[proofplan] We prove both directions. If $I$ is an ideal, then changing representatives of two cosets changes the bracket by terms lying in $I$, so the displayed formula is independent of representatives. Once well-defined, bilinearity, alternation, and the Jacobi identity descend directly from the corresponding Lie algebra identities in $\mathfrak g$. Conversely, if the displayed formula is well-defined, then the zero coset condition forces $[x,i]_{\mathfrak g} \in I$ for all $x \in \mathfrak g$ and $i \in I$, which is exactly the ideal condition. [/proofplan]

[step:Show that the bracket on cosets is independent of representatives when $I$ is an ideal]Assume first that $I \trianglelefteq \mathfrak g$. Define \begin{align*} \beta: (\mathfrak g/I) \times (\mathfrak g/I) &\to \mathfrak g/I \\ (x+I,y+I) &\mapsto [x,y]_{\mathfrak g}+I. \end{align*} We verify that $\beta$ is well-defined. Let $x,y \in \mathfrak g$, and suppose $x',y' \in \mathfrak g$ are other representatives of the same cosets. Then there exist $i,j \in I$ such that \begin{align*} x' = x+i, \qquad y' = y+j. \end{align*} Using bilinearity of $[\cdot,\cdot]_{\mathfrak g}$ in $\mathfrak g$, we compute \begin{align*} [x',y']_{\mathfrak g} - [x,y]_{\mathfrak g} &= [x+i,y+j]_{\mathfrak g} - [x,y]_{\mathfrak g} \\ &= [x,y]_{\mathfrak g} + [x,j]_{\mathfrak g} + [i,y]_{\mathfrak g} + [i,j]_{\mathfrak g} - [x,y]_{\mathfrak g} \\ &= [x,j]_{\mathfrak g} + [i,y]_{\mathfrak g} + [i,j]_{\mathfrak g}. \end{align*} Since $I$ is an ideal, $[x,j]_{\mathfrak g} \in I$ and $[i,j]_{\mathfrak g} \in I$, because $j \in I$ and $x,i \in \mathfrak g$. Also $[y,i]_{\mathfrak g} \in I$ because $i \in I$ and $y \in \mathfrak g$, and alternation gives $[i,y]_{\mathfrak g} = -[y,i]_{\mathfrak g}$, so $[i,y]_{\mathfrak g} \in I$ since $I$ is a linear subspace. Because $I$ is closed under addition, their sum lies in $I$. Hence \begin{align*} [x',y']_{\mathfrak g}+I = [x,y]_{\mathfrak g}+I. \end{align*} Therefore $\beta$ is independent of representatives.[/step]

[guided]Assume $I \trianglelefteq \mathfrak g$. We want the rule \begin{align*} (x+I,y+I) \mapsto [x,y]_{\mathfrak g}+I \end{align*} to define a genuine function on cosets. The issue is that the same coset $x+I$ has many representatives, so we must check that replacing $x$ and $y$ by different representatives does not change the resulting coset. Define \begin{align*} \beta: (\mathfrak g/I) \times (\mathfrak g/I) &\to \mathfrak g/I \\ (x+I,y+I) &\mapsto [x,y]_{\mathfrak g}+I. \end{align*} Let $x',y' \in \mathfrak g$ be representatives of the same cosets as $x,y \in \mathfrak g$. By the definition of equality in the quotient [vector space](/page/Vector%20Space), there exist $i,j \in I$ such that \begin{align*} x' = x+i, \qquad y' = y+j. \end{align*} Now we compare the two possible bracket values. Bilinearity of the Lie bracket in $\mathfrak g$ gives \begin{align*} [x',y']_{\mathfrak g} &= [x+i,y+j]_{\mathfrak g} \\ &= [x,y]_{\mathfrak g} + [x,j]_{\mathfrak g} + [i,y]_{\mathfrak g} + [i,j]_{\mathfrak g}. \end{align*} Subtracting $[x,y]_{\mathfrak g}$, we get \begin{align*} [x',y']_{\mathfrak g} - [x,y]_{\mathfrak g} = [x,j]_{\mathfrak g} + [i,y]_{\mathfrak g} + [i,j]_{\mathfrak g}. \end{align*} Each term on the right lies in $I$, but the middle term requires one extra use of alternation. The first term $[x,j]_{\mathfrak g}$ lies in $I$ because $x \in \mathfrak g$ and $j \in I$. The third term $[i,j]_{\mathfrak g}$ lies in $I$ because $i \in I \subset \mathfrak g$ and $j \in I$. For the middle term, the ideal condition gives $[y,i]_{\mathfrak g} \in I$ because $y \in \mathfrak g$ and $i \in I$; alternation gives $[i,y]_{\mathfrak g} = -[y,i]_{\mathfrak g}$, and closure of the subspace $I$ under scalar multiplication gives $[i,y]_{\mathfrak g} \in I$. Since $I$ is a linear subspace, the sum lies in $I$. Therefore \begin{align*} [x',y']_{\mathfrak g} - [x,y]_{\mathfrak g} \in I, \end{align*} which is exactly the condition that \begin{align*} [x',y']_{\mathfrak g}+I = [x,y]_{\mathfrak g}+I. \end{align*} Thus $\beta$ is well-defined on cosets.[/guided]

[step:Verify that the descended bracket satisfies the Lie algebra axioms] We now write $[\cdot,\cdot]_{\mathfrak g/I}$ for the well-defined map $\beta$. Let $a,b \in k$ and let $x_1,x_2,y \in \mathfrak g$. Bilinearity in the first variable follows from \begin{align*} [\,a(x_1+I)+b(x_2+I),\,y+I\,]_{\mathfrak g/I} &= [\,ax_1+bx_2+I,\,y+I\,]_{\mathfrak g/I} \\ &= [ax_1+bx_2,y]_{\mathfrak g}+I \\ &= a[x_1,y]_{\mathfrak g}+b[x_2,y]_{\mathfrak g}+I \\ &= a[x_1+I,y+I]_{\mathfrak g/I} + b[x_2+I,y+I]_{\mathfrak g/I}. \end{align*} Let $a,b \in k$ and let $x,y_1,y_2 \in \mathfrak g$. Bilinearity in the second variable follows from \begin{align*} [\,x+I,\,a(y_1+I)+b(y_2+I)\,]_{\mathfrak g/I} &= [\,x+I,\,ay_1+by_2+I\,]_{\mathfrak g/I} \\ &= [x,ay_1+by_2]_{\mathfrak g}+I \\ &= a[x,y_1]_{\mathfrak g}+b[x,y_2]_{\mathfrak g}+I \\ &= a[x+I,y_1+I]_{\mathfrak g/I} + b[x+I,y_2+I]_{\mathfrak g/I}. \end{align*} For every $x \in \mathfrak g$, alternation in $\mathfrak g$ gives \begin{align*} [x+I,x+I]_{\mathfrak g/I} = [x,x]_{\mathfrak g}+I = 0+I. \end{align*} Finally, for all $x,y,z \in \mathfrak g$, the Jacobi identity in $\mathfrak g$ gives \begin{align*} &[x+I,[y+I,z+I]_{\mathfrak g/I}]_{\mathfrak g/I} +[y+I,[z+I,x+I]_{\mathfrak g/I}]_{\mathfrak g/I} +[z+I,[x+I,y+I]_{\mathfrak g/I}]_{\mathfrak g/I} \\ &= [x,[y,z]_{\mathfrak g}]_{\mathfrak g}+I + [y,[z,x]_{\mathfrak g}]_{\mathfrak g}+I + [z,[x,y]_{\mathfrak g}]_{\mathfrak g}+I \\ &= \bigl([x,[y,z]_{\mathfrak g}]_{\mathfrak g} + [y,[z,x]_{\mathfrak g}]_{\mathfrak g} + [z,[x,y]_{\mathfrak g}]_{\mathfrak g}\bigr)+I \\ &= 0+I. \end{align*} Thus $[\cdot,\cdot]_{\mathfrak g/I}$ is a Lie algebra bracket on $\mathfrak g/I$. [/step]

[step:Recover the ideal condition from well-definedness of the quotient formula]Conversely, assume that the displayed formula defines a Lie algebra bracket on $\mathfrak g/I$. In particular, the formula is well-defined. Let $x \in \mathfrak g$ and $i \in I$. Since $i+I = 0+I$ in $\mathfrak g/I$, well-definedness of the formula implies \begin{align*} [x,i]_{\mathfrak g}+I = [x+I,i+I]_{\mathfrak g/I} = [x+I,0+I]_{\mathfrak g/I} = [x,0]_{\mathfrak g}+I = 0+I. \end{align*} Therefore $[x,i]_{\mathfrak g} \in I$. Since $x \in \mathfrak g$ and $i \in I$ were arbitrary, $[\mathfrak g,I]_{\mathfrak g} \subset I$, so $I$ is an ideal of $\mathfrak g$.[/step]

[guided]Now assume the formula already defines a Lie algebra bracket on $\mathfrak g/I$. The important consequence is not any deep Lie algebra identity; it is simply that the formula is well-defined. Equal inputs in the quotient must produce equal outputs. Let $x \in \mathfrak g$ and $i \in I$. In the quotient vector space, the element $i+I$ is the zero coset because $i \in I$. Hence \begin{align*} i+I = 0+I. \end{align*} Since the bracket formula is well-defined on cosets, replacing the second representative $i$ by the representative $0$ of the same coset cannot change the resulting coset. Therefore \begin{align*} [x,i]_{\mathfrak g}+I = [x+I,i+I]_{\mathfrak g/I} = [x+I,0+I]_{\mathfrak g/I} = [x,0]_{\mathfrak g}+I. \end{align*} Bilinearity of the bracket in $\mathfrak g$ gives $[x,0]_{\mathfrak g}=0$, so \begin{align*} [x,i]_{\mathfrak g}+I = 0+I. \end{align*} Equality with the zero coset means precisely that $[x,i]_{\mathfrak g} \in I$. Since $x \in \mathfrak g$ and $i \in I$ were arbitrary, we have shown \begin{align*} [\mathfrak g,I]_{\mathfrak g} \subset I. \end{align*} This is the ideal condition for a Lie algebra subspace. Thus $I \trianglelefteq \mathfrak g$.[/guided]

[step:Conclude the equivalence] The first two steps show that if $I$ is an ideal of $\mathfrak g$, then the displayed formula defines a Lie algebra structure on $\mathfrak g/I$. The third step shows that if the displayed formula defines a Lie algebra structure on $\mathfrak g/I$, then $I$ is an ideal of $\mathfrak g$. Therefore the quotient bracket is well-defined and makes $\mathfrak g/I$ a Lie algebra if and only if $I \trianglelefteq \mathfrak g$. [/step]