Read: Finish Section 6.2 and start Section 7.1.
Turn in: 6.40, 6.43, 7.2
- 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.
- For 6.43, you are assuming that $G$ is cyclic, and want to prove that $G/H$ is cyclic. You should first try to guess a generator $aH$ for $G/H$ (using what you know about $G$), and then show an arbitrary coset $gH$ can be written as $gH = (aH)^k$ for some $k$.
Extra practice: 6.46
- 6.46 is a really fun one. You are assuming that \(G/Z(G)\) is cyclic, so \(G/Z(G)\) is generated by one of its elements, which is of the form \(aZ(G)\). Thus, for all \(g\in G\), we have that \(gZ(G) = a^kZ(G)\) for some \(k\in \mathbb{Z}\), and this implies (by properties of cosets) that \(g\in a^kZ(G)\). So, for all \(g\in G\), \(g\) can be written as \(g = a^kz\) for some \(k\in \mathbb{Z}\) and some \(z\in Z(G)\). Now use this idea to prove that \(G\) is abelian.
Remark about aritmetic in quotient groups (one more 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 $a^kH = H \iff a^k\in H$.