Examples

Consider H = \(\{e , (1,2)\} \subseteq S_3\). We know \(S_3\) has six elements. Let \(\pi \in S_3\) and compute \(\pi H = \{ \pi \sigma \ : \ \sigma \in H \}\). So

$$ \begin{align*} S_3 = \{ e, \begin{pmatrix}1 & 2 \end{pmatrix}, \begin{pmatrix}1 & 3 \end{pmatrix}, \begin{pmatrix}2 & 3 \end{pmatrix}, \begin{pmatrix}1 & 2 & 3 \end{pmatrix}, \begin{pmatrix}1 & 3 & 2 \end{pmatrix} \} \end{align*} $$

For example \(\begin{pmatrix}1 & 3 \end{pmatrix} H\) is

$$ \begin{align*} \begin{pmatrix}1 & 3 \end{pmatrix} H &= \{ \begin{pmatrix}1 & 3 \end{pmatrix}e, \begin{pmatrix}1 & 3 \end{pmatrix} \begin{pmatrix}1 & 2 \end{pmatrix} \} \\ &= \{ \begin{pmatrix}1 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} \} \end{align*} $$
$$ \begin{align*} eH &= (1 \quad 2) H = H \\ \begin{pmatrix}1 & 3 \end{pmatrix} H &= (1 \quad 2 \quad 3)H = \{ \begin{pmatrix}1 & 3 \end{pmatrix}, \begin{pmatrix}1 & 2 & 3 \end{pmatrix} \} \\ \begin{pmatrix}2 & 3 \end{pmatrix} H &= (1 \quad 3 \quad 2)H = \{ \begin{pmatrix}2 & 3 \end{pmatrix}, \begin{pmatrix}1 & 3 & 2 \end{pmatrix} \} \\ \end{align*} $$

Example 2

Suppose \(G = (\mathbb{Z}_4, +)\) and \(H = \{0, 2\}\). Then we have the following cosets

$$ \begin{align*} H+0 &= \{0,2\} \\ H+1 &= \{1,3\} \\ H+2 &= \{2,0\} \\ H+3 &= \{3,1\} \end{align*} $$

We can see here that we have two distinct right cosets. Moreover, the cosets are disjoint and partition the group. Next, we’ll formalize these properties about cosets!

Proposition 2.5.4
Let \(H\) be a subgroup of a group \(G\).
  1. Let \(a\) and \(b\) be elements of \(G\). Either \(aH = bH\) or \(aH \cap bH = \emptyset\)
  2. Each left coset \(aH\) is nonempty and the union of left cosets is \(G\).


Proof
For \((a)\), Suppose \(aH \cap bH \neq \emptyset\) and let \(c \in aH \cap bH\). Since \(c \in aH\), then \(aH = cH\) by Proposition 2.5.3. Similarly since \(c \in bH\), then \(bH = cH\). Therefore, \(aH = bH\).
For \((b)\), \(aH\) is nonempty since \(H\) is a subgroup and can’t be empty. Furthermore, for each element \(g \in G\), \(g \in gH\). By \((a)\) then the union of all the left cosets must be \(G\).

Proposition 2.5.5
Let \(H\) be a subgroup of a group \(G\) and let \(a\) and \(b\) be elements of \(G\). Then \(x \longmapsto ba^{-1}\) is a bijection between \(aH\) and \(bH\).


Proof Let \(f : aH \rightarrow bH\) where \(f(x) = ba^{-1}x\). We want to show that \(f\) is a bijection by showing that it’s one-to-one and onto. Suppose \(a_1, a_2 \in aH\) then we can write \(a_1 = ah_1\) and \(a_2 = ah_2\) for some \(h_1, h_2 \in H\). Moreover, suppose that \(f(a_1) = f(a_2)\). Observe that

$$ \begin{align*} ba^{-1}a_1 &= ba^{-1}a_2 \\ b^{-1}ba^{-1}a_1 &= b^{-1}ba^{-1}a_2 \quad \text{($G$ is a group so the inverse exists)} \\ aa^{-1}a_1 &= aa^{-1}a_2 \\ a_1 &= a_2 \\ \end{align*} $$

From this we see that \(f\) is one-to-one or injective. To see that it’s surjective, we need to show that for any element \(y \in bH\), that there exists an element \(x \in aH\) such that \(f(x) = y\). Since \(y \in bH\), then we can write \(y = bh\) for some \(h \in H\). Now, let \(x = ah \in aH\) and observe that

$$ \begin{align*} f(x) &= ba^{-1}x \\ &= ba^{-1}(ah) \\ &= b(a^{-1}ah) \\ &= bh = y. \end{align*} $$

\(ah\) is in \(aH\) by definition. Therefore \(f\) is surjective. From this we see that \(f\) is a bijection from \(aH\) to \(bH\). \(\ \blacksquare\)

References