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Let $X = \{G_1,G_2,...,G_n\},$ where $G_i$ are groups under some operation $*$. Is $X$ a group under $\oplus$?

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You can make a group out of any finite set. In fact you can make any group of order $n$ from any set with $n$ elements. You just define the operation to be what it has to be to get the group you want. –  fretty Oct 15 '12 at 8:09
    
What exactly do you mean by $oplus$? –  Ittay Weiss Oct 15 '12 at 8:39
    
direct sum operator –  Euler Oct 15 '12 at 8:44
    
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A passing remark: In K-theory, the direct sum does form the group operation for an abelian group, though the objects you are 'summing' are not groups, but vector bundles. –  Sebastian Oct 15 '12 at 9:41
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4 Answers

up vote 3 down vote accepted

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 $\oplus$.

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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.

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I meant the last sentence you said. Think of $X$ as a set consisting of groups. $G_i$ are elements of $X$ –  Euler Oct 15 '12 at 9:35
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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.

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