)(bH) := (ab)H$
- $(aH)^{-1} := (a^{-1})H$
- the identity of $G/H$ is $eH = H$
So, how do we make other computations in $G/H$? For example, suppose you have a coset $aH$ and want to determine the order of $aH$ in the group $G/H$. One way to do this is to compute $aH$, $(aH)^2$, $(aH)^3$, $(aH)^4, \ldots$ until you find that $(aH)^k = H$, which is to say you keep going until $(aH)^k$ equals the identity of $G/H$. Remember that $(aH)^k$ is computed as $a^kH$. Then to determine when $a^kH = H$, look back at our notes from Section 5.2 about cosets, and you should see that $a^kH = H \iff a^k\in H$.