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In Fraleigh, there is a true/false problem in the chapter on "Direct Products" that states the following three assertions:

  1. Every abelian group of order divisible by 5 contains a cyclic subgroup of order 5.
  2. Every abelian group of order divisible by 4 contains a cyclic subgroup of order 4.
  3. Every abelian group of order divisible by 6 contains a cyclic subgroup of order 6.

I know the second statement is false because, for example, the Klein-4 group does not contain a subgroup of order 4. The solutions key says that 1 and 3 are true, but I'm not sure why (there is no explanation). Is this somehow a consequence of the Fundamental Theorem of Finitely Generated Abelian Groups?

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2 Answers 2

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Every Abelian group can be decomposed into into the direct sum of cyclic groups of prime power order.

A group of order divisible by $5$ has some cyclic subgroup of order $5^n$ take the generator of that subgroup call it $a$ and $a^n$ is the generator of a group of order $5$

A group of order divisible by $6$ has some subgroup of order $2$ and a subgroup of order $3$.

Take the generator of both and multiply them together. Since $gcd(2,3) = 1$ this will generate a group of order $6.$

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  • $\begingroup$ So for an Abelian group decomposed into, say, Zp1 X Zp2, both Zp1 and Zp2 would be subgroups? $\endgroup$
    – ponchan
    Jun 29, 2017 at 3:28
  • $\begingroup$ yes ............ $\endgroup$
    – Doug M
    Jun 29, 2017 at 3:31
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By Cauchy's theorem, since $5$ divides the order of your group, there exists an element of the group denoted $g$ with order $5$. So $\langle g\rangle$ is a subgroup of your group and thus cyclic.

For $(3)$, Cauchy's theorem furnishes elements of order $2$ and $3$ denoted $a,b$ respectively. The subgroup $\langle ab\rangle$ is cyclic of order $6$.

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