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Prove that a Lie algebra g is associative iff the derived subalgebra of g is contained in the centre of g, that is $g^{(1)} \subset c(g)$.

So we have the derived sub algebra is in the centre of g. We need to show that $(uv)w=u(vw)$.

So take $u,v,w \in g$. Since $[u,[v,w]]=0$. I was thinking that you need to use jacobian identity on that. However, I'm stuck on what to do next.

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In the "if" direction, isn't $[u,[v,w]]=[[u,v],w]$ trivially true because each side is, separately, zero? You will need the Jacobi identity for the "only if" direction. – Henning Makholm Apr 4 '12 at 0:15
@HenningMakholm Yes, I've got that. I was thinking that u(vw)=0. I suppose. Like from the other I should get (uv)w=0. So they equal as they are 0? – simplicity Apr 4 '12 at 0:22
Yes, if $a=0$ and $b=0$ then certainly $a=b$. (For $(uv)w$ you just do $(uv)w = -w(uv) = -0 = 0$). – Henning Makholm Apr 4 '12 at 0:29
up vote 1 down vote accepted

You have one direction nearly finished:

The Jacobi identity says that $[u,[v,w]]+[v,[w,u]]+[w,[u,v]]=0$ thus $[u,[v,w]]=-[w,[u,v]]-[v,[w,u]]$. By skew-symmetry and linearity, $[u,[v,w]]=[[u,v],w]+[v,-[w,u]]$ then skew symmetry again $[u,[v,w]]=[[u,v],w]+[v,[u,w]]$. This is the derivation form of the Jacobi identity (which is usually more useful when proving things or thinking about properties). Notice this looks like the product rule for derivatives: $\frac{d}{dx}\left[vw\right] = \frac{dv}{dx}w+v\frac{dw}{dx}$.

Now your proposition follows since $[u,w]$ is in the derived subalgebra and thus also the center, $[v,[u,w]]=0$. Therefore, $[u,[v,w]]=[[u,v],w]$. Thus $\mathfrak{g}$ is associative.

For the other direction, suppose that $\mathfrak{g}$ is associative. Let $\sum\limits_{i=1}^n [a_i,b_i] \in \mathfrak{g}^{(1)}$. Then for $v\in\mathfrak{g}$,

By linearity and the derivation form of the Jacobi identity, $$ \left[v,\sum\limits_{i=1}^n [a_i,b_i]\right] = \sum\limits_{i=1}^n [v,[a_i,b_i]] = \sum\limits_{i=1}^n \left([[v,a_i],b_i]+[a_i,[v,b_i]] \right)$$

Now use associativity, skew-symmetry, and linearity $$ = \sum\limits_{i=1}^n \left([[v,a_i],b_i]+[[a_i,v],b_i] \right) = \sum\limits_{i=1}^n \left([[v,a_i],b_i]+[-[v,a_i],b_i] \right) = \sum\limits_{i=1}^n \left([[v,a_i],b_i]-[[v,a_i],b_i] \right) = 0$$

Thus $\sum\limits_{i=1}^n [a_i,b_i] \in \mathfrak{z}(\mathfrak{g})$ (the center of $\mathfrak{g}$).

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I was thinking when it means associative, that it means that the elements are associative i.e. $u(vw)=(uv)w$. Not that [,] are associative. – simplicity Apr 4 '12 at 0:28
Ah, but if $\mathfrak{g}$ is a Lie algebra (and that's all), the only product you have to work with is the bracket! $ab$ is meaningless unless it is interpreted as shorthand for $[a,b]$. If you assume $\mathfrak{g}$ is the Lie algebra formed by giving an associative algebra the commutator bracket, nothing can be said about the derived subalgebra being contained in the center. In fact, if this were true for Lie algebras coming from assoc. algebras, it would need to be true of all Lie algebras since every Lie algebra can be embedded in an assoc. algebra (its universal enveloping algebra). – Bill Cook Apr 4 '12 at 0:34
For "the other direction", isn't it enough to prove that $[a,[b,c]]=0$ and then just appeal to linearity to generalize that from $[b,c]$ to an arbitrary element of the derived algebra? Wouldn't have to drag that summation though everything, then. – Henning Makholm Apr 4 '12 at 0:49
@HenningMakholm Sure. Given simplicity seems to be quite new to this, I was trying to give a somewhat careful, detailed account. I certainly wasn't aiming toward efficiency. :) – Bill Cook Apr 4 '12 at 2:14

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