Subgroups of index $p$ in additive group $\mathbb{Z}^2$? How many subgroups of index $p$, where $p$ is a prime, are there in the additive group $\mathbb{Z}^2$?
 A: Let $G=\langle a,b\colon ab=ba\rangle\cong \mathbb{Z}^2$, and $\tilde{G}=\langle x\colon x^p=1\rangle\cong \mathbb{Z}/p$. 
If $H$ is a subgroup of index $p$, then we get a surjective homomorphism $G\rightarrow \tilde{G}$ (with kernel $H$). Conversely, every surjective homomorphism $G\rightarrow \tilde{G}$ gives a subgroup of index $p$ (namely kernel).
So, let us see how many surjective homomorphisms are there from $G$ to $\tilde{G}$. 
Any homomorphism $G\rightarrow \tilde{G}$ is uniquely determined by images of $a$ and $b$, so for $i,j\in\{0,1,\cdots,p-1\}$
$$a\mapsto x^i, \,\,\, b\mapsto x^j$$
uniquely determines a homomorphism from $G$ to $\tilde{G}$, and is surjective if and only if $(i,j)\neq (0,0)$. 
Thus, there are $p^2-1$ surjective homomorphisms from $G$ to $\tilde{G}$. They certainly give subgroup of index $p$, but not necessarily distinct.
If $f,g$ are two surjective homomorphisms from $G$ to $\tilde{G}$ with same kernel then $f$ and $g$ differ by an automorphism $\sigma \colon \tilde{G}\rightarrow \tilde{G}$ (i.e $f=\sigma\circ g$), and converse is also true, which can be easily checked (i.e if $f=\sigma\circ g$ where $\sigma$ is automorphism of $\tilde{G}$ then $f,g$ have same kernel).
Since $|Aut(\tilde{G})|=|Aut(\mathbb{Z}/p)|=p-1$, thus, the number of surjective homomorphisms $G\rightarrow \tilde{G}$ with different kernels is 
$$\frac{p^2-1}{p-1}=p+1.$$
This is the number of subgroups of index $p$ in $\mathbb{Z}^2$. They are 
$$\langle a^p,b\rangle, \langle a^p,ab\rangle, \langle a^p,a^2b\rangle, \cdots, \langle a^p,a^{p-1}b\rangle, \langle a,b^p\rangle.$$
A: More generally, the number of subgroups of index $n$ is $\sigma_1(n) = \sum_{d | n} d$. You can find a proof here (skip down to "some examples"). 
