What is the order of $\wedge^2(\mathbb Z/p\mathbb Z \oplus\mathbb Z/p\mathbb Z)$? I am learning wedge product and face a question to find out what is the wedge square of the group  $\mathbb Z/p\mathbb Z \oplus\mathbb Z/p\mathbb Z$, $$\wedge^2(\mathbb Z/p\mathbb Z \oplus\mathbb Z/p\mathbb Z)=?$$
Actually I have a solution saying it has order $p$. But I am wondering how to compute this. Any hint is appreciated.
 A: In general, if $V$ is a finite dimensional $k$-vector space with $n = \dim_k V$, one can show that the wedges $v_{i_1} \wedge v_{i_2} \wedge \cdots \wedge v_{i_k}$ for each $i_1 < \cdots i_k$ form a basis for $\bigwedge^kV$, hence $\dim \bigwedge^kV = {n \choose k}$.
In your example, you are dealing with a vector space over $\Bbb Z_p$ of dimension $2$ and $k = 2$, hence
$$
\dim_{\Bbb Z_p} \bigwedge^2 \Bbb Z_p \oplus \Bbb Z_p = {2 \choose 2} = 1
$$
and this implies $|\bigwedge^2 \Bbb Z_p \oplus \Bbb Z_p| = p$.
A: If the vector space setting is anything to go by, just expand in terms of a basis:
Write $a,b$ for the generators of $\mathbb Z/p \mathbb Z$.
Then, you can certainly expand any individual product, by linearity (coefficients all integers):
$$\begin{align}
x \wedge y &= (λ\, a + ν\, b)\wedge(\eta\,a + ρ\, b)
%
\\
&= λ\, a \wedge η \, a + λ\,a \wedge ρ \,b + ν \,β \wedge \eta\,a + ν\, b \wedge ρ\,b
\\
&=λη\, (a \wedge a) + λρ(a \wedge b) + νη(β \wedge a) + νρ(b \wedge b).
%
\end{align} $$
Now, if you apply the antisymmetry relation: the first and the last product must equal zero, and the third wedge is  $(-1)$ the second wedge. I.e., your product is really a multiple of $a \wedge b$.
Therefore (since this applies to sums of $x\wedge y$'s as well) $a \wedge b$ is a generator of the group.
As for its order:
$$n (a \wedge b) = (na) \wedge b = 0 \iff p |n,$$
as the order of $a$ is $p$. I.e., the order of the generator, hence the group, is $p$.
A: I'm assuming this $\wedge^2 ( \mathbb{Z}/p \oplus \mathbb{Z}/p)$ is the $k$th exterior power of a vector space, $\Lambda^k V^*$ being the alternating $k$-tensors on the vector space $V$.
Then $V = \mathbb{Z}/p \oplus \mathbb{Z}/p$ is a two dimensional $\mathbb{Z}/p$ vector space, so $\Lambda^2 V^*$ is $1$-dimensional, and is isomorphic to the ground field, $\mathbb{Z}/p$, which has order p.
We can see this by general considerations: when dim$V = n$, $\Lambda^k V^*$ has dimension $\binom{n}{k}$. Proving this isn't too hard. Giving $V^*$ basis $\epsilon^1 , \cdots, \epsilon^n$, the exterior power has a basis consisting of all wedge products $\epsilon^{i_1} \wedge \cdots \wedge \epsilon^{i_k}$, where the indices $i_1 < \cdots < i_k$ are increasing.
