If $G$ and each $G_i$ are abelian, prove that if $G= G_1+G_2+...+G_n$, then $pG=pG_1+pG_2+...+pG_n$ for a prime $p$.
Note that "$+$" denotes the direct sum which means each element $G$ is expressed in one and only one way as sum of the subgroups $G_i$ and the intersection of $G_i$ and $G_j$ is only the identity. Also, I am using additive notation here. Some text may use multiplicative notation. Also, it is previously proven in a different problem that $pG$ is a subgroup of $G$.
Now I actually had no problem with most of this problem except the final step where I had to show that elements of $pG$ can be expressed uniquely. It is not too hard to show an element in $pG$ can be expressed as the sum of elements in $pG_1+pG_2+...+pG_n$ so I only need to show uniqueness. I tried to show this by contradiction:
Suppose, we have two diffent expressions of $pg$ so $pg=pg_1'+pg_2'+...+pg_n'=pg_1+pg_2+...+pg_n.$ From here, I am not sure how to show that each $pg_i'=pg_i$. Of course, it is not a valid group operation to just divide by $p.$
Instead of showing uniqueness of decomposition it is easier to show that for each $i$ we have $pG_i \cap \sum_{j\neq i} pG_j = \{0\}$. To do that you can argue with contradiction. Let $0 \neq z \in pG_i \cap \sum_{j\neq i} pG_j$. So $z = pg_i = \sum_{j\neq i} pg_j$ for some $g_k \in G_k$ for $k=1,\dots, n$. Since $pg_k = g_k + g_k + \dots + g_k$ for each $k$ and $G_k$ is a subgroup of $G$ we have $pg_k \in G_k$ for each $k=1,\dots, n$. This implies that $z \in G_i \cap \sum_{j\neq i} G_j = \{0\}$, a contradiction!