How to show that these groups are not isomorphic? Let $G$ is an abelian group and $tG$ is its torsion subgroup. If $p$ is a prime, how to show that: 

$$t\bigg(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\bigg)\ncong\sum_{n=1}^{\infty}\mathbb Z_{p^n}$$

I just know that $\sum_{n=1}^{\infty}\mathbb Z_{p^n}< t\bigg(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\bigg)$. Thanks
 A: They cannot be isomorphic, because they are of different cardinality.
To see this, we will find uncoutably many elements of $t\bigg(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\bigg)$. To do this, for an arbitrary sequence $\tau:\mathbb N\to\{0,1\}, \tau(i)=\tau_i$, define an element $$g_\tau=(\tau_11,\tau_2p,\tau_3p^2,\tau_4p^3,\tau_5p^4,\ldots)$$ of the torsion group. Every such element has order $p$ and for distinct sequences $\sigma,\tau:\mathbb N\to\{0,1\}$, we get distinct elements $g_\tau$, $g_\sigma.$ The set $2^{\mathbb N}$ of all such sequences is uncountable and $\tau\mapsto g_\tau$ is an injection $2^{\mathbb N}\to t\bigg(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\bigg)$. Therefore this torsion group is uncountable.
On the other hand, $\sum_{n=1}^{\infty}\mathbb Z_{p^n}$ only has countably many elements. (Since it can be written as a countable union of countable sets $\sum_{n=1}^N\mathbb Z_{p^n}$.)
A: One element that belongs to $t\big(\prod_{n=1}^{\infty}\mathbb Z_{p^n}\big)$ but not to $\sum_{n=1}^{\infty}\mathbb Z_{p^n}$ is
$$
(1,p,p^2,p^3,\ldots).
$$
That shows that the obvious inclusion is not an isomorphism. To see that there cannot be an abstract isomorphism think about cardinality.
