# Center of a lie Algebra is an ideal

Let g be a lie algebra, show that the center of g is an ideal in g.

attempt at proof: Let $\ \mathfrak g$ be a lie algebra. Define $\mathfrak h$ to be the center of $\ \mathfrak g$. Suppose that $\kappa$ is a sub-algebra of the lie algebra $\ \mathfrak g$ with $\ \kappa \subseteq \mathfrak g$. Then Letting $H = Y$ in the definition of center yields $[X,H]$ which is an element of $\ \kappa$ thus $\ \mathfrak h$ is an ideal of $\ \mathfrak g$.

I'm sort of lost with how to properly construct this proof, I'm not quite sure if I starting out correctly or need to use another approach.

## Definition of Center

The Center of a Lie Algebra $\ \mathfrak g$ is the set of all $X \in \ \mathfrak g$ for which $[X,Y]=0$ for all $Y \in \ \mathfrak g$.

## Definition of Ideal

A subalgebra $\ \mathfrak h$ of a Lie Algebra $\ \mathfrak g$ is said to be an ideal in $\ \mathfrak g$ if $[X,H] \in \ \mathfrak h$ for all $X \in \ \mathfrak g$ and $H \in \ \mathfrak h$.

## 2 Answers

Your problem solving is a little confused. Here's how the proof should look:

Let $\mathfrak{h}$ be the center of $\mathfrak{g}$. Let $X \in \mathfrak{g}$ and $H \in \mathfrak{h}$. Then, $[X, H] =$ ______, which is in $h$, because _________. Thus, $\mathfrak{h}$ is an ideal.

Filling in the blanks will give you a complete proof.

• Why does $0 \in \mathfrak{h}$? – Hendrra Jan 18 '18 at 10:05
• @Hendrra Let $X$ be any element of our Lie Algebra. So, $[X, 0] = [X, 0 + 0] = [X, 0] + [X, 0]$, so $[X, 0] = 0$. – Duncan Ramage Jan 18 '18 at 16:37

First, clearly $0$ lies in the center of the lie Algebra. It seems that if $H$ is in the center of the Lie Algebra, then $[X,H]=0$ which is in the center of the Lie Algebra. Therefore, the center is an ideal by definition.