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Here is the beginning of a proof that

Every abelian finite group $(B,+)$ with $|B|=p^r$ ($p$ prime, $r\in\mathbb{N}$) is isomorphic to a direct product of cyclic groups.

By induction on $r$. If $r=1$, then $B$ is cyclic.

Else let $b\in B$ of maximal order and $B_1$ the cyclic group generated by $b$. We use the induction hypothesis on \bar B=B/B': $\bar B\cong \bar B_1\times \bar B_2\times\cdots \times \bar B_k$ where $\bar B_1, \dots, \bar B_k$ are cyclic groups.

Then the proof uses that

$\bar B_1+ \bar B_2+\cdots+ \bar B_k\cong \bar B_1\times \bar B_2\times\cdots\times \bar B_k \ \small (1)$ to prove that there exists cyclic sub-groups of $B$, $B_1, \dots, B_k$ such that $B_1+\cdots+B_k\cong B_1\times\cdots\times B_k$

The isomorphism $\small (1)$ should be obvious as it is given without any details, but I am unable to understand why it holds.

Could someone enlighten me ?

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    Yes, sorry. It is fixed, thank you.2011-02-24

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You are identifying the cyclic group $\overline{B}_i$ with the subgroup $\{(\overline{b}_1,\ldots,\overline{b}_k)\in \overline{B}\mid b_j=0\mbox{ for }j\neq i\}.$ That is, you are going from the "external" direct product to the "internal" direct product.

The corresponding subgroups of $\overline{B}$ are disjoint ($\overline{B}_i\cap\mathop{\oplus}\limits_{j\neq i}\overline{B}_j = {0}$), and span $\overline{B}$. So you get that the sum equals $\overline{B}$, and hence is isomorphic to the direct product/sum of the groups.

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    Again: if you point me to the book you are getting this from, I might be able to see the background in question. It's hard to work on a vacuum, but I suspect you are missing the forest for the trees.2011-02-25