Is the group direct sum operation itself a group operation? Let $X = \{G_1,G_2,...,G_n\},$ where $G_i$ are groups under some operation $*$. Is $X$ a group under $\oplus$?
 A: You must understand incorrectly.
$$G_{1}\oplus...\oplus G_{n}$$ is indeed a group and is called the
direct sum of $G_{1},...G_{n}$ but $\oplus$ is not the operation
of the group.
Denote $*_{i}$ as the operation of group $G_{i}$ and let $$(g_{1},...g_{n}),(g_{1}',...g_{n}')\in G_{1}\oplus...\oplus G_{n}$$
(that is $g_{i},g_{i}'\in G_{i})$. 
The operation of $G$ that will denoted by $*$ is defined by $$(g_{1},...g_{n})*(g_{1}',...g_{n}'):=(g_{1}*_{1}g_{1}',...,g_{n}*_{n}g_{n}')$$
If you think $\oplus$ is the operation it means that $\forall i,j:\, G_{i}\oplus G_{j}:=G_{k}$
for some $1\leq k\leq n$ and I doubt that is what you meant. 
A: What's going to be the identity element for your group, the group $E$ such that $E\oplus G=G$ for all $G$? I guess it has to be the one-element group. 
But then where will you find inverses? Given a group $G\ne E$, how can there be a group $H$ such that $G\oplus H=E$? 
It appears there will only be some very special cases where $X$ is a group under $\oplus$. 
A: To add a bit to what others have already said: No, in general what you have is not a group. The "first thing" that you would need for this to be a group is that $G_i\oplus G_j \in X$ for all $i,j$. So you cannot just take an arbitrary set of groups.
One simple example of when it would work is when $X = \{E \}$, where $E = \{1\}$. Here $E\oplus E \simeq E$. 
As you think about this, remember that if the $G_i$s are finite then the order of $G_i\oplus G_j$ is the product of the orders of the two groups.
A: See http://crazyproject.wordpress.com/2010/06/14/basic-properties-of-the-direct-sum-of-groups/
