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Among finite abelian groups, only the cyclic groups are indecomposable.

As per the title of the question, I am trying to disprove the above statement. On a recent test, I marked this "True" and found that I was incorrect. Still, I am confused and cannot think of a counterexample.

Moreover, take a finite abelian group $G$, then isn't $G$ isomorphic to the direct product of cyclic groups by the Fundamental Theorem for Finite Abelian Groups?

What am I missing here?

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The statement is true if it is taken to mean "any indecomposable finite abelian group must be cyclic", however, it is not true if it is taken to mean "the indecomposable finite abelian groups are precisely the finite cyclic groups". Indeed, $ C_6 \cong C_2 \times C_3 $ is not indecomposable.

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  • $\begingroup$ In short, tell your instructor not to word things so horribly. It's one thing to test one's knowledge of facts and their logical consequences. It's another to test one's ability to properly parse unnecessarily ambiguous statements. $\endgroup$ – Jeremy West May 26 '17 at 1:55
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The indecomposable finite abelian groups are exactly the cyclic $p$-groups.

Indeed, here is a key step: if $G$ is a finite abelian group of order $n=ab$, with $a,b >1$ and $\gcd(a,b)=1$, then $G \cong A \times B$, where $A= \{ x \in G : ax = 0 \}$ and $B= \{ x \in G : bx = 0 \}$. The map $A \times B \to G$ is given by $(\alpha,\beta) \mapsto \alpha+\beta$. Its inverse is given by $x \mapsto (bvx, aux)$ where $au+bv=1$.

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