How can we show that $\mathbb{Q}$ as an additive group is not isomorphic to $\mathbb{Q}^+$ as a multiplicative group?  Both have a countable number of elements, neither is cyclic, neither has an element $x\neq e$ such that $x^2=e$, both are abelian ... I don't know what to use.
26441
A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{&gt;0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} , \times )$, not required to be positive, one can argue the group contains elements with order 2 (namely all negatives). In the case of the requirement for all rationals to be positive this argument does not fly. What trivial fact am I missing here?
81674
How do I prove that a finite ring $R$ of order 10 is isomorphic to the ring $\mathbb{Z}/10 \mathbb{Z}$?    I know that as a group under addition, $(R,+)$ is isomorphic to the group $(\mathbb{Z}/10 \mathbb{Z}, + )$, but the multiplication is rather mysterious to me.
66503
Let $\mathbb Q$ be group of rational numbers under addition.  Then I want to find out the automorphism group of $\mathbb Q$. In a book it is given that Aut($\mathbb Q$) is isomorphic to $\mathbb Q^*$, the  multiplicative group of rational numbers. I stuck in this.
45169
Can anyone please help me out on how to use the fundamental theorem of finitely generated abelian groups to prove that $({\mathbb{Q}},+)$ is not finitely generated?
32316
So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses.  The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
20701
From  I see that $\mathbb{Z}^{\mathbb{N}}$ is not a free abelian group. Someone can explain me why?  Thanks a lot!
21587
I want to prove that $\mathbb{Q}$ is countable. So basically, I could find a bijection from $\mathbb{Q}$ to $\mathbb{N}$. But I have also recently proved that $\mathbb{Z}$ is countable, so is it equivalent to find a bijection from $\mathbb{Q}$ to $\mathbb{Z}$?
23752
The question (from my exam) quoted verbatim:  Show that the additive group of $\mathbb{Z}[x]$ is isomorphic to the group of positive rational numbers under multiplication.  First of all, is this $\mathbb{Z}[x]$ the same as $\mathbb{Z}_x$ (addition modulo $x$)? Secondly, how do I still get the isomorphism? Thanks
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.
98793
Question is to Prove that $(\mathbb{Z}/2^n \mathbb{Z})^*$ is not  cyclic Group for $n\geq 3$.  Hint : Find two subgroups of order $2$.  I somehow feel that a cyclic group can not have two distinct groups of same order. but, I am not sure about the proof.  I have no idea how to proceed for this.   any hint would be appreciated.  Thank You.
37754
Here's the problem: if $S = \mathbb{Q}[x]/(x^2)$, compute the group  of automorphisms $\mbox{Aut}(S)$.   First off, I have the solution and I know it's isomorphic to the multiplicative group $\mathbb{Q}\backslash\{0\}$, so I'm trying to understand it. It uses what is called the substitution principle and the fact that the restriction of any endomorphism $\phi: \mathbb{Q}[x]\to \mathbb{Q}[x]$ to $\mathbb{Q}$ is the identity. It says that these two facts implies that the endomorphisms of $\mathbb{Q}[x]$ are in 1-1 correspondence with $h \in \mathbb{Q}[x]$, where the endomorphism corresponding to $h$ is $\phi_h : \mathbb{Q}[x] \to \mathbb{Q}[x]$ given by $\phi_h(f) = f(h)$ for $f \in \mathbb{Q}[x]$. I guess I just don't understand how it applied the substitution principle? It's probably obvious, but for some reason I am not getting it.  An explanation would be greatly appreciated! Or if you have a different solution, that'd be cool too. Thanks!!
91535
It's true that the irrationals do not form a group under addition or multiplication, but I want to find a binary operation "∗" such that ($\mathbb R \setminus \mathbb Q,∗$) is a group. It is possible by transpose of operation from $\mathbb R$, but I want a explicit example.
107062
I would like to prove that the additive quotient group $\mathbb{Q}/\mathbb{Z}$ is isomorphic to the multiplicative group of roots of unity.  Now every $X \in \mathbb{Q}/\mathbb{Z}$ is of the form $\frac{p}{q} + \mathbb{Z}$ for $0 \leq \frac{p}{q} &lt; 1$ for a unique $\frac{p}{q} \in \mathbb{Q}.$ This suggest taking the map $f:\mathbb{Q}/\mathbb{Z} \mapsto C^{\times}$ defined with the rule $$f(\frac{p}{q} + \mathbb{Z}) = e^{\frac{2\pi i p}{q}}$$  where $\frac{p}{q}$ is the mentioned representative.   Somehow I have problems showing that this is a bijective function in a formal way. I suspect I do not know the properties of the complex roots of unity well enough.  Can someone point me out (perhaps with a hint) how to show that $f$ is injective and surjective?
51006
I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z$. This is abelian and has order $p^2$. I think the other is $\Bbb Z/p^2 \Bbb Z$.   Now, it should follow from the fact that there is only one cyclic group of order $n$ up to isomorphism that these two are unique up to isomorphism. All I need to show is these two are in fact not isomorphic. It suffices to show that $G$ as before is not cyclic. But this is easy to see, since we cannot generate any $(x,y)$ with $x\neq y$ by repeated addition of some $(z,z)$.  Now, it suffices to show that any other group of order $p^2$ is isomorphic to either one of these two groups. If the group is cyclic, we're done, so assume it is not cyclic. One can see that $G=\langle (1,0) ,(0,1)\rangle$. How can I move on?
86069
What is an example of an abelian group homomorphism from $\mathbb{R}$ to $\mathbb{R}$, respecting addition, that is NOT a linear transformation?  I'm trying to understand to what extent the scaling condition for the linear transformation is necessary. For $\mathbb{Q}$ vector spaces, maps that preserve addition also preserves scaling.  It seems the situation is more complicated for $\mathbb{R}$.  I guess I can "choose" a basis of $\mathbb{R}$ as a $\mathbb{Q}$ vector space and go from there, but are there any explicit examples, preferably without using the Axiom of Choice?   More generally, is there a description of the group of abelian group homomorphisms from $\mathbb{R}^n$ to $\mathbb{R}^m$?
7492
I've tried, but I can't solve the question. Please help me prove that:  $\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to  $U_n$.
111385
Let G be the multiplicative group $\mathbb{Z}_n^*$ for $n = 2^k$ and $k \ge 3$.  Can we prove that no element has order bigger than $2^{k-2}$ ?  My solution (not really a solution) :  Since $n=2^k$, I thought that this group is cyclic and the generator is 2. I'm not sure if the group order can't be bigger than $2^{k-2}$?