A word problem in abelian $p$-groups Let $G=\langle a,b,c\rangle$ an abelian $p$-group of rank $3$ with only following more relations between $a,b,c$: 
$$a^{p^2}=b^{p^2}=c^{p^2}\neq 1, \mbox{ and } a^{p^3}=1.$$
Then, consider the elements 
$$b_1=ba^{-1} \mbox{ and } c_1=ca^{-1}.$$
It is easy verify that 
$$b_1^{p^2}=c_1^{p^2}=1.$$
Question: Can we prove (if true) that 
$$\langle b_1\rangle \cap \langle c_1\rangle=\{1\}?$$

I'd like to consider this problem as a word problem. The problem I am facing is : it suffices to show $\langle(ba^{-1})^p\rangle\cap \langle (ca^{-1})\rangle =\{1\}$; but assuming that this is false, I'm unable to get a contradiction.
 A: My interpretation of your description of $G$ is that it is the abelian group defined by the presentation $\langle a,b,c \mid a^{p^2}=b^{p^2}=c^{p^2}, a^{p^3}=1 \rangle$, so I will take that as the definition. It is true that $c^{p^2}\ne 1$ but that is something that you can prove, and should not be part of the definition.
Since $G$ is an abelian group generated by the elements $a,ba^{-1},ca^{-1}$ which have orders $p^3,p^2,p^2$, it is clear that $|G| \le p^7$.
Define the group $H$ to be the direct product $\langle x \rangle \times \langle y \rangle \times \langle z \rangle$ of cyclic groups of order $p^3,p^2,p^2$. So $|H| = p^7$.
Now, since $x^{p^3}=1$ and $x^{p^2} = (xy)^{p^2} = (xz)^{p^2}$, the map $a \mapsto x$, $b \mapsto xy$, $c \mapsto xz$ extends to a homomorphism $\theta:G \to H$ and, since $H = \langle x,xy,xz \rangle$, it is a surjective. But, since we have proved that $|G| \le |H| = p^7$, $\theta$ must be an isomorphism. 
Now $\theta(b_1) = y$ and $\theta(b_2) = z$, so $\langle \theta(b_1) \rangle \cap \langle \theta(c_1) \rangle = 1$, and hence $\langle b_1 \rangle \cap \langle c_1 \rangle = 1$.
