If $\frak{g}$ = $span\lbrace{e_{1},e_{2}}\rbrace$ is a Lie algebra such that $[e_{1},e_{2}] = e_{1}$, show that $Der(\frak{g})$ = $ad(\frak{g})$.

Question

If $\frak{g}$ = $span\lbrace{e_{1},e_{2}}\rbrace$ is a Lie algebra such that $[e_{1},e_{2}] = e_{1}$, show that $Der(\frak{g})$ = $ad(\frak{g})$.

Could someone give me a suggestion to solve this problem?

Answer 1

This is linear algebra. We can write $D(e_i)=a_{i1}e_1+a_{i2}e_2$ for $i=1,2$, since $(e_1,e_2)$ is a basis. Then, by definition of a derivation $$ D([e_i,e_j])=[D(e_i),e_j]+[e_i,D(e_j)]=[a_{i1}e_1+a_{i2}e_2,e_j]+[e_i,a_{j1}e_1+a_{j2}e_2] $$ Expand this to the basis and compare coefficients. We may assume that $i=1$ and $j=2$. Then the claim follows.