The Set of Left (Right) Cosets of a Subgroup Partitions the Whole Group

# The Set of Left (Right) Cosets of a Subgroup Partitions the Whole Group

Recall from the Left and Right Cosets of Subgroups page that if $(G, \cdot)$ is a group, $(H, \cdot)$ is a subgroup, and $g \in G$ then the left coset of $H$ with representative $g$ is the set:

(1)
\begin{align} \quad gH = \{ gh : h \in H \} \end{align}

The right coset of $H$ with representative $g$ is the set:

(2)
\begin{align} \quad Hg = \{ hg : h \in H \} \end{align}

We will now look at a nice theorem which tells us that the set of all left cosets of a subgroup $(H, \cdot)$ actually partitions $(G, \cdot)$. The proof below can be mirrored to analogously show that the set of all right cosets of a subgroup $(H, \cdot)$ also partitions $(G, \cdot)$.

 Theorem 1: Let $(G, \cdot)$ be a group and let $(H, \cdot)$ a subgroup. Then the set of all left cosets of $(H, \cdot)$ partitions $(G, \cdot)$.

Recall that a partition of a set $A$ is a collection of nonempty subsets of $A$ that are pairwise disjoint and whose union is all of $A$.

• Proof: We first show that any two distinct left cosets of $H$ are disjoint. Let $g_1, g_2 \in G$, $g_1 \neq g_2$ and assume the left cosets $g_1H$ and $g_2H$ are distinct. Suppose that $g_1H \cap g_2H \neq \emptyset$. Then there exists an $x \in g_1H \cap g_2H$. So $x \in g_1H$ and $x \in g_2H$ and there exists $h_1, h_2 \in H$ such that:
(3)
• So using $(*)$ and $(**)$ we see that $g_1h_1 = g_2h_2$. So $g_1 = g_2h_2h_1^{-1}$. But $h_2h_1^{-1} \in H$ since $(H, \cdot)$ is a group and is hence closed under $\cdot$. So $g_1 \in g_2H$. But this means that $g_1H = g_2H$, a contradiction since $g_1H$ and $g_2H$ distinct. So the assumption that $g_1H \cap g_2H \neq \emptyset$ was false. So:
• Now if $i$ is the identity element for $(G, \cdot)$ then $i \in H$ since $(H, \cdot)$ is a subgroup and must contain the identity element. So $g = gi \in gH$ for all $g \in G$. So:
• Therefore the left cosets of $(H, \cdot)$ partition $(G, \cdot)$.