# Semisimple lie algebra has semisimple ideals

Let $g$ be a semisimple Lie Algebra, and let $h \subset g$ an ideal in $g$. Show that $h$ is semisimple.

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What have you tried? –  draks ... Jul 19 '12 at 18:47
Nothing with much success... I was thinking suppose that $i$ is an ideal of $h$, then let $a$ be an abelian ideal of $h/i$; maybe this could correspond to some abelian ideal of $g/h$? (since g/h is semi-simple) –  bijection Jul 23 '12 at 16:37
Hint: Use Cartan Killing criterion and the fact $\langle\cdot,\cdot\rangle_\mathfrak{h}=\langle\cdot,\cdot\rangle|_\mathfrak{h}$‌​, where $\langle\cdot,\cdot\rangle_\mathfrak{h}$ is the Cartan Killing form of $\mathfrak{h}$ and $\langle\cdot,\cdot\rangle|_\mathfrak{h}$ is the Cartan Killing form of $\mathfrak{g}$ restrict to $\mathfrak{h}$. –  Yuki Jul 24 '12 at 16:30
@Yuki How do you directly show that $\langle\cdot,\cdot\rangle$ is nondegenerate then $\langle\cdot,\cdot\rangle_{|\mathfrak{h}}$ is nondegenerate? –  mez Jan 7 at 1:56
@mezhang Let $B=\{X_1,\cdots,X_m,Y_{m+1},\cdots,X_n\}$ be a basis of $\mathfrak{g}$, with $X_1,\cdots,X_m$ a basis of $\mathfrak{h}$. Since $\langle\cdot,\cdot\rangle$ is nondegenerate, for each nonzero $X\in\mathfrak{h}$, there exists at least one $Z\in B$ such that $\langle X,Z\rangle\ne0$ (right?). We can calculate $\langle X,X_i\rangle$ directly, computing $Z_j:=\text{ad}(X)\text{ad}(X_i)X_j$, for $j=1,\cdots,n$, and summing the coeficient of $X_j$ of $Z_j$. But, since $\mathfrak{h}$ is an ideal, $Z_j\in h$ and the coefficient of $X_j$ of $Z_j$ is $0$ for $j>m$. So $Z\in\mathfrak{h}$. –  Yuki Feb 3 at 13:07