Find all subgroups of $\mathbb{Z}_3\oplus \mathbb{Z}_3\oplus \mathbb{Z}_3\oplus \mathbb{Z}_3$ Let $G$ denote the group $\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus \mathbb{Z}_3\oplus\mathbb{Z}_3$, try to find all subgroups of it
I think it may be quite similar to the extension of finite fields since $\mathbb{Z}_3$ is also a finite field, so we can regard $G$ as a four dimensional vector space over $\mathbb{Z}_3$, so all subgroups are exactly subspaces?
I wonder whether it’s correct or false, and how to use it to solve this problem.
 A: You're correct that every additive subgroup of $G$ is also a vector space over $\Bbb Z_3$.  That's because the only non-zero scalars are $1$ and $2=1+1$, so closure under addition implies closure under scalar multiplication, and it's not hard to verify that the necessary distributive properties also hold.
Each non-zero element $g \in G$ defines a subspace of dimension $1$ and a subspace of dimension $3$.  The latter subspace is the set of elements perpendicular to $g$ using the obvious dot product.  Both $g$ and $2g$ define the same two subspaces, so there are $40$ subgroups of order $3$ and $40$ subgroups of order $27$.
Now choose $g, h \in G$ such that $S= \{ g, h \}$ is linearly independent.  There are $\frac{80 \cdot 78}{2}=3120$ such subsets  Then $S$ defines a subgroup of order $9$.  However, each subgroup of order $9$ has $\frac{8 \cdot 6}{2} = 24$ different bases, so there are $130$ different subgroups of order $9$.  Thus, $G$ has a total of $210$ non-trivial proper subgroups.  Listing them is left as an exercise for the reader.
