Consider the group $U(20)$ of units of $\mathbb{Z}_{20}$ under multiplication mod $20$.
- List the elements of U(20) and find the inverse of 7.
- Show that $H = \{1, 9, 11, 19\}$ is a subgroup of $U(20)$.
Sketch of Solution
$U(20) = \left \{ k \in \mathbb{Z}_{20} \mid \mathrm{gcd}(k,20) = 1 \right \}$. This gives us: $$\left \{ 1, 3, 7, 9, 11,13,17,19 \right \}$$ . $7*3 = 21$. Towards subgroup, see that $19^2 = 1$, $11^2 = 1$, and $9^2 = 1$ so it's closed under inverses. Products are a bit more work, but $19*11 \equiv 9$, $9 * 11 \equiv 19$, and $19*9 \equiv 11$.
Let $G$ be a group such that for all $g \in G$, $g^2 = e$. Show that $G$ must be abelian.
Sketch of Solution
See that for any $a,b \in G$ multiplying on the left by $a$ and right by $b$ in: $$e = (ab)^2 = abab \quad \text{ gives } \quad ab = a(abab) b = \cancelto{e}{a^2} ba \cancelto{e}{b^2} = ba$$
Let $H$ and $K$ be subgroups of $G$. Prove that $H \cup K$ is a subgroup if and only if either $H \subset K$ or $K \subset H$.
Sketch of Solution
The converse direction is trivial. We show forwards; suppose for contradiction that neither $H \subset K$ nor $K \subset H$; then there exists $h \in H \setminus K$ and $k \in K \setminus H$; the product $$hk \in H \cup K \quad \text{ implying either } \quad hk \in H \text{ or } hk \in K$$ Suppose the former, then $h^{-1} hk = k \in H$, contradiction. Mutatis mutandis for the latter.
Denote by $\sigma = (1357)(246)$ and $\tau = (12)(34)(567)$ in $S_{7}$
- Compute $\sigma \circ \tau$ in terms of disjoint cycles.
- What is the order of $\sigma$, $\tau$, and $\sigma \circ \tau$?
- Which of $\sigma$, $\tau$, and $\sigma \circ \tau$ are elements of $A_{7}$?
Sketch of Solution
We show 1: $$\sigma \tau = (145236)(7)$$ Compute orders by computing cycle types: $\sigma$ is of type $(3,4)$, $\tau$ is of type $(2,2,3)$, and $\sigma \tau$ is of type $(6)$. The orders are, respectively, 12, 6, and 6. Finally, see that $\tau$ is the only one of even sign.
Prove that every group of prime order is cyclic
Sketch of Solution
Let $G$ be a group of prime order. By Lagrange's theorem, if we pick any $g \in G$, the subgroup $H = \langle g \rangle = \left \{ e, g, g^2, \dots \right \}$ must satisfy $$|H| \mid |G| = p \quad \text{ implying } \quad |H| \in \left \{ 1,p \right \}$$ $|H| = 1$ exclusively in the case $g = e$; if $|H| = p$, then $g$ is a generator for $G$ and we're done.