Proving the existence of exactly six subspaces of $(\mathbb{Z}/5\mathbb{Z})^2$ I am trying to prove that $(\mathbb{Z}/5\mathbb{Z})^2$ has exactly six one-dimensional subspaces.
$\mathbb{Z}/5\mathbb{Z}$ has $5$ elements, so $\left(\mathbb{Z}/5\mathbb{Z}\right)^2$ has $25$ elements, $24$ of which are non-zero. Every subspace must contain $(0,0)$, so I'm left with $24$ non-zero elements. Furthermore, if $W_1, W_2$ are distinct such subspaces and they share a non-zero element $v$ in common, then $W_1 = W_2$, as the subspaces are one-dimensional and closed under scalar multiplication.
This tells me that because $24/4 = 6$, there are at most six such subspaces, but it doesn't tell me that there are exactly six such subspaces. What am I missing?
 A: Every non-zero element is in a unique non-zero subspace. If $W_1,W_2,\dots, W_k$ are your non-zero proper subspace. Let $U_i=W_i\setminus \{0\}.$

*

*We know that each $|U_i|=4.$

*We know the union $U_1\cup\cdots\cup U_k=(\mathbb Z/5\mathbb Z)^2\setminus\{0\}.$

*We know when $i\neq j,$ $U_i\cap U_j=\emptyset.$
by the uniqueness assured in the first sentence.
Or you can find particular elements that generate different subspaces.
If $$W_{(x,y)}=\{(ax,ay)\mid a\in\mathbb Z/\mathbb Z\}$$
Show $$W_{(1,0)},W_{(1,1)},W_{(1,2)},W_{(1,3)},W_{(1,4)},W_{(0,1)}$$
are distinct subspaces.
Given any field $k,$ the non-zero proper subspaces are either $W_{(1,x)}$ for each $x\in k$ and $W_{(0,1)}.$
In particular, if $k$ is finite, then there are $|k|+1$ subspaces.
A: We can explicitly construct the six desired subspaces; for them to be distinct their elements must be distinct, and since each subspace must contain exactly four non-zero elements these subspaces will perfectly partition the $24$ non-zero elements of $\mathbb Z_5^2$.
These subspaces are explicitly given by the generators
$$(1,0),(0,1),(1,1),(1,2),(2,1),(1,4)$$
