This question is not a duplicated of  . It is a follow-on, asking about the distinction between the subject of that post, and the subject of this post.  In The Number System by Thurston the author introduces an algebraic structure he calls a .  The laws of a hemigroup are:   (i)   $\left(x\odot y\right)\odot z=x\odot \left(y\odot z\right),$ (ii)  $\left(x\odot y\right)=\left(y\odot x\right),$ (iii) $\left(x\odot y\right)=\left(x\odot z\right)\implies{y=z},$ (iv)  $\exists_e e\odot e=e.$   I have been told this defines a .  I will retain the uncommon designation hemigroup for now.  Thurston then introduces entities he calls dyads which are sets of ordered pairs formed of elements of the hemigroup constituting equivalence classes of the ordered pairs.  I write this as  $$\left[\![a,b\right]\!]\equiv\{ \left&lt;x,y\right&gt;\backepsilon{x\odot b=y\odot a}\}.$$  The algebra of these dyads satisfies the laws of a commtative group.  One such commutative group is what  call the module of integers.  In that source, however, this module is constructed of residue classes of ordered pairs from an algebraic structure which amounts to Thurston's hemigroup without the existence of an identity element.  That is   (i)   $\left(x\odot y\right)\odot z=x\odot \left(y\odot z\right),$ (ii)  $\left(x\odot y\right)=\left(y\odot x\right),$ (iii) $\left(x\odot y\right)=\left(x\odot z\right)\implies{y=z}.$   But those authors do not formally nominate this structure.  It is the additive algebraic structure of the natural numbers, which by traditional American definition are $\mathbb{N}\equiv\{1,2,3,\dots\}$.  Thurston's hemigroup is the additive structure of the whole numbers, which by traditional American definition are $\mathbb{N}_0\equiv \mathbb{N}\cup\{0\}.$    Both developments produce the same integral domain, but from different underlying algebraic structures.  I am asking if the algebraic structure consisting of a hemigroup sans identity has a common mathematical name.  If so, what is that name?
122240
In The Number System by Thurston the author introduces an algebraic structure he calls a "hemigroup".  It doesn't appear to be a very common usage.  The laws of a hemigroup are:   (i)   $\left(x*y\right)*z=x*\left(y*z\right)$ (ii)  $\left(x*y\right)=\left(y*x\right)$ (iii) $\left(x*y\right)=\left(x*z\right)\implies{y=z}$ (iv)  $\exists_e e*e=e$   Apparently there are other definitions for the same term.  Is there a more common term than "hemigroup" for this kind of structure?
91590
Find $|\mathbb{Z} \oplus \mathbb{Z} / \langle(3,6)\rangle|$ and determine if its cyclic.   $\mathbb{Z} \oplus \mathbb{Z} = \{(1,1), (1, 2),... (2, 1),...(3,1),...\}$  $\langle(3,6)\rangle = \{(0,0),(3,6),(6,12),...\}$  $\mathbb{Z} \oplus \mathbb{Z} / \langle(3,6)\rangle = \{(a,b) + \langle(3,6)\rangle : a,b \in \mathbb{Z}\}$  How would I go by finding the order of the quotient group? Would it be infinite? Since both sets that make up the quotient group are infinite, does that mean that the order of the quotient group is also infinite? What are some things I'm not noticing that could be helpful?
54678
I'd like to see a proof why $\varphi \in \operatorname{Hom}{(S^1, S^1)}$ looks like $z^n$ for an integer $n$.  At first I thought I could argue that if I have a homomorphism that maps $e^{ix}$ to some $e^{iy}$ for $x,y \in \mathbb R$ then $z = rx$ for some real number $r$. But on second thought I'm not sure why I can't have $\varphi (e^{ix} ) = e^{g(x)i}$ for a $g$ other than $g(x) = \lambda x$.   I'm also interested in seeing different proofs. I'm sure there are several ways to prove this.  Thanks for your help.
38110
I see this notation a lot, but I can't seem to find an answer in my google searches. In particular I am looking at the question      Let $\alpha:S^n\rightarrow S^n$ be the antipodal map. Prove that $n$ is odd, then $\alpha\simeq \operatorname{id}.$    If it is an arbitrary symmetric group, then I understand, but I have seen $S^1$ be used for the circle.
79181
I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).  I've already constructed two non-isomorphic total orderings. They are $$ 1 &lt; 2 &lt; 3 &lt;\ldots\\ 1 &gt; 2 &gt; 3 &gt;\ldots $$ where "$&lt;$" and "$&gt;$" are  strict total ordering relations.  Thanks in advance for your help.
107012
I know books have slightly different ways of presenting the axioms, but I think they tend to go like this: Group is a set with a law of composition with closure that satisfies the following properties (i) associativity, i.e. a(bc) = (ab)c (ii) identity, i.e. 1a = a1 = a (iii) inverse, i.e. every element a has an inverse such that ab = ba = 1.  Michael Artin's "Algebra" made no mention of this, but Fraleigh's "A First Course in Abstract Algebra" implied that it was important that the axioms were in that specific order, and wouldn't work any other way.  Why is this?  Thanks in advance.
64640
We know that every group $G$ of order $n$ can be embedded into $\mathrm{GL}_n(\mathbb{Z})$, because   $G \hookrightarrow S_n$ (Cayley theorem) $S_n\hookrightarrow \mathrm{GL}_n(\mathbb{Z})$ (via permutation matrices)   Furthermore we can show that $S_n \hookrightarrow \mathrm{GL}_{n-1}(\mathbb{Z})$, but we can't improve Cayley's theorem in general (i.e. we don't have $G \hookrightarrow S_{n-1}$).  For $n&gt;2$, let $k(n)$ the smallest integer $k$ such that all groups of order $n$ can be embedded in $\mathrm{GL}_{k}(\mathbb{Z})$.  We can show easily that $k(4)=2$, $k(5)=4$, $k(6)=2$. From the initial remark we have $k(n)\leqslant n-1$. Let $u_n=\frac{k(n)}{n}$. What can we say about $u_n$ ? It seems hard to fully determine $u_n$. For instance:   Can one extract a subsequence $u_{\varphi(n)}$, such that $\lim_{n\to \infty}u_{\varphi(n)} =1$? Or even stronger: is there infinitely many $n$ such that $k(n)=n-1$? What are the accumulation points of $u_n$? etc
71505
Consider ${\bf Z}$ as a first-order structure in the language $\{0,+,-\}$ of abelian groups.  Is ${\bf Z} \times {\bf Z}$ elementarily equivalent to ${\bf Z}$ and why?  I think they are not elementarily equivalent.
61948
I am trying to show that if for a group $G$ we have $Z(G)=1$, then $Z(\operatorname{Aut}(G))=1$.  I have tried everything I can think of, but to no avail.  The first part of the problem was to show that $\operatorname{Inn}(G) \triangleleft  \operatorname{Aut}(G)$ - does this have any connection?  I couldn't find one.  I know that $\operatorname{Inn}(G) \cong G/Z(G)$, so in this case $\operatorname{Inn}(G) \cong G$ and $Z(\operatorname{Inn}(G))=1$.  Is there any way to get from this to $Z(\operatorname{Aut}(G))=1$?  How do we go about showing this?  Thanks.
58037
Lets view groups as consisting of a binary operation, a distinguished element $e$, and unary operation $x \mapsto x^{-1}$. Then the group axioms can be stated as follows.   $(xy)z=x(yz).$ $xe=ex=x.$ $xx^{-1}=x^{-1}x=e.$   Supposing we dropped axiom 2 and forgot about $e$, we would need to rephrase 3. One possibility would be  3'. $xx^{-1}y=x^{-1}xy=yxx^{-1}=yx^{-1}x=y.$  So my first question is, can 3' be shortened and simplified?  Now suppose we dropped associativity, as well. Then 3' needs to be further rephrased. One possibility would be  3''. $(xx^{-1})y=(x^{-1}x)y=y(xx^{-1})=y(x^{-1}x)=y,$ and $x(x^{-1}y)=x^{-1}(xy)=(yx)x^{-1}=(yx^{-1})x=y.$  Again, my question is, can 3'' be shortened and simplified?
93443
$$(\mathbb{Z}/(n))[x] \simeq \mathbb{Z}[x] / (n)$$  I've shown this by showing that the map that sends $\overline{1} \mapsto [1+(n)]$ (where the bar denotes the congruence class mod $n$) and $x \mapsto [x+(n)]$ is a homomorphism that is injective and bijective, which is a bit cumbersome. Is there an easier way to show this? I can't tell if this is trivial or not.
48535
Let $G$, $H$, and $K$ be abelian groups. Assume $G \oplus H \cong G \oplus K$. If each are finitely generated, then the structure theorem easily gives us that $H \cong K$. However, there are counterexamples in the general case, namely $H=\mathbb{Z}$, $K=\mathbb{Z} \oplus \mathbb{Z}$, and $G=\bigoplus_{i=1}^\infty\mathbb{Z}$.   I am wondering what goes wrong with the following argument:   Form the split short exact sequence  $$0 \to G \xrightarrow{i} G \oplus H \xrightarrow{\pi} H \to 0$$  where $i$ and $\pi$ are the inclusion and projection, respectively. Then don't we immediately get from the first isomorphism theorem that $H \cong (G \oplus H)/G$, which would contradict the counterexample above?   If we run this for the counterexample given, we get the sequence   $$0 \to \bigoplus_{i=1}^\infty\mathbb{Z} \to \bigoplus_{i=1}^\infty\mathbb{Z} \to \mathbb{Z} \to 0$$  which seems to still be short exact, even though here $G \cong G \oplus H$. But of course here we can see directly that $(G \oplus H)/G \cong G/G$, which is trivial. So what went wrong?   It seems to me to be something along these lines: In this case, the kernel of the projection is not the same as the original group, it is just, say, $0 \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \dots$, i.e. nontrivial from the second coordinate on, which just happens to be isomorphic to the original group. So if we form the other short exact sequence with $K$ in place of $H$, then the difference seems to be that we have two different projection maps with different (but isomorphic) kernels.   But I'm not really sure how to explain this in the general case, if my reasoning is correct to begin with.   Thanks in advance for any help!
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.
123173
The set $Z_n = \{ 0, 1, . . . , n-1 \}$ for $n ≥ 1$ is a group under addition modulo n. For any $j &gt; 0$ in $Z_n$, the inverse of $j$ is $n - j$. This group is usually referred to as the group of integers modulo n.  My understanding of this is as following, please let me know if I have gotten the main idea.  Let's pick $Z_3$ as an example.  Then the members of this group is $0,1,2$ let $j=2$ then $n-j=3-2=1$  it says the inverse of $j$ is $n - j$, by definition, an inverse in a group is such ab=identity. And here the identity is 0 under addition.   therefore, 2+1=3 and 3 mod 3 =0 which is the identity.   This applies to any n in $Z_n$  So, it satisfies as a group?
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.
70385
The reasoning my book uses is as follows:   Since $(3+i)(3-i) = 10$, $10 + \langle 3+i\rangle = 0 + \langle 3+i\rangle $. Also, $i + \langle 3+i\rangle = -3 + \langle 3+i\rangle = 7 + \langle 3+i\rangle$. So, $\mathbb{Z}[i]/\langle 3+i\rangle = \{k + \langle 3+i\rangle \mid k = 0,1,\ldots ,9\}$, since $1 + \langle3+i\rangle$ has additive order $10$.   I do not understand why the additive order of $1 + \langle 3+i\rangle$ is $10$, or why the fact that $(3+i)(3-i) = 10$, shows the additive order or $1 + \langle 3+i\rangle$  is $10$. If someone could please explain this I would greatly appreciate it.   Note: I apologize for not using MathJax to write this post. I intend to learn it over the coming weekend.
27737
I was trying to give the wording "isomorphic groups have the same structure" a precise interpretation, at least for the finite order case. A positive answer to my final question would reach the goal, as it would translate a "$\cong$" between groups into a "$=$" between sets.    For $n$ positive integer, let be:   $I_n:=\{1,\dots,n\}$; $G$, $\overline G$ groups of order $n$; $\psi\colon G \rightarrow \overline G$ isomorphism; $f$, $\bar f$ bijections; $\theta$, $\bar \theta$ embeddings; in general, $\varphi^{(\alpha)}$ the isomorphism between symmetric groups on sets of the same cardinality, defined by $\sigma \mapsto (g \mapsto (\alpha\sigma\alpha^{-1})(g))$, where $\alpha$ is a bijection between the sets; $S_n$ the symmetric group of degree $n$.   Visually:  $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c}    I_n     &amp;   &amp;    I_n      \\     \da{f}         &amp;           &amp;     \da{\bar f}          \\    G &amp; \ras{\psi} &amp;    \overline G  \\     \da{\theta}         &amp;           &amp;     \da{\bar \theta}          \\    \operatorname{Sym}(G) &amp; \ras{\varphi^{(\psi)}} &amp;    \operatorname{Sym}(\overline G)  \\     \da{\varphi^{(f)}}         &amp;           &amp;     \da{\varphi^{(\bar f)}}          \\    S_n &amp;  &amp;    S_n  \\ \end{array} $$     Is it $(\varphi^{(f)}\circ\theta)(G)=(\varphi^{(\bar f)}\circ\bar \theta)(\overline G)$?