"Almost Retractible" Abelianizations of Groups

Question

I have two related questions.

1. Is there a name for a nonabelian group $G$ whose abelianization is $\bigoplus_{i=1}^n \mathbb{Z}/p_i\mathbb{Z}$ such that for each $i$ there is an element $g$ whose order in $G$ is $p_i$? (If it's easier, we could assume each $p_i$ to be prime.)

2. Is there a way to "create" (rather than find in GAP) preferably finite groups $G$ that do not satisfy this property, particularly in a way as far off from that as possible, for example to have a factor $\mathbb{Z}/p\mathbb{Z}$ in the abelianization, say with $p$ prime, such that every representative of $\bar{1}$ has order $p^k$ for $k = 2, 3, \dots$ (presumably the construction getting more complicated for larger $k$)?

Examples:

• For a dihedral group $D_{2n}$, the abelianization is either one or two copies of $\mathbb{Z}/2\mathbb{Z}$, depending on whether $n$ is odd or even, resp., and in either case we can find a representative of any nontrivial coset of $[D_{2n}, D_{2n}]$ which is a flip, so dihedral groups do satisfy the property.

• For a symmetric group $S_n$, $(12)$ is a representative of the nontrivial coset of $A_n$ so it does satisfy this property.

• For a Heisenberg group $H$ of order $p^3$ ($p$ an odd prime), the abelianization is $\mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Z}/p\mathbb{Z}$, and every element of $H$ has order $p$ so it does satisfy the property.

• If the sequence $1 \to [G, G] \to G \to G^{\text{ab}} \to 1$ splits (i.e. the map $G \twoheadrightarrow G^{\text{ab}}$ has a retraction) then $G^{\text{ab}}$ is a subgroup of $G$ and certainly does satisfy the property, hence the name I'm thinking of for this property.

• For the quaternion group $Q_8$, the abelianization is $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ and the representatives of the nontrivial cosets are $\pm i, \pm j, \pm k$, all of which have order 4 so $Q_8$ does not have this property.

Possible direction / further question: in trying to create a group that doesn't satisfy the property, say with just $k = 2$, does anyone know of a way to make a quotient of a free group on two generators, say, that is finite and maybe doesn't have the property? e.g. $\langle x, y \mid x^9 = y^9 = 1, (\text{other relations})\rangle$ so that in the abelianization $\bar x$ has order 3 but in the group it still has order 9 and there's no other generator for $\bar x$ of lower order?

Related: here, @IAAW asked essentially if we can get every abelian group as the abelianization of a nonabelian group.

Any help would be greatly appreciated! Thank you in advance!

Answer 1

Here's an answer for #2: Let $p$ be an odd prime, and consider ${\mathbb Z}_{p^n} \rtimes {\mathbb Z}_{p^{n-1}}$, say $\langle x \rangle \rtimes \langle y \rangle$, where the action is given by $y^{-1} x y = x^{p+1}$. (It's well-known that $p+1$ has order $p^{n-1}$ mod $p^n$, so this action is well-defined.) Then in the abelianization, this relation yields $x^p = 1$, and we end up with ${\mathbb Z}_{p} \times {\mathbb Z}_{p^{n-1}}$. Now the preimage of any generator of $\mathbb Z_p$ has order $p^n$, so you can easily create groups that are as "far away" from your desired property as you want.

For $p = 2$, basically the same thing works but with ${\mathbb Z}_{p^n} \rtimes {\mathbb Z}_{p^{n-2}}$ instead.