In Herstein, the defination of isomorphism is a homomorphism which is one-one and the definition of isomorphic is that it's the onto isomorphism (i am sure it's not a printing mistake because he used this consistently) but some other sources on net show that isomorphism is a bijective homomorphism? So which definition is true?
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ProofWiki  as a bijective homomorphism. In Topics in Algebra 2$\varepsilon$, Herstein defines a group isomorphism as an injective homomorphism:     Definition. A homomorphism $\phi$ from $G$ into $\bar{G}$ is said to be an isomorphism if $\phi$ is one-to-one.   Herstein does not define a homomorphism to be surjective, so the definition on ProofWiki and the definition in Topics in Algebra 2$\varepsilon$ contradict. Which definition is standard?
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It is in my understanding that to prove that the stabilizer of $a \in A=\{1,...,n\}$ over the permutation group $S_n$ is isomorphic to the permutation group $S_{n-1}$ one needs to show that there exists a bijection between these two groups, which is easy. However, the other step is to show that a homomorphism defined on this bijection exists.  I would think that one such homomorphism can be defined as follows: for every permutation $\sigma \in S_n$, redefine $\sigma(a)$ such that $\sigma(a)=a$, then redefine the element which was initially pointing to $a$ in $\sigma$ to point to the element to which $a$ was initially pointing in $\sigma$.  But the question is: how to define this homomorphism formally?
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I know ring homomorphisms map units to units, which made me curious about the following. Suppose $f:R\to R'$ is a surjective ring homomorphism, mapping $1$ to $1'$. Is it necessarily surjective from $U(R)\to U(R')$?  I know if $f(u)$ is a unit in $R'$ with $f(w)$ its inverse, then $f(uw)=f(wu)=1'$, but I see no reason to conclude $uw=wu=1$ in $R$ without assuming $f^{-1}(1')=\{1\}$. But I can't find a counterexample, so I'm not sure whether it's true or not.
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Let $G = (G,\times)$ be a group, and $G^{op} = (G,o), a \ o \ b = b \times a$ be it's opposite group.  I need to prove that there is a group isomorphism $f: G \to G^{op}$ Obviously, $id_G: G \to G$ is a bijection, but it's a group homomorphism if and only if $G$ is commutative.
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How can we prove that any bijection on any set is  a composition of 2 involutions ?  Since involutions are bijections mapping elements of a set to elements of the same set, I find it weird that this applies to any bijection.  Thanks for your help !
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I am interested in the following question:  Does $G \times H \cong G$ necessarily imply that $H$ is trivial?  In order to demonstrate it is not, I attempted to construct an example, but I am not entirely sure it is valid:  Let $G = \langle \mathbb{Z}; [a,b] \ (\forall a,b \in \mathbb{Z})\rangle$.  Then $G \times G = \langle \mathbb{Z} \cup \mathbb{Z}; \ \{[a_1, b_1]\} \cup \{ [a_2, b_2]\} \cup \{[a_1,b_2]\}\rangle$, where my notation is intent to convey that the direct product presentation has the abelian relations from the "first" copy of G, the abelian relations from the "second" copy of G, and the required abelian relations which commute generators of the "first" copy with those of the "second" copy.    In this example, since our cardinalities are equal, $|\mathbb{Z} \times \mathbb{Z}| = |\mathbb{Z}|$, and since this implies that commuting all elements of $G$ and commuting all elements of the same cardinality generating set $G \times G$ requires a bijection between the relator sets, then we should have a bijective group homomorphism $\phi : G \rightarrow G \times G$ so that we have isomorphism, but obviously $G$ is not trivial.  Therefore this provides the necessary counterexample to the original statement.  My first question is: is there anything wrong with this proof?  My second question is: if there is nothing wrong with the proof, is there a more standard way of constructing this type proof?  In general, it would seem that whenever you have isomorphic generating sets and isomorphic relator sets, then this should be enough to describe any number of counterexamples to the claim that $H$ must be trivial.  Is this true?  Finally, since direct products can be viewed as the products in the category Ab, and since all final objects must be isomorphic, would there be an easier way (or maybe just an insightful way) of construct this argument entirely in a category theoretic context?  Thanks!
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I am trying to teach myself category theory and I am reading the section for Equivalence of categories. After some reading the question of "if there is a property of a category that is preserved by category isomorphism, but not equivalence" came to mi mind. The definitions are kind of similar and I simply cannot come up with an example about it.   any help will be appreciated since I cannot concentrate or move into the reading since my mind cannot rest due to the question above.
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It is commonly said that group homomorphisms "preserve the structure of the group", e.g., :     The purpose of defining a group homomorphism as it is, is to create functions that preserve the algebraic structure.   Now, my default notion of structure is the one that holds that isomorphisms are structural identities---two isomorphic mathematical objects have the same structure. But that notion of structure requires preserving not just the operations on the elements of the domain but also requires the two sets of objects be equipollent.  But homomorphisms aren't, in general, reversible---or else they'd be isomorphisms. Since homomorphisms don't preserve size, in what sense does it preserve "structure"? Is there any more to the notion of structure here than the simple definition of algebraic structure as a set endowed with certain operations?  Toy example illustrating the problem: suppose you have a group homomorphism between groups $G$ and $H$ (but no isomorphism). Suppose also that you have an isomorphism between $G$ and some further group $J$. Since there is a homomorphism between $G$ and $H$, the structure in $G$ is preserved in $H$. Since there is an isomorphism between $G$ and $J$ they have the same structure. But presumably if $H$ preserves the structure of $G$ then they have the same structure. That can't be true, though, since they aren't isomorphic. It seems then that either "structure" means two different things when discussing isomorphisms and homomorphisms, or "preserves" doesn't mean that the homomorphic sets have the same structure (and then I have no idea what "preserves" actually means in this context).  I'm clearly confused about something. I'm just not sure what.   TL;DR What is the difference between the notions of "structure" when you say that isomorphic structures have the same structure and when you say that a group homomorphism preserves the group structure?
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Why is an image of a homomorphism a subgroup of the codomain? I know that the image of any function is a subset of the domain, but I don't know how to proceed with the former statement.
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Suppose that we are given two graphs on a relatively small number of vertices. Here's an example:    Are these two graphs isomorphic? (That is: is there a bijection $f$ from $\{A,B,\dots,I\}$, the vertex set of the first graph, to $\{1,2,\dots,9\}$, the vertex set of the second graph, such that $vw$ is an edge of the first graph if and only if $f(v)f(w)$ is an edge of the second graph?)   If not, how can we tell? If so, how can we tell—and how can we find an isomorphism?     This question comes up a lot, usually asking about specific examples of two graphs which may or may not be isomorphic. Sometimes, the fiendish teacher of graph theory will even give three or more graphs and ask "Which of these are isomorphic?"  I'm asking this question so that we can provide  to questions of this type which would be worth linking to when specific instances of this question come up. So all improvements and/or suggestions are welcome!
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How do I show that $\text{Hom}(\mathbb{Z},G)$ and $G$ are isomorphic?  The method suggested by the book is to define a map $k$: $\text(\mathbb{Z},G) \to G$ by $f \to f(1)$.   I am stuck on how exactly to show that it is an isomorphism. Could anyone shed some light into this? Is $f(1)$ an element of G?  (For those who wanted to know, this is not homework)
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So imagine you have a group $G$ and we consider the set of group homomorphisms from $\mathbb{Z}$ to  $G$ specified by  $\forall g$ $\in G$ $\exists$   $\phi(1)=g$. Each of these homomorphisms is in 1-1 relation with the elements of $G$.  Consider groups for a moment in the context of category theory, viewing the groups as one object categories and thereof group homomorphism as functors. What would a natural isomorphism mean in group theory terms for G?  My thoughts are that the natural transformations equate to maps between homomorphisms and since the homomorphisms are in 1-1 relation with the elements of G, then a natural transformation is a map between group elements, i.e another element of the group. I'm not entirely convinced by this.   Anyway continuing my reasoning a natural isomorphism is a map between the homomorphisms that has an inverse, the thing is in relation to the elements of G this is every element? Furthermore I'm looking to find an equivalence class defined by this natural isomorphism, is this just a collection of the elements that are each others inverses?  Thanks in advance for any assistance.
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So I believe I've finished the proof, but it all amounts to case-work and I really think (and hope) there's a better approach than this. Here's what I've done: Attempt: Within $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ identities must be mapped to identities thus what defines each isomorphism is the bijective function between its last three elements: \begin{matrix} (0,0) &amp; \mapsto &amp; (0,0) \\ (0,1) &amp; &amp; (0,1) \\(1,0) &amp; &amp; (1,0) \\(1,1) &amp; &amp; (1,1) \\ \end{matrix} If every bijective function between these last three elements constitutes a homomorphism (including the first unchanged mapping of identities), then $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3$ in that isomorphisms in $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ can be identified with permutations of three elements, which is exactly what $S_3$ is. I then, essentially go case-by-case showing every one of the six different bijections is a homomorphism. For two of those bijections (which are just compositions of other bijections I've already checked) I apply the following lemma:  If $\phi :G \to H$ and $\psi : H \to K$ are bijective homomorphisms, then $(\psi \circ \phi)$ is a homomorphism. As proof, assume the premise. We then have $\forall a,b \in G:$ $$\psi (\phi (ab))= \psi(\phi(a) \phi(b))=\psi (\phi(a))\psi (\phi(b))$$ $\implies (\psi \circ \phi)$ is a homomorphism.  So my questions are:  Is the proof that I have sufficient? Is there a better proof for this problem? (A hint towards it rather than complete give-away would be much more appreciated)  EDIT: I have indeed looked at , but the approach carried out there uses vector spaces which I have not been introduced to or am supposed to use to solve this problem.
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Consider $\mathbb{Z}_{15}$, and $\mathbb{Z}_{18}$. Let's say I want to find all homomorphisms $f:\mathbb{Z}_{15}\rightarrow \mathbb{Z}_{18}$.  I'm not interested in the answer in particular, mostly I'm concerned about understanding the properties of homomorphism, so I can answer these kind of questions myself.    So, first of all, I know that homomorphism of cyclic group is completely determined by it's generator. But, will any mapping do?    For example, the easiest one to find is $f(1)=0$, where $Imf=\left\{0\right\}$, and $Ker=G$  (correct me if I'm wrong, this kind of $f$ can be defined between any two groups). Now, what things do I need to consider, when trying to find another one (if it exist)? Can I decide that $f(1)=1$? (It is not onto, but that shouldn't bother me) And what about $f(1)=2$? and so on...  My second question is: what about non-cyclic groups? Consider $D_{10}$ and $\mathbb{Z}_{18}$, for example. Do I need to go and define $f$ for each and every $g\in D_{10}$? (it doesn't have a generator)  A link to a useful (and simple) summary regarding homomorphisms properties will also be great. Thank you in advance.
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I am revising for my Rings and Modules exam and am stuck on the following two questions:  $1.$ Let $M$ be a noetherian module and $ \  f : M \rightarrow M \  $  a surjective homomorphism. Show that $f : M \rightarrow M $ is an isomorphism.  $2$. Show that if a semi-simple module is noetherian then it is artinian.  Both these questions seem like they should be fairly straightforward to prove but I cannot seem to solve them.
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If $f: \mathbb Q\to \mathbb Q$ is a homomorphism, prove that $f(x)=0$ for all $x\in\mathbb Q$ or $f(x)=x$ for all $x$ in $\mathbb Q$.  I'm wondering if you can help me with this one?
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This is a very simple question but I believe it's nontrivial.  I would like to know if the following is true:      If $R$ and $S$ are rings and $R[x]$ and $S[x]$ are isomorphic as rings, then $R$ and $S$ are isomorphic.    Thanks!  If there isn't a proof (or disproof) of the general result, I would be interested to know if there are particular cases when this claim is true.