Cosets
We will see soon that subgroups "split" groups into "even bins." This is confirmed by a famous theorem of Lagrange:
Let $G$ be a group, $H \leq G$, $g \in G$. A left coset of $H$ in $G$ is $$gH = \left \{ gh \mid h \in H \right \}$$ In this case, we call $g$ the coset representative.
Let $G = \mathbb{Z}$, $H = 3\mathbb{Z} \leq \mathbb{Z}$. If we pick $2 \in \mathbb{Z}$, see that $$2 + 3\mathbb{Z} = \left \{\dots -1,2,5,8,\dots \right \}$$ We will notice that this is not all of $\mathbb{Z}$. If we pick: $$1 + 3\mathbb{Z} = \left \{\dots, -2,1,4,7,\dots \right \} \quad \text{ and } \quad 0 + 3\mathbb{Z} = \left \{ \dots, -3,0,3,6,9,\dots \right \}$$ then every element $z \in \mathbb{Z}$ sits in exactly one of the cosets.
As a brief remark, notice that the identity is always an option.
Let $G$ be an abelian group, then left cosets and right cosets are the same.
Sketch
From the definitions, we can see that $gh = hg$ implies $gH = Hg$.
The cosets of $H$ in $G$ form a partition
Proof
$\forall, g_1, g_2 \in G$, $g_1H \cap g_2H = \emptyset$ or $g_1H = g_2H$. If nonempty, then $\exists, a \in G$ w/ $$a = g_1h_1 = g_2h_2 \quad \leadsto \quad g_1 = g_2h_2h_1^{-1} \quad \leadsto \quad g_1 \in g_2H. \qquad \square$$
Let $G$ be a finite group, $H \leq G$. Then $$[G : H] = \frac{|G|}{|H|}.$$
This means the order of any element must divide the order of the group. The converse is false!
Let $p$ prime, $p \nmid a$, then $$a^{p-1} \equiv 1 \pmod{p}.$$
Proof Sketch
$a$ has order $p - 1$ in $U(p)$, the gp. of units in $\mathbb{Z}/p\mathbb{Z}$.
Compute the left cosets of $A_4$ in $S_4$. Then do this for $D_4$.