# Is the (un-)even part of the exterior algebra an (anti-)commutative sub-algebra?

It is well known that the exterior algebra of an $$n$$-dimensional vector space $$V$$ has the following decomposition: $$\Lambda V=\bigoplus_{k=0}^n\Lambda^kV=\bigoplus_{\text{even }k}\Lambda^kV\oplus \bigoplus_{\text{uneven }k}\Lambda^kV$$ I am wondering if the (un-)even part of the exterior algebra is a (anti-)commutative sub-algebra. It is clear that it is a subalgebra$$^1$$, but I'm not sure if is really (anti-)commutative, since I would be surprised if this was true and nobody told me...Here's my attempt of a proof:

Proof: Suppose $$v_1,\ldots,v_k,w_1,\ldots,w_l\in V$$ and consider $$v:=v_1\wedge\cdots\wedge v_k$$ and $$w:=w_1\wedge\cdots w_l$$. Because of bilinearity of the wedge-product it suffices to show $$v\wedge w=\begin{cases}w\wedge v&\text{if k and l are both even}\\-w\wedge v&\text{if k and l are both uneven}\end{cases}$$ Case 1: $$\{v_1,\ldots,v_k\}\cap\{w_1,\ldots,w_l\}\neq\emptyset$$

In this case, $$v\wedge w=w\wedge v=-w\wedge v=0$$.

Case 2: $$\{v_1,\ldots,v_k\}\cap\{w_1,\ldots,w_l\}=\emptyset$$

In this case, $$v\wedge w=(-1)^{k\cdot l}w\wedge v$$

$$^1$$ This is only true for the even part. As was pointed out in the answer, the odd part is NOT a subalgebra since $$\Lambda^k\wedge\Lambda^l\subset\Lambda^{k+l}$$

The odd (what you call uneven) part of the exterior algebra is not a subalgebra as it is not closed under wedge product: the wedge of two odd forms is an even form. Moreover, it isn't even a subspace of the exterior algebra because it doesn't contain $$0$$ (which has degree $$0$$).
On the other hand, the even part of the exterior algebra is a subalgebra. Moreover, for $$v \in \bigwedge^kV$$ and $$w \in \bigwedge^lV$$, we have $$v\wedge w = (-1)^{kl}w\wedge v$$, so if $$k$$ and $$l$$ are even, we see that $$v\wedge w = w\wedge v$$. It follows by linearity that $$\bigwedge^{\text{even}}V$$ is a commutative subalgebra of $$\bigwedge^*V$$.