[proofplan] We characterize an element $x \in G$ by whether it fixes every left coset $gH$. Fixing the coset $gH$ is equivalent to the conjugacy condition $g^{-1}xg \in H$, which is the same as $x \in gHg^{-1}$. Since this must hold for every $g \in G$, the kernel is exactly the intersection of all conjugates of $H$. [/proofplan] [step:Translate kernel membership into fixing every left coset] Let \begin{align*} K := \ker \lambda. \end{align*} For an element $x \in G$, by the definition of the kernel of the homomorphism $\lambda: G \to \operatorname{Sym}(G/H)$, we have \begin{align*} x \in K \iff \lambda(x) \text{ is the identity permutation on } G/H. \end{align*} The identity permutation on $G/H$ fixes every element of the set $G/H$. Since every element of $G/H$ has the form $gH$ for some $g \in G$, this is equivalent to \begin{align*} (xg)H = gH \quad \text{for every } g \in G. \end{align*} Thus \begin{align*} x \in K \iff (xg)H = gH \text{ for every } g \in G. \end{align*} [/step] [step:Convert fixed cosets into membership in every conjugate of $H$] Fix $x \in G$ and $g \in G$. We first record the coset equality criterion in this special case. The equality \begin{align*} (xg)H = gH \end{align*} holds if and only if $xg \in gH$, because two left cosets of $H$ are equal exactly when one representative belongs to the other coset. This condition is equivalent to the existence of an element $h \in H$ such that \begin{align*} xg = gh. \end{align*} Multiplying the equality $xg = gh$ on the left by $g^{-1}$ gives \begin{align*} g^{-1}xg = h. \end{align*} Therefore \begin{align*} (xg)H = gH \iff g^{-1}xg \in H. \end{align*} Equivalently, \begin{align*} g^{-1}xg \in H \iff x \in gHg^{-1}. \end{align*} Combining this with the preceding step gives \begin{align*} x \in K \iff x \in gHg^{-1} \text{ for every } g \in G. \end{align*} [guided] Fix $x \in G$ and one coset $gH \in G/H$. We want to translate the statement “$x$ fixes the coset $gH$” into a statement about membership in a conjugate of $H$. The element $x$ fixes $gH$ under the left coset action exactly when \begin{align*} \lambda(x)(gH)=gH. \end{align*} By the definition of $\lambda$, this means \begin{align*} (xg)H = gH. \end{align*} Now we analyze this equality of left cosets. The coset $(xg)H$ equals $gH$ if and only if the representative $xg$ lies in the coset $gH$. This means precisely that there exists an element $h \in H$ such that \begin{align*} xg = gh. \end{align*} Multiplying on the left by $g^{-1}$ gives \begin{align*} g^{-1}xg = h. \end{align*} Since $h \in H$, this proves \begin{align*} (xg)H = gH \implies g^{-1}xg \in H. \end{align*} Conversely, suppose that \begin{align*} g^{-1}xg \in H. \end{align*} Define $h \in H$ by \begin{align*} h := g^{-1}xg. \end{align*} Then $xg = gh$, so $xg \in gH$. Hence the left coset represented by $xg$ is the same as the left coset represented by $g$: \begin{align*} (xg)H = gH. \end{align*} We have proved the equivalence \begin{align*} (xg)H = gH \iff g^{-1}xg \in H. \end{align*} Finally, the condition $g^{-1}xg \in H$ can be rewritten by multiplying the inclusion relation through the conjugation map. It is equivalent to saying that $x$ has the form $ghg^{-1}$ for some $h \in H$, namely \begin{align*} x \in gHg^{-1}. \end{align*} Thus $x$ fixes the particular coset $gH$ if and only if $x$ lies in the particular conjugate subgroup $gHg^{-1}$. [/guided] [/step] [step:Intersect the conjugate conditions to identify the kernel] From the previous steps, for every $x \in G$, \begin{align*} x \in \ker \lambda \iff x \in gHg^{-1} \text{ for every } g \in G. \end{align*} By the definition of set-theoretic intersection, the right-hand condition is equivalent to \begin{align*} x \in \bigcap_{g \in G} gHg^{-1}. \end{align*} Using the definition \begin{align*} \operatorname{Core}_G(H) := \bigcap_{g \in G} gHg^{-1}, \end{align*} we obtain \begin{align*} x \in \ker \lambda \iff x \in \operatorname{Core}_G(H). \end{align*} Since this equivalence holds for every $x \in G$, the two subsets of $G$ are equal: \begin{align*} \ker \lambda = \operatorname{Core}_G(H). \end{align*} [/step]