I have a problem with this question  is $(\mathbb{R},*)$, when $a*b=ab+a+b$ a group? If not, can you skip any element $a \in\mathbb{R} $ in that way, that $(\mathbb{R}$\{a}$,*)$ is a group?  I can prove, that $(\mathbb{R},*)$ is a binary operation, associative and it's neutral element is $0$. But I couldn't find an inverse element.  And I have no idea which element I can skip to get the other group. I tried to skip $a=0$ but it wasn't a good tip.  Thank you for your time.
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.
24596
I am trying to prove that $(G, *)$ is an abelian group with $G=(-1,1)$ and $a*b=$$\frac{x+y}{1+xy}$.  Thus far I have found that the identity element $e=0$. From here, I set $a*b=0$ and found $a^{-1}$ to be $-a$.   My work for trying to prove closure and that the set is abelian is:  Let $a,b \in G$, where $a=e$ and $b=a^{-1}$. Does  $a*b=b*a$?  Evaluating, I see that $a^{-1}=a^{-1}$ which then proves that G is closed and G is an abelian group.  Did I go about this proof correctly and efficiently?
109899
$\mathbb R$ is an extension of $\mathbb Q$.  Why the elements of $\mathbb Q(2^{1/3})$ have the form $a+b\sqrt[3]2$?  More specifically why $\mathbb Q(2^{1/3})=\{a+b\sqrt[3]2+c\sqrt[3]4:a,b,c \in\mathbb Q\}$?
49696
Find all the group homomorphisms from $(\mathbb{Q}, +)$ into $(\mathbb{R}, +)$.  My attempt:  If $\mathbb{Q}$ were a cyclic group, I could tell that any homomorphism will be determined by the image of generator. But here $\mathbb{Q}$ is not a cyclic group, so there's no generator. All one can say is that:   if $f$ is a homomorphism then $f(0)=0$.   But this doesn't help me to solve this problem. So how should it be tackled?
82315
What do we mean when we say &quot;Let $x$ be an element of the set $\mathbb{R}$&quot;?  Does $x$ represents only a single  element of set $\mathbb{R}$? or does $x$ represent all the element of set $\mathbb{R}$ simultaneously at the same time? People say that if $x\in\mathbb{R}$ then $x$ is any real number; that means $x$ represents all real numbers. But if $x$ is any real number then let's say $x=1$; so $x$ is one, then how it can represent all real numbers? Please help me I am very confused.
29488
Is easy to show that $\mathbb{R}$ is isomorphic to $\mathbb{R}^n$ as $\mathbb{Q}$-vector spaces and then $(\mathbb{R},+)$ is isomorphic to $(\mathbb{R}^n,+)$ as abelian groups. This result might suggest that we also have $(\mathbb{Q},+) \cong (\mathbb{Q}^n,+)$ as abelian groups, but a similar approach to this dosen't work.  How can I prove or disprove this?
2008
I. Let $G$ be a nonempty set closed under an associative product, which in addition satisfies:  (a) There exists an $e\in G$ such that $a\cdot e=a$ for all $a\in G$.  (b) Give $a\in G$, there exists an element $a^{-1}\in G$ such that $a\cdot a^{-1}=e$.  Prove that $G$ must be a group under this product.  II. Prove, by an example, that right indentity element and left inverse does not imply that $G$ is group.  My solution:      I. Since $G$ is closed  set under an associative product, i.e. if $a,b,c\in G$ then $(a\cdot b)\cdot c=a\cdot (b\cdot c)\in G$. Taking $c=e$ we get $(a\cdot b)\cdot  e=a\cdot (b\cdot e)=a\cdot b \in G$. We have shown that $\cdot$ is binary operation.       Since $a\in G$ then $a^{-1}\in G$ and we have the following identities $$a^{-1}=a^{-1}\cdot e=a^{-1}\cdot (a\cdot a^{-1})=(a^{-1}\cdot a)\cdot a^{-1} $$   Then    $$ \begin{align} e &amp;= a^{-1}\cdot (a^{-1})^{-1} \\ &amp;= \left( \left( a^{-1}\cdot a \right) \cdot a^{-1} \right) \cdot \left( a^{-1}\right)^{-1} \\ &amp;= \left( a^{-1}\cdot a \right) \cdot \left( a^{-1}\cdot \left( a^{-1} \right)^{-1} \right) \\ &amp;= \left( a^{-1}\cdot a \right) \cdot e \\ &amp;=a^{-1}\cdot a.  \end{align} $$   Thus we have shown that    $$ a\cdot a^{-1}=a^{-1}\cdot a=e. $$   Then we see that    $$ e\cdot a= \left( a\cdot a^{-1} \right) \cdot a = a \cdot \left( a^{-1} \cdot  a \right) = a \cdot e = a. $$       We have shown that for this set $G$ and the associative binary operation assumed to be defined on $G$, the properties of the existence of a two-sided identity element in $G$ and the existence  in $G$ of a two-sided inverse for each element of $G$ are satisfied. Therefore $G$ is indeed a group.   II. But II indeed is true. Lets take the set $G=\{a,b,e\}$ and define the product $\cdot$ by the following identities: $e\cdot e=a\cdot e=b\cdot e=e$ and $a^{-1}=b, \ b^{-1}=a$ and consider the following multiplication table for our set $G$  $\begin{array}{c | c c c c c} \hline\hline  &amp; e &amp; a &amp; b  \\ \hline e &amp; e &amp; b &amp; b &amp; \\ a &amp; a &amp; a &amp; e &amp; \\ b &amp; b &amp; e &amp; a &amp; \\ \hline  \end{array} $  It's easy to verify that conditions of second problem hold for our $G$, however, $G$ is not group since we can show that $b=a$.  Is my reasoning above correct?  EDIT: Maybe this is a duplicate but I would like to know if my solution is true since I have solved it by myself. Especiaaly I am interested in the solution of the second problem.
118281
[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
86031
Is it correct to say that $\mathbb{R}$ is a subset of $\mathbb{R}^2$? Or, put more generally, given $n,m\in\mathbb{N}$, $n&lt;m$, is $\mathbb{R}^n$ a subset of $\mathbb{R}^m$?  Also, strictly related to that: what is then the "relationship" between the set $\{(x,0)\in\mathbb{R}^2,x\in\mathbb{R}\}\subset\mathbb{R}^2$ and $\mathbb{R}$? Do they coincide (I would say no)? As vector spaces, do they have the same dimension (I would say yes)?  If you could give me a reference book for this kind of stuff, I would really appreciate it.  Thank you very much in advance.  (Please correct the tags if they are not appropriate)
3922
G is the set of matrices of the form $G=$$\begin{pmatrix} x &amp; x \\ x &amp; x \end{pmatrix}$. So for this set to be a group I know it needs to be:    Closed under matrix multiplication The Associative Property holds Contains an Identity Element Every element needs to have an inverse   So the form of the matrices is such that all the elements are the same but not 0. How do I go about proving these?  Working through this problem, I seem to have hit a contradiction. Since G is a subgroup of the bigger $2x2$ nonsingular matrices group why does G not have the same identity element as its parent group? Namely \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1 \end{bmatrix}  Isn't the subgroup supposed to have the same identity element as its parent group?
97559
how to prove that if $A$ and $B$ are open in $(\mathbb{R},|.|)$ then  $A+B$ is open ?  Where   $A+B=\{a+b\mid a\in A,b\in B\}$   Thank you
54372
I am presented with the following task:     Show that if $U$ is the collection of all units in a ring $\langle R, +, \cdot\rangle$ with unity, then $\langle U, \cdot\rangle$ is a group.   I am still not confident with proofs in Abstract Algebra, and I'd like to receive some verification/slaughter/constructive critique on my reasoning here. I do not know if there are any mistakes, or how serious they are.     We know that there is an identity element, since the unity itself is a unit, and therefore in the group. It follows directly from the definition of a unit that there exists an inverse for every element in the group, and associativity follows directly from the Ring-axioms. As for closure, we need to show that the product of two units is also a unit. Let $a, b \in U$ be two elements with the property that $a\cdot b \ne e$, where $e$ is the identity element in $\langle U, \cdot \rangle$. We know that there exists elements $a', b' \in U$ with the property that $a' \cdot a = e = a' \cdot a$ and $b' \cdot b = e = b \cdot b'$. Let $a', b'$ be two such elements with the property that $a' \cdot b' \ne e$. We may now compute $(a\cdot b) \cdot (a' \cdot b')$, and from the Ring-axioms we have associativity under multiplication, so we may write the expression as $a \cdot (b \cdot a' \cdot b') = a \cdot (b \cdot b' \cdot a') = a \cdot (e \cdot a') = a \cdot a' = e$. Thus we have shown that the product of any two elements is a unity, which leads to closure. All the group-axioms are met and $\langle U, \cdot \rangle$ is a group.    I have to apologize if you find the argument needlessly long; I've been having some trouble with assuming things I am not supposed to assume (like what I'm going to prove), so I wanted to stay clear of that.
108895
Let $G$ be a group and $A$ &amp; $B$ be two sets s.t. $G$ acts transitively on each of $A$ &amp; $B$. Choose some $\alpha$ and $\beta$ in $A$ &amp; $B$ respectively then prove that if $G=G_\alpha G_\beta$ then $G=G_x G_y$ for all $x$ $\in A$ &amp; $y$ $\in B$   where $G_x$ is stabilizer of $x$ in $G$ and similarly $G_\alpha$ be the stabilizer of $\alpha$ and $G_y$ is the stabilizer of $y$.  I know we have to choose some elements and apply some manipulative trick. it is hardly 2 liner but just can't think of right elements. help!
123815
I have some difficulties with a task in algebra. I guess it's trivial and really easy but I can't figure out how to solve it.  I have a set $G$ and a binary operation on it, let it be $\circ$. I have that the operation is associative and that the equations $a\circ x = b$ and $x\circ a = b$ have unique solutions. I have to prove that $(G, \circ)$ is a group.  I already have that the operation is binary and associative, so I have to prove that there is unique identity element and unique inverse element and it will come from the equations, but how exactly?
110811
Is my proof correct? I have made use of the fact isomorphism preserves order of elements, which I proved couple of exercises back. I am also interested in other ways of proving it. Is there a more explicit way or is this explicit enough?    Problem Prove that the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.  Solution Recall that isomorphism preserves order of elements and hence if there exists an isomorphism from $\phi: \mathbb{C}-\{0\} \mapsto \mathbb{R}-\{0\}$, then $x \in \mathbb{C} - \{0\}$, and $\phi(x) \in \mathbb{R} - \{0\}$, then $\vert x \vert = \vert \phi(x) \vert$. Now note that the element $i \in \mathbb{C} - \{0\}$ has order $4$. However, no element in $\mathbb{R}-\{0\}$ has order $4$. Hence, no isomorphism can exist. Hence, the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.    Thanks
33550
Is there a better (or other) way(s) to prove the following statement? Also, the same argument works for multiplicative groups $\mathbb{R}-\{0\}$ and $\mathbb{Q}-\{0\}$, right?       Problem Prove that the additive groups $\mathbb{R}$ and $\mathbb{Q}$ are not isomorphic.   Solution By cantor's diagonal argument, there is no possible bijection between $\mathbb{Q}$ and $\mathbb{R}$. Since an isomorphism needs to be a bijection, there is no possible isomorphism between the additive groups $\mathbb{R}$ and $\mathbb{Q}$.    Thanks
21427
Let $R$ and $R'$ be rings (with 1 but no further assumptions) and $n \in \mathbb{N}$. Does the following implication hold?     If $M_n(R) \simeq M_n(R')$ then $R \simeq R'$.   If the rings are commutative then it follows from considering the centers of $M_n(R)$ and $M_n(R')$. If the rings are division it also holds and even stronger, it holds even if the matrices are not of the same size.  I thought about considering the embeddings of $R$ in $M_n(R)$ as diagonal matrices, but since I do not want to asume anything about the isomorphism I was not able to conclude that these subrings of matrices in $M_n(R)$ and $M_n(R')$ are isomorphic. I hope someone can help me or with a proof, an idea or a counterexample. Thank you.