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(I apologize, if you find my question bad, but I m struggling to understand the definition of free abelian groups.)

Definition. An abelian group $G $ is called Free Abelian Group with rank $ n\in \Bbb{N}$, if $G $ is the direct sum of $ n $ infinite cyclic groups. I.e. $$G=\langle x_1 \rangle \oplus...\oplus\langle x_n \rangle $$ where $\mathcal{U}_i=\langle x_i\rangle,\ (i=1,...,n )$ infinite cyclic groups.

My Questions:

1) I found in some books that $\mathcal U_i $ are abstract groups, and in other that $\mathcal{U_i} $ are subgroups of $G $. I suppose that these are both true. But why?

2) In the first case, we have that the every element $ g\in G$ can be uniquely written as $g=a_1+...+ a_n, a_i \in \mathcal{U_i} $. Could this happen in external?

3) Do we loose this property if your sum is external direct sum? (More generally, in external direct product of abstract groups. Do we have uniqueness?)

Thank you.

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2 Answers 2

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An abelian group $G$ is the internal direct sum of subgroups $H_i$ if and only if $G$ is isomorphic to the external direct sum of the $H_i$ seen as (abstract) groups.

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  • $\begingroup$ Thank you for your answer. Could you please explain why the inverse is true? How do we take the internal direct sum from isomorphism? $\endgroup$
    – Chris
    Commented May 4, 2018 at 17:35
  • $\begingroup$ Quite simole: if the $Hi$ are subgroups of G, and if $u:G\longrightarrow\bigoplus_{i\in I}H_i$ is the isomorphism from $G$ onto the (external) direct sum, the image of a family $(h_i)_{i\in I}$ with finite support, take the image $u^{-1}\bigl((h_i)\bigr)$ by the inverse isomorphism. The internal direct sum of the submodules $K_i=u^{-1}\bigl(\{0\}\times\dots\times H_i\times\dots\times\{0\}\bigr)$ is equal to $G$. $\endgroup$
    – Bernard
    Commented May 4, 2018 at 18:03
  • $\begingroup$ Could we prove it without modules? $\endgroup$
    – Chris
    Commented May 4, 2018 at 18:37
  • $\begingroup$ I don't understand: abelian groups are $\mathbf Z$-modules… Just replace the word ‘submodules’ with ‘subgroup’. $\endgroup$
    – Bernard
    Commented May 4, 2018 at 18:50
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1) There are the notions of the internal direct sum and external direct sum which are equivalent. For abelian groups, say $G = G_1\oplus\cdots\oplus G_n$ as an internal direct sum if every $g\in G$ can be written uniquely as a sum $g_1+\cdots+g_n$ where $g_i\in G_i$. For external direct sums, $G_1\oplus\cdots\oplus G_n$ consists of all ordered pairs $(g_1,\cdots,g_n)$ with addition component wise. The internal sum is isomorphic to the external direct sum with an isomorphism $\phi:G \to G_1\oplus\cdots\oplus G_n$ given by $g_1+\cdots+g_n\mapsto(g_1,\cdots,g_n)$. This is a canonical isomorphism.

2) and 3) In the external direct sum, you can write elements uniquely as sums $(g_1,0\cdots,0)+(0,g_2,0,\cdots,0)+\cdots+(0,\cdots,0,g_n)$

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  • $\begingroup$ Thank you for your very clear answer. $\endgroup$
    – Chris
    Commented May 3, 2018 at 22:16

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