I'm trying to think of examples to these last two questions
For each part below, provide a group $G$ and a proper, non-trivial subgroup $H$ of $G$ according to the different criteria. Provide a different group $G$ in each case.
$G$ is infinite and abelian and $H$ is infinite and not cyclic
The second example
$G$ is infinite and non-abelian and $H$ is infinite and cyclic
As the question wants a different $G$ for each, I've already used $\mathbb{C}$ and $\mathbb{Z}$ under addition and $SL(n,\mathbb{R})$ (special linear group under multiplication)
For the last one I was thinking perhaps using $G = M_n(\mathbb{R})$ under multiplication but couldn't think of a suitable subgroup of the square matrices which is cyclic.
Thanks for any suggestions!
$\endgroup$ 15 Answers
$\begingroup$- $(\mathbb{Q},+)$ is a subgroup of $(\mathbb{R},+)$ which is not cyclic (in fact not finitely generated).
- The group of bijections $\mathbb{Z} \to \mathbb{Z}$ contains the element $x \mapsto x+1$. It generates an infinite cyclic subgroup, consisting of of translations.
Here's two hints: use $G = \mathbb{Z} \times \mathbb{Z}_2$ for the first question and $G = \mathbb{Z} \times Sym(3)$ for the second. If you know the infinite dihedral group that will also work for question two.
$\endgroup$ $\begingroup$For 1, take $G=\mathbb{Z}\times \mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ and $H=\{0\}\times \mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$.
For 2, pick $G_0$ any non-Abelian group. Then set $G=\mathbb{Z}\times G_0$ and $H=\mathbb{Z}\times\{0\}$.
$\endgroup$ 0 $\begingroup$For part one, how about $G=\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$ and then you can take $H=G$ (or if you want a proper subgroup, take $H=\mathbb{Z}\times\mathbb{Z}$.
For part two, you could take $G=\operatorname{Mat}(2,\mathbb{Z})$ (under addition) and $H=\left\langle\left(\begin{array}{cc}1 & 0 \\ 0 & 1 \end{array}\right)\right\rangle$.
$\endgroup$ $\begingroup$If $t(G)$ shows the subgroup of all torsion elements of $G$, so $t(\mathbb R/\mathbb Z)=(\mathbb Q/\mathbb Z)$. Both of $\mathbb R/\mathbb Z$ and $\mathbb Q/\mathbb Z$ are infinte and $\mathbb Q/\mathbb Z$ is not cyclic. This is a bit similar to what @Martin noted. From this point of view, we can see that $$(\mathbb Q/\mathbb Z)[p]=\mathbb Z_p, ~~ p~~\text{is prime}$$ so if we take $G=\mathbb Q/\mathbb Z\times S_3$, then $$H=\mathbb Z_n\times\{id\}$$ will be the abelian finite cyclic subgroup of non-abelian infinite $G$. You didn't ask this latter example and I hope you use it somewhere.
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