Theorem
Prove that if \(H\) is a subgroup of a group \(G\) and \(S\) is the set of all distinct left cosets of \(H\) in \(G\), then there exists a homomorphism \(\psi: G \to A(S)\) such that \(\ker \psi \subseteq H\).
Proof
Step 1
Define the Mapping
Let \(S = \{gH : g \in G\}\) be the set of all left cosets of \(H\) in \(G\). Define \(\psi: G \to A(S)\) by
\[
\psi(g)(xH) = (gx)H \quad \text{for all } xH \in S.
\]
Step 2
To prove the mapping is well-defined
For each \(g \in G\), the function \(\psi(g)\) sends any coset \(xH\) to \((gx)H\). Since left multiplication by \(g\) is a bijection on \(G\), the induced function on the set of cosets is well-defined and a permutation of \(S\).
Step 3
To show that \(\psi\) is a homomorphism
For any \(g_1, g_2 \in G\) and for all \(xH \in S\), we compute:
\[
\begin{align}
\psi(g_1g_2)(xH) &= \big((g_1g_2)x\big)H \nonumber\\
&= g_1\,(g_2x)H \nonumber\\
&= \psi(g_1)\big((g_2x)H\big) \nonumber\\
&= (\psi(g_1) \circ \psi(g_2))(xH)\nonumber
\end{align}
\]
Hence, \(\psi(g_1g_2) = \psi(g_1) \circ \psi(g_2)\), so \(\psi\) is a group homomorphism.
Step 4
To determine the Kernel of \(\psi\)
The kernel of \(\psi\) is defined as:
\[
\ker \psi = \{g \in G \mid \psi(g)(xH) = xH \text{ for all } xH \in S\}.
\]
For \(g \in \ker \psi\) and for every \(x \in G\), we have:
\[
\psi(g)(xH) = (gx)H = xH. \tag{5}
\]
In particular, taking \(x = e\) (the identity in \(G\)), we obtain:
\[
(g \cdot e)H = gH = eH = H. \tag{6}
\]
Thus, \(gH = H\) and so \(g \in H\). Therefore, \(\ker \psi \subseteq H\).
