Corollary 5

*Let $h$ be a subalgebra of the Lie algebra $g$ and $U'$ its enveloping algebra. Suppose that the $K$-modules $h$ and $g/h$ are free (for example if $K$ is a field). Let $(x_\alpha)_{\alpha \in L}$ be a basis of $h$ and $(y_\beta)_{\beta \in M}$ a family of elements of $g$ whose canonical images in $g/h$ form a basis of $g/h$.

(a) The canonical homomorphism of $U'$ into $U$ is injective.

(b) If $M$ is totally ordered, the elements $y_{\beta_1} \ldots y_{\beta_q}$, where $\beta_1 \leq \cdots \leq \beta_q$, form a basis of $U$ considered as a left or right module over $U'$.

We give $L \cup M$ a total ordering such that every element of $L$ is less than every element of $M$. The elements $x_{\alpha_1} x_{\alpha_2} \ldots x_{\alpha_p}$ calculated in $U'$ (where $\alpha_1 \leq \cdots \leq \alpha_p$) form a basis of $U'$ (Corollary 3). The elements

calculated in $U$ (where $\alpha_1 \leq \cdots \leq \alpha_p \leq \beta_1 \leq \cdots \leq \beta_q$) similarly form a basis of $U$. Hence the canonical homomorphism of $U'$ into $U$ maps the elements of a basis of $U'$ to linearly independent elements of $U$ and is therefore injective. It is moreover seen that the $y_{\beta_1} \ldots y_{\beta_q}$ (where $\beta_1 \leq \cdots \leq \beta_q$) form a basis of $U$ considered as a left $U'$-module. Ordering $L \cup M$ so that every element of $M$ is less than every element of $L$, it is similarly seen that the $y_{\beta_1} \ldots y_{\beta_q}$ (where $\beta_1 \leq \cdots \leq \beta_q$) form a basis of $U$ considered as a right $U'$-module.

Under the conditions of Corollary 5, $U'$ is identified with the subalgebra of $U$ generated by $h$ by means of the canonical homomorphism of $U'$ into $U$.