I've read in many books, that $\Bbb Z/3\Bbb Z \times\Bbb Z/5\Bbb Z\cong \Bbb Z/15\Bbb Z$   and $\Bbb Z/3\Bbb Z ⊗ \Bbb Z/5\Bbb Z\cong 0$ .  How can I prove the isomorphism of them?  For  $\Bbb Z/3\Bbb Z ⊗ \Bbb Z/5\Bbb Z\cong 0$,   I let B: $\Bbb Z/3\Bbb Z \times\Bbb Z/5\Bbb Z \to A $ , a bilinear form  and $3B(a,b)=B(3a,b)=B(0,b)=0; $  and $5B(a,b)=B(a,5b)=B(a,0)=0.$ Is it right? How can I prove the first one?  Any help would be appreciated!
26849
I've just started to learn about the tensor product and I want to show: $$(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.$$  Can you tell me if my proof is right:  $\mathbb{Z}/m\mathbb{Z}$ and $\mathbb{Z} / n \mathbb{Z}$ are both finite free $\mathbb{Z}$-modules with the basis consisting of one single element $\{ 1 \}$. So $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$ has the basis $\{ 1 \otimes 1 \}$.  Therefore, any element in $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$  is of the form $(ab) 1 \otimes 1$ and any element in $\mathbb{Z}/ \gcd(m,n)\mathbb{Z}$ is of the form $k 1 = k$ where $k \in \{ 0, \dots , \gcd(n,m) \}$.  I would like to construct an isomorphism that maps $ab$ to some $k$. Let this map be $ab (1 \otimes 1) \mapsto ab \bmod \gcd(n,m)$.  This is a homomorphism between modules: it maps $0$ to $0$ because it maps the empty sum to the empty sum. It also fulfills $f(a + b) = f(a) + f(b)$ because there is only one element, $a = 1$.   It is surjective. So all I need to show is that it is injective. But that is clear too because if $ab \equiv 0 \bmod \gcd(m,n)$  then both $a \equiv 0 \bmod n$ and $b \equiv 0 \bmod m$ so the kernel is trivial.  Many thanks for your help!!
2197
Let $a, b$ be elements of a group $G$. Suppose that a has order $5$ and that $a^3b = ba^3$. I want to show that that $ab = ba$.  Here is what I think:   We know that we have $a^1, a^2, a^3, a^4, a^5 = 1$. So, $a^4 = a^{-1}$ and $a^3 = a^{-2}$ and $a^{2} = a^{-4}$. Now,   $a^3b = ba^3$ then $a^{-2}b = ba^{-2}$ then $ba^2 = a^2b$ then $ba^{-4} = a^{-4}b$ then $a^{-1}b = ba^{-1}$ then multiply on the left and right by $a$ to get $ba=ab$. Is this correct or is there a simpler approach?
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.
109849
Show that cyclic group of finite order $n$ is isomorphic to $(\mathbb Z_n,+)$ Actually, the answer to this question is available in MSE but I am not getting it,and one answer is not in the English language.Please kindly give a proof in easy language. Thank you.\ Here is a solution I don't understand:
21732
How many isomorphisms are there from $\Bbb Z_{12}$ to $\Bbb Z_{4} \oplus \Bbb Z_{3}$?  I can see that if $f$ is the isomorphism, the isomorphism is determined by the value of $f(1)$ because $f(1) = (1,1)$ and $f(k) = f(k) + f(1) = f(k) + (1,1)$.  But since $\Bbb Z_{12}$ has $4$ generators, there are $4$ different ways to set $f(1)$ equal to a generator.  Therefore there are $4$ isomorphisms.  Is this correct?
28394
Consider the quotient $\mathbb{Z}-\text{module}$ $$M=\frac{\mathbb{Z}^3}{\langle (3,3,1),(2,2,2)\rangle}$$ Prove that $M$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_4$.  I think the first step is to find a basis of $M$. Clearly $(1,0,0)$ and $(0,0,1)$ are in a basis of $M$. I don't know how to find the third element in that basis. Also the basis in $\mathbb{Z} \oplus \mathbb{Z}_4$ are $(1,0)$ and $(0,1)$.
101238
Prove that $\mathbb Z^n$ is not isomorphic to $\mathbb Z^m$ for $m\neq n$.   My try:  Let $\mathbb Z^n\cong \mathbb Z^m $. To show that $m=n$.  Case 1: Let $m&gt;n$. Now that $\mathbb Z^m$ has $m$ generators whereas $\mathbb Z^n$ has $n$  generators and an isomorphism takes a generator to generator ; that is the contradiction.  Please correct me if I am wrong.  The case  $m&lt;n$ also follows similarly.
44993
In every Modern Algebra book I've read, I've seen that the groups $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}_n$ are isomorphic, but not equal. I understand the difference between "isomorphic" and "equal," but this particular example raises a couple of questions for me.  I know the first group consists of cosets $k + n\mathbb{Z}$, and the second group consists of equivalence classes $[k]_n$, but isn't it true that $k + n\mathbb{Z} = [k]_n$ for every integer $k$? (They are both sets containing the same elements from $\mathbb{Z}$.) And if so, then can't we say that $\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n$? Thanks in advance!
46925
Find a subgroup of $S_4$ isomorphic to $\mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$.  Here is what I have so far: We need to find a subgroup of $S_4$ that is isomorphic to $\{0, 1 \} \times \{0, 1 \}$. Now, $\{0, 1 \} \times \{0, 1 \} = \{(0,0), (0,1), (1,0), (1,1) \}$. The pairs $(0,0)$ and $(1,1)$ tell us that the identity map, $e$, is present. The pairs $(0,1)$ and $(1,0)$ tell us that the transposition map, $\tau$ is present. So a subgroup of $S_4$ isomorphic to $\mathbb{Z} / 2\mathbb{Z} \times \mathbb{Z} / 2\mathbb{Z}$ must be $\{e, \tau \}$.  Comment: This seemed too easy and I'm thinking there must be more. Thank you for your help!
109391
I am working through Paolo Aluffi's new GSM text on my own (self-study). On page 63, he asks the reader to     Prove that $\Bbb{Q}$ is not the direct product of two nontrivial groups.   For some context, this is an exercise following a section entitled "The category Grp".  I am assuming that he means "is not isomorphic to the direct product of two nontrivial groups", and I can see two possible ways to proceed with this proof, but have been unsuccessful with either approach.    Approach 1:  Show that the additive group of rationals has a property that is preserved by isomorphism that the direct product of two nontrivial groups does not have or vice-versa.  This seems challenging unless I can significantly narrow down the properties that a direct product of two nontrivial groups that was isomorphic to $\Bbb{Q}$ would necessarily have.  Approach 2:  Considering the section in which this question occurs, show that if $G$ and $H$ are nontrivial groups and $\Bbb{Q} \cong G \times H$, then there are homomorphisms $\varphi_{G}:\Bbb{Q} \rightarrow G$ and $\varphi_{H}:\Bbb{Q} \rightarrow H$ which do not factor or do not factor uniquely through the product $G \times H$.  This would be a contradiction, as $G \times H$ is a final object in the category Grp.  I would greatly appreciate suggestions on how to proceed further with either of these approaches or with alternate approaches.
94389
One can show that $\mathbb{P}^n\times \mathbb{P}^m$ is birational to $\mathbb{P}^{n+m}$ by making note of the Zariski topologies and the canonical isomorphism between affine spaces $\mathbb{A}^n\times \mathbb{A}^m\to \mathbb{A}^{n+m}$.   Question: Does it follow that there is isomorphism $\mathbb{P}^n\times \mathbb{P}^m\to \mathbb{P}^{n+m}$? If so, what is an example of such an isomorphism?   Unfortunately I am having a hard time coming up with one, and haven't seen too many examples thus far.
67241
I'm interested in the following problem from Artin's Algebra text:     Determine the structure of the ring $\mathbb Z[x]/(x^2 + 3,p)$, where (a) p = 3, (b) p = 5.   I know that by the isomorphism theorems for rings we can take the quotients successively, and so  $$\mathbb{Z}[x]/(p) \cong (\mathbb{Z}/p \mathbb{Z})[x] $$  as the map $\mathbb{Z}[x] \to (\mathbb{Z}/p \mathbb{Z})[x]$ defined by $\sum_{n} a_n x^n \mapsto \sum_{n} \overline{a_n} x^n$ is a surjective ring homomorphism with kernel $(p)$. Thus it remains to study the quotients  $$(\mathbb{Z}/p \mathbb{Z})[x]/(x^2+3) $$ for $p \in \{3,5\}$.    If $p=3$, $(x^2+3)=(x^2)$ in $(\mathbb{Z}/3 \mathbb{Z})[x]$, and by using polynomial division all distinct coset representatives can be reduced to the following list of 9 elements  $$\{0,1,2,x,1+x,2+x,2x,1+2x,2+2x\}. $$  Moreover, it can shown that the list above gives 9 distinct cosets, as no difference of two distinct elements of the list is a multiple of $x^2$. Since $1$ and $x$ generate two distinct additive groups of order $3$, the additive group of our quotient ring is not cyclic. Elementary group theory then shows   $$(\mathbb{Z}/3 \mathbb{Z})[x]/(x^2)^+ \cong (\mathbb{Z}/ 3 \mathbb{Z})^2 $$  as additive groups.   I was then about to conclude that the multiplication on the quotient is then compatible with the usual one in $(\mathbb{Z}/3\mathbb{Z})^2$, but this is wrong!  It can be seen that the quotient is not isomorphic to $(\mathbb{Z}/3\mathbb{Z})^2$ as a ring, because the former contains a nonzero element (represented by $x$) whose square is zero, while the latter contains no such elements.    If $p=5$, a full list of coset representatives is of length 25  $$\{0,1,2,3,4,x,1+x,2+x,3+x,4+x,2x,1+2x,2+2x,3+2x,4+2x,3x,1+3x,2+3x,3+3x,4+3x,4x,1+4x,2+4x,3+4x,4+4x \} .$$  And once again, one can see that these represent 25 distinct cosets. Similarly to the $p=5$ case, I've managed to prove that the additive group of this ring is isomorphic to $(\mathbb{Z}/5 \mathbb{Z})^2$.    My questions:   Have I made any mistakes in my argument? What exactly am I supposed to do in this question? determine the number of elements? Write down the tables for addition and multiplication?   Any further information about these quotients will be appreciated, thanks!
96660
I have a problem with something in abstract algebra. The problem is the following:  Prove or disprove that the mapping $\theta: \mathbb{C} \to \mathbb{C}$ defined by $\theta(a+bi)=a-bi$ is a ring isomorphism.  I know that it should be an isomorphism. I know how to show that the mapping is one-to-one, and I know how to show that the mapping respects both addition and multiplication as defined in $\mathbb{C}$. But how would you clearly show that $\theta$ is onto?  Thanks in advance for any help!
98374
The map $\phi: \mathbb{Z} \to \mathbb{Z}$ defined by $\phi (n) = n + 1$ for $n \in \mathbb{Z}$ is one to one and onto $\mathbb{Z}$. Give the definition of a binary operation $*$ on $\mathbb{Z}$ such that $\phi$ is an isomorphism mapping   a) $\langle\mathbb{Z},+\rangle$ onto $\langle\mathbb{Z}, *\rangle$    I absolutely do not understand this. What is the solution doing? Why are we starting out with $m*n$? And why do we have $\phi(m-1) * \phi(n-1)$?
109350
I am trying to determine if $\Bbb Z_6 \cong   \Bbb Z_3 \times \Bbb Z_2 $. I noticed that $\Bbb Z_6$ has a generator $1$ and $\Bbb Z_3 \times \Bbb Z_2$ has generator $(1,1)$. Now I set up the bijection $f :\Bbb Z_6 \to   \Bbb Z_3 \times \Bbb Z_2 $ where \begin{eqnarray} f(1)&amp;=&amp;(1,1)\\ f(1+1)&amp;=&amp;(1,1)+(1,1) = (2,0)\\ f(2+1)&amp;=&amp;(2,0)+(1,1) = (0,1)\\ f(3+1)&amp;=&amp;(0,1)+(1,1) = (1,0)\\ f(4+1)&amp;=&amp;(1,0)+(1,1) = (2,1)\\ f(0)&amp;=&amp;(1,1)+(2,1) = (0,0)\\ \end{eqnarray} So basically $f(1^n) = (1,1)^n$ where $1^n$ means we add the $1$ to itself $n$ times and take the result mod 6, so as per usual $1^n$ means applying the group operation on 1 to itself for $n$ times. So this is a bijection but I am stuck on showing that $f(ab)=f(a)f(b)$ for all $a,b \in \Bbb Z_6$. I could check all possibilities but there must be a better way to show this. I was thinking something like $f(1^n1^m) = f(1^{n+m})=(1,1)^{n+m}=(1,1)^n(1,1)^m = f(a)f(b) $ because every element in $\Bbb Z_6$ can be expressed as some power of $1$.  Is this right? And is it enough to show $\Bbb Z_6 \cong   \Bbb Z_3 \times \Bbb Z_2 $?
103007
Question. Let $M = \left\{ \begin{bmatrix} a &amp; b \\ -b &amp; a \end{bmatrix} \,\middle\vert\,  a, b \in \mathbb R \right\}$.  Show that $(M,+)$ and $(\mathbb C,+)$ are isomorphic.  Show that $(M^{*},*)$ and $(\mathbb C^{*},*)$ are isomorphic.    (Here, $M^{*}$ is defined as excluding the zero-element) I understand that there exists an isomorphism $\phi:M \rightarrow\mathbb C$ defined by $\begin{bmatrix} \alpha &amp; \beta \\ -\beta &amp; \alpha \end{bmatrix}\mapsto \alpha+i\beta$, ie. I can show that $\phi(M_{1}+M_{2}) = \phi(M_{1})+\phi(M_{2})$ and $\phi(M_{1}*M_{2}) = \phi(M_{1})*\phi(M_{2})$. However, I'm unsure how to go about the two subquestions seperately.
81649
I found an argument online that the two fields are not isomorphic but I can't make sense of the argument. You find it .  These are the things I'm confused about when it comes to the argument.     It says that the field $\mathbb Q[x]/(x^2+1)$ contains a root of the polynomial $x^2+1$. I suppose this is the congruence class $[x]$? But this is not the same as the zero element of the field $\mathbb Q[x]/(x^2+1)$, right?   It says that every isomorfism takes $0$ to $0$. I interpret this as the zero-element in $\mathbb Q[x]/(x^2+1)$ (which I interpret to be $[0]$) is taken to the zero-element in $\mathbb Q[x]/(x^2+2)$ denoted $[0]_2$. But in the argument it says that this means that if $\mathbb Q[x]/(x^2+1)$ has a root of the polynomial, then also $\mathbb Q[x]/(x^2+2)$ must have a root of this same polynomial. Why is that?     Greatful for any help!