[Math] How many distinct subgroups of order 10 are there in a non-cyclic abelian group of order 20

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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.

Best Answer

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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