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We are currently working with free abelian groups and finitely generated groups. The homework problem asks us to find the number of distinct subgroups of order 10 in a non-cyclic abelian group of order 20. I know there has to be at least 1, generated by an element of order 5 and an element of order 2, but I don't know how to find all of them. A subgroup of order 10 is not a Sylow p-subgroup, so I can't say that they are all conjugates.

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You can see that if G is noncyclic abelian then $G$ be must be $Z_{10}\times Z_2$.

Claim: $H=Z_5\times1$ is uniqe subgroup of $G$ with order $5$.

if there is also $K$, $HK$ has $25$ elements which is impossible.

Now,Any subgroup of order $10$ must include $H$ and $G/H\cong Z_2\times Z_2$ Since $G/H$ has three subgroup of index $2$, $G$ has three subgroup of order $10$.

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  • $\begingroup$ How is G/H isomorphic to Z2*Z2 $\endgroup$ – UNM Oct 12 '14 at 19:49
  • $\begingroup$ @UNM: Why do you think so ? $\endgroup$ – mesel Oct 12 '14 at 19:53
  • $\begingroup$ You tell me why G/H is isomorphic to Z2*Z2 $\endgroup$ – UNM Oct 12 '14 at 19:54
  • $\begingroup$ If you believe that G/H is not $Z_2\times Z_2$ then you must believe that $G/H= Z_4$. But $G/H$ is not cyclic. $\endgroup$ – mesel Oct 12 '14 at 19:56
  • $\begingroup$ weel,if it is $Z_4$ there must be an element of order $4$ right ? $\endgroup$ – mesel Oct 12 '14 at 19:58
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I'm new so I can't comment, but here's a reason: If $H_1 \unlhd G_1$ and $H_2 \unlhd G_2$, then

$(G_1 \times G_2) / (H_1 \times H_2) \cong (G_1 / H_1) \times (G_2 / H_2)$,

so in this case

$(Z_{10} \times Z_2) / (Z_5 \times Z_1) \cong (Z_{10}/Z_5) \times (Z_2/Z_1) \cong Z_2 \times Z_2$.

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Since G is abelian so must its subgroup of order 10. Now for order 10 there exists only one subgroup Z5*Z2 which is abelian. Others Z10*{0} and <(1,1)>.

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