Read: Section 3.2.
Turn in: 3.26, 3.27, 3.30, 3.33. Please also revisit HW09.
- On 3.26, you will first show that $H\cap K$ is a subgroup; for that, you should use the Two-Step Subgroup Test (3.6). To show that $H\cap K$ is the “largest” subgroup contained in both $H$ and $K$, your proof should begin by assuming that $L$ is contained in both $H$ and $K$; then explain why $L$ must also be contained in $H\cap K$.
- 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.
Extra practice: 3.25, 3.29, 3.31, 3.32, , 3.34
- 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: $(S\subseteq H\le G)\implies (S\subseteq \langle S \rangle \le H\le G)$
