[proofplan] We first verify that $\mathfrak a + I$ is closed under the [vector space](/page/Vector%20Space) operations and under the Lie bracket, using that $\mathfrak a$ is a subalgebra and $I$ is an ideal of $\mathfrak g$. We then check the two ideal assertions directly from the definitions. Finally, we construct the canonical quotient map from $\mathfrak a/(\mathfrak a \cap I)$ to $(\mathfrak a + I)/I$ and prove that it is a well-defined bijective Lie algebra homomorphism. [/proofplan] [step:Show that $\mathfrak a + I$ is a Lie subalgebra of $\mathfrak g$] Since $\mathfrak a$ and $I$ are $k$-linear subspaces of $\mathfrak g$, their sum $\mathfrak a + I$ is a $k$-linear subspace of $\mathfrak g$. It remains to prove closure under the Lie bracket. Let $x,y \in \mathfrak a$ and let $u,v \in I$. By bilinearity of the Lie bracket on $\mathfrak g$, \begin{align*} [x+u,y+v] = [x,y] + [x,v] + [u,y] + [u,v]. \end{align*} Because $\mathfrak a$ is a Lie subalgebra of $\mathfrak g$, we have $[x,y] \in \mathfrak a$. Because $I \trianglelefteq \mathfrak g$, we have $[x,v] \in I$, $[u,y] \in I$, and $[u,v] \in I$. Hence \begin{align*} [x+u,y+v] \in \mathfrak a + I. \end{align*} Therefore $\mathfrak a + I$ is a Lie subalgebra of $\mathfrak g$. [guided] To prove that $\mathfrak a + I$ is a Lie subalgebra, we must verify two things: it is a vector subspace of $\mathfrak g$, and it is closed under the Lie bracket of $\mathfrak g$. First, since $\mathfrak a$ and $I$ are both $k$-linear subspaces of $\mathfrak g$, their sum \begin{align*} \mathfrak a + I = \{x + u \in \mathfrak g : x \in \mathfrak a,\ u \in I\} \end{align*} is a $k$-linear subspace. Indeed, if $x_1,x_2 \in \mathfrak a$, $u_1,u_2 \in I$, and $\lambda,\mu \in k$, then \begin{align*} \lambda(x_1+u_1)+\mu(x_2+u_2) = (\lambda x_1+\mu x_2)+(\lambda u_1+\mu u_2), \end{align*} where $\lambda x_1+\mu x_2 \in \mathfrak a$ and $\lambda u_1+\mu u_2 \in I$. Now take arbitrary elements $x+u,y+v \in \mathfrak a + I$, where $x,y \in \mathfrak a$ and $u,v \in I$. The Lie bracket is bilinear, so \begin{align*} [x+u,y+v] = [x,y] + [x,v] + [u,y] + [u,v]. \end{align*} The term $[x,y]$ belongs to $\mathfrak a$ because $\mathfrak a$ is a Lie subalgebra. The remaining three terms belong to $I$ because $I$ is an ideal of $\mathfrak g$ and $x,y,u,v$ are all elements of $\mathfrak g$. Thus the whole bracket is the sum of one element of $\mathfrak a$ and one element of $I$, so \begin{align*} [x+u,y+v] \in \mathfrak a + I. \end{align*} This proves that $\mathfrak a + I$ is closed under the Lie bracket, and therefore it is a Lie subalgebra of $\mathfrak g$. [/guided] [/step] [step:Verify that $I$ and $\mathfrak a \cap I$ are ideals in the required Lie algebras] Since $I \subset \mathfrak a + I$, the set $I$ is a $k$-linear subspace of $\mathfrak a + I$. Let $z \in \mathfrak a + I$ and $u \in I$. Since $z \in \mathfrak g$ and $I \trianglelefteq \mathfrak g$, we have $[z,u] \in I$. Hence $I \trianglelefteq \mathfrak a + I$. Next, $\mathfrak a \cap I$ is a $k$-linear subspace of $\mathfrak a$. Let $x \in \mathfrak a$ and let $w \in \mathfrak a \cap I$. Since $\mathfrak a$ is a Lie subalgebra and $x,w \in \mathfrak a$, we have $[x,w] \in \mathfrak a$. Since $I \trianglelefteq \mathfrak g$, $x \in \mathfrak g$, and $w \in I$, we also have $[x,w] \in I$. Therefore \begin{align*} [x,w] \in \mathfrak a \cap I. \end{align*} Thus $\mathfrak a \cap I \trianglelefteq \mathfrak a$. [/step] [step:Define the canonical quotient map and prove it is well-defined] Define a map \begin{align*} \Phi: \mathfrak a/(\mathfrak a \cap I) &\to (\mathfrak a + I)/I \\ x + (\mathfrak a \cap I) &\mapsto x + I, \end{align*} where $x$ ranges over $\mathfrak a$. We prove that $\Phi$ is well-defined. Let $x,y \in \mathfrak a$ satisfy \begin{align*} x + (\mathfrak a \cap I) = y + (\mathfrak a \cap I). \end{align*} Then $x-y \in \mathfrak a \cap I$, hence $x-y \in I$. Therefore \begin{align*} x + I = y + I \end{align*} in $(\mathfrak a + I)/I$. Thus $\Phi$ is well-defined. [guided] The quotient element $x + (\mathfrak a \cap I)$ can be represented by many different elements of $\mathfrak a$, so the first task is to check that the formula for $\Phi$ does not depend on the chosen representative. Define \begin{align*} \Phi: \mathfrak a/(\mathfrak a \cap I) &\to (\mathfrak a + I)/I \\ x + (\mathfrak a \cap I) &\mapsto x + I, \end{align*} where $x \in \mathfrak a$. This formula makes sense because $x \in \mathfrak a \subset \mathfrak a + I$, so $x + I$ is a coset in $(\mathfrak a + I)/I$. Suppose $x,y \in \mathfrak a$ represent the same coset in $\mathfrak a/(\mathfrak a \cap I)$. By the definition of equality in a quotient vector space, \begin{align*} x + (\mathfrak a \cap I) = y + (\mathfrak a \cap I) \end{align*} means exactly that \begin{align*} x-y \in \mathfrak a \cap I. \end{align*} In particular, $x-y \in I$. Therefore $x$ and $y$ define the same coset modulo $I$: \begin{align*} x + I = y + I. \end{align*} So $\Phi$ is independent of the representative and is well-defined. [/guided] [/step] [step:Show that the canonical quotient map is a Lie algebra homomorphism] Let $x,y \in \mathfrak a$ and let $\lambda,\mu \in k$. Using the vector space operations in the quotient Lie algebras, \begin{align*} \Phi\bigl(\lambda(x+\mathfrak a \cap I)+\mu(y+\mathfrak a \cap I)\bigr) &= \Phi\bigl((\lambda x+\mu y)+(\mathfrak a \cap I)\bigr) \\ &= (\lambda x+\mu y)+I \\ &= \lambda(x+I)+\mu(y+I) \\ &= \lambda\Phi(x+\mathfrak a \cap I)+\mu\Phi(y+\mathfrak a \cap I). \end{align*} Hence $\Phi$ is $k$-linear. For the bracket, the quotient bracket is induced by the bracket on the ambient Lie algebra. Therefore \begin{align*} \Phi\left([x+(\mathfrak a \cap I),y+(\mathfrak a \cap I)]\right) &= \Phi\left([x,y]+(\mathfrak a \cap I)\right) \\ &= [x,y]+I \\ &= [x+I,y+I] \\ &= [\Phi(x+\mathfrak a \cap I),\Phi(y+\mathfrak a \cap I)]. \end{align*} Thus $\Phi$ is a Lie algebra homomorphism. [/step] [step:Prove that the canonical quotient map is bijective] We first prove injectivity. Let $x \in \mathfrak a$ and suppose \begin{align*} \Phi(x+(\mathfrak a \cap I)) = I, \end{align*} where $I$ denotes the zero coset in $(\mathfrak a+I)/I$. By the definition of $\Phi$, this means \begin{align*} x+I = I. \end{align*} Hence $x \in I$. Since also $x \in \mathfrak a$, we have $x \in \mathfrak a \cap I$, so \begin{align*} x+(\mathfrak a \cap I) = \mathfrak a \cap I. \end{align*} Thus $\ker \Phi = \{\mathfrak a \cap I\}$, and $\Phi$ is injective. We next prove surjectivity. Let $z+I \in (\mathfrak a+I)/I$, where $z \in \mathfrak a+I$. By the definition of $\mathfrak a+I$, there exist $x \in \mathfrak a$ and $u \in I$ such that \begin{align*} z = x+u. \end{align*} Then \begin{align*} z+I = (x+u)+I = x+I = \Phi(x+(\mathfrak a \cap I)). \end{align*} Therefore $\Phi$ is surjective. Since $\Phi$ is a bijective Lie algebra homomorphism, it is an isomorphism of Lie algebras. Consequently, \begin{align*} \mathfrak a/(\mathfrak a \cap I) \cong (\mathfrak a + I)/I. \end{align*} This completes the proof. [/step]