Let p > 0 is prime. Let $G = \mathbb Z_{p} \oplus\mathbb Z_{p^2}\oplus \mathbb Z_{p^2}$. Find number of epimorphisms $\phi: \mathbb{F_3} $ $\rightarrow G$. Find number of subgroups $N$ of free group $\mathbb{F_3}$, such $\mathbb{F_3}/N \cong G$.
I've found number of decompositions of G in in direct sum of cyclic groups: $p^2(p^3+p^2-1)(p^3+p^2)$ (that is number of $A \oplus B \oplus C \cong G$, such that $A \cong \mathbb Z_{p}$, $B \cong \mathbb{Z}_{p^2}$, $C \cong \mathbb{Z}_{p^2}$. So I should look at generators of $\mathbb{F_3}$ and they should go to generators of G? How do i count it? And i have no idea about normal subgroups.
Let $a,b,c$ be the generators of the free group.
As the images all satisfy $x^{p^2}=1$ and the image is abelian, we can first pass from the free group $F(a,b,c)$ to its quotient $F(a,b,c)/K$ where $K$ is generated by all $x^{p^2}$ and all $x^{-1}y^{-1}xy$; and then to G.
Note that $F(a,b,c)/K\simeq \mathbb{Z}_{p^2}\oplus \mathbb{Z}_{p^2}\oplus \mathbb{Z}_{p^2}$.
Deal with the epimorphisms to $F(a,b,c)/K$ first.
Then the number of possible images of $a$ is $p^6-p^3$, as that is the number of elements of $G$ of order $p^2$. Given this, the number of choices for the image of $b$ is now $p^6-p^4$ since it must be outside the subgroup generated by the image of $a$ and the elements of order $p$. Given these, the number of possibilities for the image of $c$ is $p^6-p^5$ since it must be outside the subgroup generated by the images of $a,b$ and the elements of order $p$.
Multiplying together, we have so far $$ p^{12} (p^3-1)(p^2-1) (p-1)$$ choices.
Now we need to see how many ways epimorphisms from our intermediate group to $G$. We get exactly one for each way we can factor out a cyclic subgroup of order $p$. There are $(p^3-1)$ of these in $\mathbb{Z}_{p^2}\oplus \mathbb{Z}_{p^2}\oplus \mathbb{Z}_{p^2}$, so that gives us a final tally of $$ p^{12} (p^3-1)^2(p^2-1) (p-1)$$
As to the kernels $N$, all must contain $K$. Beyond that the only choice was made when we chose a subgroup of order $p$, so the answer if $p^3-1$.