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We know that by Fundamental theorem of Finite abelian groups, any Finite abelian group $A$ can be expressed as $A\cong \langle a_1\rangle \times \langle a_2 \rangle\times \cdots \times \langle a_n \rangle$.

But do these cyclic factors intersect trivially? Can we say that $\langle a_i \rangle \cap \langle a_j \rangle=e$ for $i\neq j$. If it is can someone explain it please.

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  • $\begingroup$ They are not disjoint but they intersect trivially. $<a_i>\cap <a_j>=e$ for $i\neq j$. $\endgroup$ – mesel Jan 2 '16 at 19:30
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    $\begingroup$ If you had duplicates there wouldn't be enough elements in the product. $\endgroup$ – Mark Bennet Jan 2 '16 at 19:30
  • $\begingroup$ @mesel can you explain please $\endgroup$ – Departed Jan 2 '16 at 19:32
  • $\begingroup$ @MarkBennet : What about $G\times G$ ? $\endgroup$ – mesel Jan 2 '16 at 19:43
  • $\begingroup$ @mesel The components of the product may be isomorphic, but that doesn't mean that $e\times g$ will be equal to $g\times e$ within the product. $\endgroup$ – Mark Bennet Jan 2 '16 at 19:50
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Let $U = G \times H$ be a direct product. Then the following statements are equivalent to U being a direct product of $G$ and $H$:

1) $U = GH$

2) $G \cap H = 1$.

3) $G,H$ are normal subgroups of $U$.

If you let $G = <a_i>$ and $H = <a_j>$ then as the abelian group is the direct product of the $<a_i>$ they intersect pairwise trivially.

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