Let $p$ be a prime. For any one-to-one homomorphisms $f,g:\Bbb Z_{p^\infty}\to \Bbb T$, we have $f[\Bbb Z_{p^\infty}]=g[\Bbb Z_{p^\infty}]$, where $\Bbb T $ is the circle group.
Is this correct for all torsion groups $G$? That is, is $f[G]$ equal to $g[G]$ for one-to-one homomorphisms $f,g:G\to \Bbb T$?
Since the torsion subgroup of $\mathbb T$ is isomorphic to $\mathbb Q/\mathbb Z$ the question reduced to the following:
If $H_1,H_2$ are isomorphic subgroups of $\mathbb Q/\mathbb Z$ then are they equal?
The answer is yes. Let $H$ be a subgroup of $\mathbb Q/\mathbb Z$, and $x\in H$, $x\ne 0$. Then $x$ is the residue class modulo $\mathbb Z$ of a rational number $m/n$ with $n\ge 2$ and $\gcd(m,n)=1$. Then there exist $a,b\in\mathbb Z$ such that $am+bn=1$, so $a(m/n)+b=1/n$. This shows that the residue class of $1/n$ belongs to $H$.
Now let $h:H_1\to H_2$ be an isomorphism, and $x_1\in H_1$. Then $x_1$ is the residue class of a rational number $m_1/n_1$ with $n_1\ge 2$ and $\gcd(m_1,n_1)=1$. We can also write $h(x_1)$ as a residue class of a rational number $m_2/n_2$ with $n_2\ge 2$ and $\gcd(m_2,n_2)=1$. From $n_1x_1=0$ we get $n_1h(x_1)=0$ which leads to $n_2\mid n_1$. Since $h$ is an isomorphism we can interchange the role of $x_1$ with $x_2$ and find $n_1\mid n_2$, so $n_1=n_2$. Set $n=n_1=n_2$. From the previous reasoning we deduce that the residue class of $1/n$ belongs to $H_1\cap H_2$. In particular we get $x_1\in H_2$.