Read: Finish Section 3.1 and start Section 3.2.
Turn in: 3.27, 3.30, 3.33
- For 3.29–3.34, you could proceed as follows. Given a group $G$, list the elements as \(G = \{e,g_1,g_2,g_3,\ldots\}\).
- First, you know that \(\{e\}\) and $G$ are subgroups of $G$.
- Next, determine all subgroups generated by 1 element by computing \(\langle g_i \rangle\) for all \(g_i\).
- To reduce computations, remember that \(\langle g \rangle = \langle g^{-1} \rangle\).
- Next, determine all subgroups generated by 2 elements by computing \(\langle g_i,g_j \rangle\), for all possible \(g_i\) and \(g_j\).
- To reduce computations, use 3.25: if \(h \in \langle g \rangle\) then you know \(\langle g,h \rangle = \langle g \rangle\).
- Next, consider subgroups generated by 3 elements or explain why you already have found all subgroups.
- Continue on or explain why you already have found all subgroups. (In 3.30, you’ll find that every subgroup is generated by just one element, and in 3.33, you’ll find that every subgroup is generated by either one or two elements.)
Extra practice: 3.25, 3.29, 3.31
- There are multiple ways to prove 3.25, but one way is to make use of 3.10. Recall that 3.10 says that if $S\subseteq G$, then \(\langle S \rangle\) is the “smallest” subgroup of $G$ containing $S$. This means that if $H$ is a subgroup of $G$ with $S\subseteq H$, then $S\subseteq \langle S \rangle \subseteq H$; in symbols: \((S\subseteq H\le G)\implies (S\subseteq \langle S \rangle \le H\le G)\)