# Questions tagged [infinite-groups]

For questions about groups where the underlying set has infinite cardinality.

327 questions
124 views

### $\mathbb{Z}\times \mathbb{Z}_{2}$ is a cyclic group?

I think that $\mathbb{Z}\times \mathbb{Z}_{2}$ isn't a cyclic group becuase we don't have any $(a,b)\in \mathbb{Z}\times \mathbb{Z}_{2}$ that can create the group $\mathbb{Z}\times \mathbb{Z}_{2}$. I'...
235 views

### Can $\mathbb{R}$ be written as an ascending union of proper additive subgroups?

Can the group $\mathbb{R}$ be written as countable ascending union of proper subgroups? (i.e. does there exists a series of proper subgroups $H_1\leq H_2\leq \cdots$ such that $\cup {H_i}=\mathbb{R}$?...
129 views

### Question about Abelian group proof

I prove that if $G$ is Abelian group so if $a,b\in G$ has a finite order so $ab$ has a finite order to.. (Maybe later I'll upload here my proof to see of she is correct....) Now, I have to show that ...
77 views

### find a generator for the group $G =\{ f(x) = x+n\mid n\in \Bbb Z \}$ with the group operation being composition.

Another question from 'A book of Abstract Algebra' by Pinter. For each $n\in \Bbb Z$ define $f_n = x+n$. Then $f_n\in S_{\Bbb R}$, the symmetric set on $\Bbb R$. The group operation being composition....
214 views

### Groups with 3 conjugacy classes and finite exponent

I have seen the question on groups with two conjugacy classes, and I proved to myself that such a group must be torsion-free (if it isn't the cyclic group of order 2), but what about a group with ...
38 views

### Cyclic Groups of Infinite and Finite order [duplicate]

If a cyclic group has an element of infinite order , how many elements of finite order does it have ?
101 views

### If $G$ is a group, $H$ is a subgroup of $G$ and $g\in G$, is it possible that $gHg^{-1} \subset H$? [duplicate]
If $G$ is a group, $H$ is a subgroup of $G$ and $g\in G$, is it possible that $gHg^{-1} \subset H$ ? This means, $gHg^{-1}$ is a proper subgroup of $H$. We know that $H \cong gHg^{-1}$, so if $H$ is ...