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

Question

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.

Answer 1

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 $10$.

Answer 2

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)>.

Answer 3

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

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