# A question on finitely generated Abelian groups with a minimal number of generators

In my class on group theory I have encountered this strange looking question relating to Abelian groups in terms of generators which states:

We are to find, up to isomorphism, all Abelian groups $A$ with a minimal number of generators equal to five, and $4\operatorname{tor}(A) = \{ 0 \}$ where $\operatorname{tor}(A)$ denotes the subgroup of $A$ of the finite order elements.

I have never encountered this sort of question in the past and have no idea what to do with the minimal number of generators assumption, so I am really hoping and kindly asking for some help on this problem that I ave no idea on. All I know is Abelian groups are Z modules and I know the fundamental theorem on finitely generated Abelian groups. Thanks to all helpers

$4 \cdot tor(A)=0$ means, that the torsion part of $A$ consists of factors of the form $C_2$ and $C_4$ only. Hence you have to write down all combinations of $5$ factors of $C_2, C_4, \mathbb Z$, namely the group is of the form

$$\mathbb Z^a \times C_2^b \times C_4^c$$

with $a+b+c=5$.

• I'm guessing the $4$ isn't a typo. Jan 24, 2016 at 7:06
• No the 4 is not meant to be a typo Jan 24, 2016 at 7:08
• Ah ok. See the edit.
– MooS
Jan 24, 2016 at 7:09
• @MooS I do not understand how to do what you say could you please show this? Jan 24, 2016 at 7:10
• What precisely is out of reach for you? The conclusion about the form of the factors? Or how to actually use this to write down all possibilities?
– MooS
Jan 24, 2016 at 7:12