Read: Read Section 6.2. Also consider watching the following Socratica video:
Turn in: 6.27, 6.29, 6.34, 6.35(a,b,c), 6.36, 6.40
- Make sure to read the beginning of Section 6.2 a couple of times.
- Please read the remark below about aritmetic in quotient groups. So, for example, when you are working on 6.35(b), you are trying to determine the smallest positive integer $k$ such that $(j\langle -1 \rangle)^k = \langle -1 \rangle$, and by the remark, this is the same as determining the smallest positive integer $k$ such that $j^k \in \langle -1 \rangle$. In part 6.35(c), you are using additive notation, so you will be looking for the smallest positive integer $k$ such that $k\cdot 5 \in \langle 4 \rangle$.
- For 6.36(a) (and the remaining parts), 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.
- For 6.40, you need to work abstractly with $G/H$. To show $G/H$ is abelian, you should take arbitrary $aH,bH \in G/H$ and then prove that $(aH)(bH) = (bH)(aH)$. Remember how the operation $(aH)(bH)$ is defined.
Remark about aritmetic in quotient groups: 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 $a^kH = H \iff a^k\in H$.