Let the functor $F\colon\bf Ring\rightarrow\bf Grp$ send the ring $A$ to its group of units $A^\times,$ and the ring homomorphism $f\colon A\rightarrow B$ to the group homomorphism $f^\times\colon A^\times\rightarrow B^\times:a\mapsto f(a)$. I was curious about this functor, and in particular, whether it is essentially surjective. That is, for any group $G$ (not just finite,) is there a ring $A$ such that $G\cong A^\times$? If not, what groups $G$ satisfies this? A similar question was asked in  but what can be said for infinite groups, or groups in general? Calling groups that satisfy this condition R-groups, I have proved that any finitely generated abelian group is an R-group. I have no idea what to do from now, but I  have a conjecture that the unit group of the group ring $\mathbb F_2[G]$ is isomorphic to $G.$ If this is true, certainly, all groups are R-groups, and the functor $F$ is essentially surjective, but I am having trouble proving it. Can anyone help me?
68270
Motivated by :   Which finite groups are the group of units of some ring? Which finite groups are the group of units of some finite ring? Which finite abelian groups are the group of units of some commutative ring? Which finite abelian groups are the group of units of some finite commutative ring?   It  that even for fields this is not simple to answer.  I expect the general question (Which groups are the group of units of some ring?, with no finiteness hypotheses) is even harder.
18867
I am working on a school assignment and have been stuck on this question for some time.   Let $f : G \rightarrow G_1$ be a surjective homomorphism (also called epimorphism) from $G$ to another group $G_1$. Prove that $f(Z(G)) \subseteq Z(G_1)$  I know we start by assuming we have an element in $f(Z(G))$ then showing it is also in $Z(G_1)$. Also, $Z(G)$ is the center of $G$, so $Z(G)$ is all the elements of $G$ that commute with all elements in $G$. But how do we know that if we map the elements of $Z(G)$ to $G_1$, then their image will also commute with all the elements in $G_1$? Does it have something to do with using commutativity with the homomorphism? i.e. $f(gz) = f(g)f(z) = f(zg) = f(z)f(g)$.
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Let $R$ be a simple ring and $T$ be a non-zero ring. Let $f\colon R \rightarrow T$ be a ring homomorphism. Show $f$ is injective.  Proof: $\ker f \lhd R$, so $\ker f=R$ or $\ker f=\{0\}$. If $\ker f=R$ then $R/R \cong \{0\}$, but $T \neq \{0\}$, which is a contradiction. I can't see why does the last part means contradiction.   Thanks for any assistance!
6596
Let $G$ be a group containing a normal subgroup $H \cong \mathbb{Z_2}$ such that $G/H$ is infinite cyclic. Then $G \cong \mathbb{Z} \times \mathbb{Z_2}$.  Since $G/H$ is infinite cyclic it's isomorphic to $\mathbb{Z}$. And since $H \cong \mathbb{Z_2}$, we have $G/H \times H \cong \mathbb{Z} \times \mathbb{Z_2}$. So the question now is can we make $G/H \times H \cong G$ (without any recourse to the Isomorphism theorems please)?  Any hints appreciated.
111687
Let $\phi : G_1 \rightarrow G_2$ be a surjective group homomorphism. Let $H_1$ be a normal subgroup of $G_1$ and suppose that $\phi (H_1) = H_2$. Prove or disprove that $G_1/H_1 \cong G_2/H_2$.  I say they are indeed isomorphic. Because:  Let $f$ be the group homomorphism from $G_1$ to $G_2/H_2$ that sends $a$ to $\phi(a)$. Then the kernel of $f$ is everything that is sent to $H_2$. Well by assumption this is $H_1$. Since $\phi$ is surjective, so is $f$, so by the first isomorphism theorem, $G_1/H_1$ is isomorphic to $G_2/H_2$  Is this correct reasoning?
6226
Possible Duplicate:          $f:K\rightarrow R$ is a homomorphism from field to ring. To prove is that it is injective or zero.
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I am studying basic ring theory and I am stuck on the following (possibly easy) question.     Let $R$ be a commutative ring. Prove that there is a bijection between the set of ring homomorphisms $\mathbb{Z}[i]\to R$ and the set of $x\in R$ for which $x^2+1=0$.   Some  things I remember and can prove:   If $f$ is a ring homomorphism then $f(1)=1, f(a+b)=f(a)+f(b)$ and $f(ab)=f(a)f(b)$. Also $\mathbb{Z}[i]^* =\{1,i,-1,-i\}$ and I can prove $\mathbb{Z}[i] \cong \mathbb{Z}[X]/(X^2+1)$. My idea would be to construct a map from the set of ring homomorphisms to the set of $x \in R$ for which $x^2+1=0$ and show that it's bijective. The problem is that I don't know how to construct such a map (what does the set of ring homomorphisms look like?) and where to use that $R$ is commutative.  Any answer/big hint would be appreciated. Thank you.
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The following is the first part of Proposition 2.9 in "Introduction to Commutative Algebra" by Atiyah &amp; Macdonald.     Let $A$ be a commutative ring with $1$. Let $$M' \overset{u}{\longrightarrow}M\overset{v}{\longrightarrow}M''\longrightarrow  0\tag{1} $$ be sequence of $A$-modules and homomorphisms. Then the sequence (1) is exact if and only if for all $A$-modules $N$, the   sequence  $$0\longrightarrow \operatorname{Hom}(M'', N)  \overset{\overline{v}}{\longrightarrow}\operatorname{Hom}(M,  N)\overset{\overline{u}}{\longrightarrow}\operatorname{Hom}(M', N)  \tag{2} $$ is exact.   Here, $\overline{v}$ is the map defined by $\overline{v}(f)=f\circ v$ for every $f\in\operatorname{Hom}(M'', N)$ and $\overline{u}$ is defined likewise.   The proof one of direction, namely $(2)\Rightarrow (1)$ is given in the book, which I am having some trouble understanding. So assuming (2) is exact sequence, the authors remark that "since $\overline{v}$ is injective for all $N$, it follows that $v$ is surjective". Could someone explain why this follows?  Given that $\overline{v}$ is injective, we know that whenever $f(v(x))=g(v(x))$ for all $x\in M$, we have $f=g$. I am not sure how we conclude from this surjectivity of $v$.  Thanks!
88526
We have a function $\varphi:G\rightarrow H$ is an isomorphism, show its inverse $\varphi^{-1}:H\rightarrow G$ is also an isomorphism  I am fine with showing it to be a homomorphism and surjective, this is my attempt at showing injectivity, I'm unsure if they would be sufficient  Assume: $\forall h\in H, \exists g\in G$ s.t. $\varphi(g)=h$ and $\forall g_1, g_2\in G, \varphi(g_1)=\varphi(g_2)\Rightarrow g_1=g_2$  Injectivity - Need to show $\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=g_2$  $\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=\varphi(\varphi^{-1}(g_2))$ from our surjectivity assumption on $\varphi$ we know there exists such a $\varphi^{-1}(g_2)\in H$ and from our injectivity assumption on $\varphi$ we know that this solution is unique.   [Sufficient? What more could be added? How could I do this clearer?]
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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?
74387
I'm trying an old problem here:   Suppose $n\mid m$, and I have a natural ring homomorphism $\varphi\colon \mathbb{Z}/m\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ defined by $\varphi(j)=j\mod{n}$. I can verify that this is a ring homomorphism, but why does it induce a surjective group homomorphism on the unit groups $(\mathbb{Z}/m\mathbb{Z})^\times\to(\mathbb{Z}/n\mathbb{Z})^\times$?  My question in particular is why is it surjective? I take $k\in(\mathbb{Z}/n\mathbb{Z})^\times$. Then there exist integers $s,t$ such that $sn+tk=1$. Since $n\mid m$, I can also write $m=na$. So multiplying through I get $sm+tka=a$. I'm lost here. How can I find a unit in $(\mathbb{Z}/m\mathbb{Z})^\times$ which maps to $k$ to show surjectivity?
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I'm reading Grillet's Abstract Algebra. Let $F_X$ denote the free group on the set $X$. I noticed on wiki the claim $$F_X\cong\!\!F_Y\Leftrightarrow|X|=|Y|.$$ How can I prove the right implication (find a bijection $f:X\rightarrow Y$), i.e. that the rank is an invariant of free groups?  I am hoping for a simple and short proof, having all the tools of Grillet at hand. Rotman (Advanced Modern Algebra, p.305) proves it only for $|X|&lt;\infty$, Bogopolski's (Introduction to Group Theory, p.55) proof seems (unnecessarily?) complicated, and Lyndon &amp; Schupp's (Combinatorial Group Theory, p.1) proof I don't yet understand. It's the very first proposition in the book; in the proof, they say:     The subgroup $N$ of $F$ generated by   all squares of elements in $F$ is   normal, and $F/N$ is an elementary   abelian $2$-group of rank $|X|$. (If   $X$ is finite, $|F/N|=2^{|X|}$ finite;   if $|X|$ is infinite, $|F/N|=|X|$). $\square$   Is $N:=\langle w^2;w\in F\rangle$? What is an abelian $2$-group? Elementary? What and how does the above quote really prove? I'm guessing a free abelian group on $X$ is $\langle X|[X,X]\rangle\cong\bigoplus\limits_{x\in X} \mathbb{Z}$?  Can an isomorphism $\varphi:F_X\rightarrow F_Y$ not preserve the length of words? At least one letter words?
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If $m|n$. Why the map $f\colon \mathbb{Z}_n^\times \to \mathbb{Z}_m^\times$ given by $a \mod{n}\mapsto a \mod m$ is a surjective homomorphism of groups?  Attempt: I proved it is well a well defined homomorphism because the canonical projection $\overline{f}\colon \mathbb{Z}_n \to \mathbb{Z}_m$ is a surjective ring homomorphism, it extends $f$ and maps units to units.  But I can't prove surjectiveness. Given $a\in\mathbb{Z}$ such that $(a,m)=1$ I'm looking for a $k\in\mathbb{Z}$ such that $(a+k\cdot m,n)=1$. I don't know why one of the $n/m$ candidates for such $k$ makes the number $a+k\cdot m$ an unit modulo $n$.
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Is the following proof correct, or are certain steps illegal?  Suppose $f:G_1\to G_2$ is a surjective group isomorphism, $N_1\lhd G_1, N_2\lhd G_2$ normal subgroups, with $N_1=f^{-1}(N_2)$. Prove that $(G_1/N_1)\cong (G_2/N_2)$.  Proof:  As $f$ is surjective, we an use the first isomorphism theorem to get $G_1/\ker(f)\cong G_2$, which implies that $$(G_1/\ker f)/(N_1/\ker f)\cong G_2/(N_1/\ker f)$$  Using the first isomorphism theorem we also know that $$N_1/\ker (f)\cong f(N_1)=f(f^{-1}(N_2))=N_2$$  Combining this we can see that $(G_1/\ker f)/(N_1/\ker f)\cong G_2/(N_1/\ker f)\cong G_2/N_2$.  Using the third isomorphism theorem we have that: $$ (G_1/\ker f)/(N_1/\ker f)\cong G_1/N_1$$  And from the last two lines we have $G_1/N_1\cong G_2/N_2$ $\tag*{$\Box$}$
2529
I am having trouble with the following problem.      Let $R$ be an integral domain, and let $a \in R$ be a non-zero element. Let $D = \{1, a, a^2, ...\}$. I need to show that $R_D \cong R[x]/(ax-1)$.    I just want a hint.  Basically, I've been looking for a surjective homomorphism from $R[x]$ to $R_D$, but everything I've tried has failed. I think the fact that $f(a)$ is a unit, where $f$ is our mapping, is relevant, but I'm not sure. Thanks
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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?
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Let $G$ and $H$ be groups and assume that there exist onto homomorphisms $$\phi :G\rightarrow H,\qquad\psi :H\rightarrow G.$$ If $H$ and $G$ are finite then it is clear that $H$ and $G$ are isomorphic. It is also valid for finitly generated abelian groups. However, in general this is not true even in the abelian case.     My question is: if we know that $G,H$ are finitly generated and such $\phi ,\psi$ exist then are $G$ and $H$ isomorphic?