Solution.

\(S_3 = \{(1)\, ,\, (1\,2)\, , \, (1\, 3)\,,\, (2\,3)\,,\, (1\,2\,3)\,,\,(1\,3\,2)\}\).

Let cosets are all \(aH = \{ah:\, a\in S_n\) and \(h\in H\}\)

1.

\(a = (1) H = \{(1)\,,\,(1\,3)\} = H\)

2.

\(a = (1\,2) \implies \, (1\,2)H = \{(1\,2)\, , \, (1\,3\,2)\}\)

3.

\((1\,3)H = \{(1\,3)\, ,\, (1)\} = H \hspace {1cm} a = (1\,3)\)

4.

\((2\, 3)H = \{(2\, 3)\, , \, (1\, 2\, 3)\} \hspace {1cm} a = (2\, 3)\)

5.

\((1\, 2\,3)H = \{(1\, 2\, 3)\, , \, (2\, 3)\}\hspace {1cm} a = (1\, 2\, 3)\)

6.

\((1\, 3\, 2)H = \{(1\, 3\, 2)\, ,\, (1\, 2)\}\)

\[ 1 = 3\hspace {1cm} 2 = 6\hspace {1cm} 4 = 5\]

Note that there are only three distinct cosets of \(H\) in \(G\), since \begin {align*} (1) H & = (1\, 3)H\\ (1\, 2)H & = (1\, 3\,2)H\\ (2\,3)H & = (1\, 2\, 3)H\\\\ \end {align*}

Solution. \(\mathbb {Z}_9 = \{0, \, 1,\, 2,\, 3,\, 4,\, 5,\, 6, \, 7,\, 8\}\)

\(\mathbb {Z}_{9/3} = \{0,\, 3,\, 6\}\)

The left cosets are all \(a + H\) such that \(\, a\in \mathbb {Z}_9\)

Proof.

(i)

Assume \(a H = b H\), then we show that \(b^{-1}a\in H\) and assume \(b^{-1}aH\) and then show that \(aH = bH\).
Suppose \(a H = b H\), then \(ah_1 = bh_2\) for some \(h_1\, , \, h_2\in H\). Then \(a = bh_2h^{-1}_1 = bh_0\) where \(h_0 = h_2h^{-1}_1\in H\). Since \(a = bh_0\), then \(b^{-1}a = h_0\in H\).
Conversely, suppose \(b^{-1}a\in H\), we need to show that \(a H = b H\). We have that \(b^{-1}a\in H\) implies that \(a = bh_0\) for some \(h_0\in H\). If \(x\in aH\), then \(x = ah = bh_0\in bH\implies aH \subset bH\cdots \cdots (*)\). If \(y\in H\), then \(y = bh\implies y = ah^{-1}_0h\in aH\). (Since \(a = bh_0\)
\(\implies b = ah_0^{-1}).\, \implies bH\subset aH\cdots \cdots (**)\).
By \((*)\) and \((**),\, \, aH = bH\).

(ii)

If \(x\in aH \cap bH\), then \(x\in aH\) and \(x\in bH. \implies x = ah_1\), and \(x = bh_2\implies (ah_1 = bh_2)\) for some \(h_1\, , \, h_2\in H\). We get \(ah_1 = bh_2\implies b^{-1}a = h_2h_1^{-1}\in H\). So by (i) \(\, aH = bH\).

(iii)

To show that \(\begin {vmatrix} aH\\ \end {vmatrix} = \begin {vmatrix} H\\ \end {vmatrix}\), we need a bijection \(f: \, H \longrightarrow aH\). For each \(h\in H\), define \(f(h) \implies ah\). We need to show that \(f\) as defined is one-one and onto. Suppose
\(f(h_1) = f(h_2)\) in \(aH\). \begin {align*} \implies \, & ah_1 = ah_2\\ \implies \, & a^{-1}ah_1 = a^{-1}ah_2\\ \implies \, & h_1 = h_2\hspace {0.5cm} \text {Hence one-one} \end {align*}

Now for each element \(y\in aH\), there must be some \(h'\in H\) such that \(y = ah'\). Thus \(f(h') = ah' = y\). Hence we have that for onto. So that \(f\) is a bijection. We have \(\begin {vmatrix} aH\\ \end {vmatrix} = \begin {vmatrix} H\\ \end {vmatrix}\).