Show that there are only two types of subgroups in $\mathbb{R}$ , either Discrete or Dense?
37977
Let's say $G$ is some additive subgroup of $\mathbb{R}$ that has at least two elements.   From what I understand, $G$ is then either dense in $\mathbb{R}$, or has some least positive element. What is the reason for this?
54677
How can I show that if some group $G$ has only one subgroup $K$ of order $n$, then $K$ is a normal subgroup? Would that mean that it only has one subgroup total? If so then I guess that makes sense.
64647
It's problem #1 of Chapter 2 in Pugh's book.  I think I got the first part by taking $r=1$ and pointing out that $d(p,q) &lt; 1 \implies q \in (0,1)$ for all $p \in (0,1)$ but I have no idea how to show it isn't a subset of $\mathbb{R}^2$.  I tried proof by contradiction and got no where.  I've been working at this problem for hours and not getting anywhere.  How should I do it?
32866
Show that a discrete and compact subset $D \subset \mathbb{C}$ must be finite. Does this conclusion hold if $D$ is just discrete and bounded? How about discrete and closed?  Compact is the usual (for these simple spaces), closed and bounded, where closed is contained under the limit operation/contains all limit points.  Bounded it can be contained in a ball of some radius around the origin.  $D \subset \mathbb{C}$ is a discrete subset if $\forall z \in D$ there exists a ball of radius $r&gt;0$ such that $D \cap B_r(z)$ = $\{z\}$.  Okay, for bounded set: why is discrete required? Can you give mme an example of a bounded set that is NOT finite?
67390
Show that any finite $T_1$-space $X$ is discrete.     My thoughts:- Let $X=${$a_1 , a_2, …,a_n$} and $U$ be any subset of $X$. Let $U=${$a_1 , a_2, …,a_k$} where$k \le n$ thenwe need to show that $U$ is open that is $X \setminus U$ is closed. Now $X \setminus U =$ { $a_{k+1} , …,a_n$ }$ = ⋃_{(k-1)}^{n}a_i$ now each $a_i$ is closed. So the result follows. Does my approach correct?
123438
Let $H$ and $K$ denote two subgroups of a group $G$.  Prove that the union $H \cup K$ is a subgroup of $G$ if and only if $H \subset K$ or $K \subset H$.
96242
Say I have a finite group and I tell you how many elements and how many subgroups it has. Do you now know what group it is? Or could two groups which are not isomorphic to each other have the same amount of elements and subgroups? In other words, does the amount of elements and subgroups of a group define it completely? It is interesting because if I give you just the number of elements there are several possibilities and if I give you the number of subgroups also but perhaps both combined are enough to define the group.
56971
Does there exists an uncountable family $\mathcal{A}$ consisting of uncountable compact subsets of $\mathbb{R}$ and pairwise disjoint?
109136
Prove that the additive groups of $\Bbb Z$ and $\Bbb Q$ are not isomorphic.    It is hard to create a map that will show homomorphism.
38028
Let $K$ be a subfield of $\mathbb{C}$ not contained in $\mathbb{R}$. Show that $K$ is dense in $\mathbb{C}$.     completely stuck on it. can I get some help please.
110867
Is the set of integers $ \mathbb{Z}$ closed in $\mathbb{R}$ equipped with the usual topology?
67157
Does an uncountable and discrete subspace of the reals exist?
31385
I need to show that there are only 2 groups of order 6 up to isomorphism.  I did prove it, but the proof is quite cumbersome. I wonder if there is a very concise proof.  My proof outline: Suppose $G$ is of order 6 and is not $\mathbb{Z}_6$. Then any element that is not identity must have order 2 or 3.  I went on to show that there must be a element of order 2 and another one of order 3, and the intersection of the cyclic subgroups generated by the two is the identity. Then I'm able to show it is isomorphic to $S_3$.
92119
Why does $PSL(2,\mathbb C)\cong PGL(2,\mathbb C)$ but $PSL(2,\mathbb R) \not\cong PGL(2,\mathbb R)$?
122250
This is exercise 1.9 in Lie Groups, Lie Algebras and Representations - Hall.     Suppose $a$ is an irrational real number.  Show that the set $E_a$ of the numbers of the form $e^{2\pi i n a}$, $n \in \mathbb{Z}$, is dense in the unit circle $S_1$.  Hint: Show that if we divided $S^1$ into $N$ equally sized "bins" of length $2\pi/N$, there is at least one bin that contains infinitely many elements of $E_a$.  Then use the fact that $E_a$ is a subgroup of $S^1$.   My proof of this proposition is as follows.  Since $a$ is irrational, you can determine that the set of rotations $E_a$ is infinite.  Since $S^1$ is compact we can find two $r_1, r_2$ that are within $\epsilon$ of each other.  Then $r_1^{-1}r_2$ is a small rotation of size $\epsilon$.  Now, $r_1^{-1}r_2$ generates rotations that are within $\epsilon$ distance of any point of $S^1$.    That said, I don't believe that the hint suggested in the problem uses that technique.  My knowledge of algebra is not all that strong so I was hoping someone could shed some light on what is being suggested there.
2919
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?  Why isn't there other possibility?  Thanks :)
102088
I am trying to understand subgroups.I know a given a group G under a binary operation ∗, a subset H of G is called a subgroup of G if H also forms a group under the operation ∗.   So if I where to list 5 subgroups of $\mathbb{Z}$ could I simply say, $\mathbb{Z}_2$, $\mathbb{Z}_4$, $\mathbb{Z}_6$, $\mathbb{Z}_7$, $\mathbb{Z}_9$, are all subgroups of $\mathbb{Z}$?