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 $\Lambda^k\wedge\Lambda^l\subset\Lambda^{k+l}$
Best Answer
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 $\bigwedge^*V$.