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In Serge Lang's Algebra, Chapter 1, exercise 54, the problem statement begins with:

Let $G$ be a group and $\{G_{i}\}_{i\in I}$ be a family of subgroups generating $G$.

Does this mean that the union of all the $G_{i}$'s generates $G$? or that each $G_{i}$ on its own generates $G$?

Edit: Please correct me if this is wrong: (yes it is wrong)

So given that this means the union of the $G_{i}$'s generates $G$, I must in fact have that the union of the $G_{i}$'s is equal to $G$, since the union of subgroups is a subgroup.

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    Thanks for pointing this out, I forgot to check back but my friend pointed this out as well with the example of subgroups $2\mathbb{Z}$ and $3\mathbb{Z}$ of $\mathbb{Z}$. I spent all summer thinking about convergence and was too eager to apply properties of open sets to groups. :)2012-09-25

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Well, it's definitely not the latter, because proper subgroups are closed under multiplication, and there is no way they will individually generate the entire group.

Your first guess is right (assuming that you properly understand what it means for a set generates a group.) Generally when you say a subset $S\subseteq G$ generates $G$, it just means that "the smallest subgroup of $G$ containing $S$ is $G$."

In your context, this just means that any subgroup of $G$ containing all of those groups has to be $G$ itself, not some smaller subgroup of $G$.

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    Yes thanks, for some reason I was thinking of each $G_{i}$ as a set rather than as a subgroup, even though I wrote subgroup in my question... (I can't explain that one). Anyways, thanks for clarifying.2012-09-21
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A subgroup generates itself, so the second option is not an option. Thus the intended meaning seems to be that all the subgroups together generate $G$.

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    No, you don't look silly. Keep asking until you understand it! Good question!2012-09-21