Non-Trivial Problem: Determine, up to isomorphism, the following abelian groups:

Question

I'm practicing for an upcoming exam and I have the following exercise:

Determine, up to isomorphism, all the abelian groups $A$ that satisfies the 3 following conditions:

1. $A$ has a subgroup $B$ with order 40 such that $A/B$ is free abelian with finite rank.
2. $20x=0$ for all $x \in B$.
3. There exist $x,y,z\in A$ such that for all $a\in A$ we can write $a=nx+my+kz+b$ for some $n,m,k\in \mathbb{Z}$ and some $b\in B$. Remark: We don't assume that $x,y,z$ are different each other.

My attemps (small advances): The first observation is that $B$ is non-cyclic, for if $B$ were cyclic, then $B$ would contain an order 40 element, which is not possible because the condition 2 ($|x|\leq 20$ for all $x\in B$).

Then, by the Classification Theorem of Finitely Generated Abelian Groups we note that $|B|=40=2^3\cdot 5$ implies $B\cong \mathbb{Z}_2\oplus\mathbb{Z}_4\oplus \mathbb{Z}_5$ or $B\cong (\mathbb{Z}_2)^3\oplus \mathbb{Z}_5$.

Also, we know the following result: If $B$ is a subgroup of an abelian group $A$ and $A/B$ is free abelian, then $A\cong B\oplus (A/B)$. We have all the conditions to use that theorem, so is another advance.

Now I would like to use condition 3 to try to determine $A/B$, hence $A$, but I am stuck.

Any suggestions on how to move forward on this problem would be great for me. Thank you all.

Answer 1

We're all but done.

Between what you and @Arturo did, we've got $6$ possibilities: $3$, namely $\Bbb Z,\Bbb Z^2,\Bbb Z^3$ for $A/B$; and $2$ for $B$. (I didn't include the trivial group as free abelian, but probably should.)

So :

• $ \Bbb Z×(\Bbb Z_2)^3×\Bbb Z_5$

• $ \Bbb Z×(\Bbb Z_2)^2×\Bbb Z_2×\Bbb Z_5$

• $ \Bbb Z^2×(\Bbb Z_2)^3×\Bbb Z_5$

• $\Bbb Z^2×(\Bbb Z_2)^2×\Bbb Z_2×\Bbb Z_5$

• $\Bbb Z^3×(\Bbb Z_2)^3×\Bbb Z_5$

• $\Bbb Z^3×(\Bbb Z_2)^2×\Bbb Z_2×\Bbb Z_5$