Read: Finish Section 3.1 and start Section 3.2.
Turn in: 3.23, 3.24(b,d,e,i,j), 3.26(just show \(H\cap K\le G\))
- There are two things to prove for 3.23: that \(Z(G)\) is a subgroup and that it is abelian (commutative). To show it is a subgroup, you will want to use Theorem 3.6, and for that there are three things to show:
- You need to prove that \(Z(G)\) is nonempty. To do this, prove that \(e\in Z(G)\).
- Next you need to show that $Z(G)$ is “closed under inverses”. To do this, assume \(z\in Z(G)\), and prove that \(z^{-1} \in Z(G)\). Remember, you know that \(z^{-1} \in G\) (because $G$ is a group), so to show \(z^{-1} \in Z(G)\), you need to show that \(z^{-1}g = gz^{-1}\) for all $g\in G$. You are assuming that $z\in Z(G)$, so can you find a way to manipulate \(z^{-1}g = gz^{-1}\) so that $z$ is involved?
- You also need to show that $Z(G)$ is closed under the operation of $G$. For this, let \(z_1,z_2\in Z(G)\), and prove that \(z_1z_2 \in Z(G)\) by showing \(z_1z_2g = gz_1z_2\) for all \(g\in G\).
- Two remarks about 3.24.
- The book highlights that if $G$ is abelian then $Z(G) = G$. Make sure you believe (and could prove this); it makes many parts of 3.24 easy using what we’ve learned before.
- If a group is not abelian, it may be most efficient to first think about which group elements are not in $Z(G)$. The idea is as follows. Suppose $g$ is an element of your group. If you can find some other group element $h$ such that \(gh\neq hg\) then you know $g\notin Z(G)$ (and also \(h\notin Z(G)\)). For example, using our previous notation for \(D_3 = \{e,r,r^2,s,sr,sr^2\}\), notice how \(rs\neq sr\) since \(rs\) moves the top of the triangle to lower right while \(sr\) moves the top of the triangle to lower left. Thus, we see that \(r,s\notin Z(D_3)\). On the other hand, we know \(e\in Z(D_3)\). So now you just need to determine if each of \(r^2,sr,sr^2\) are in \(Z(D_3)\) or not.
Extra practice: 3.24(remaining parts)