Let G be $\underbrace{\mathbb Z_p\times\dots\times \mathbb Z_p}_{n \text{ times}}$. Find $A(G)$ (the Automorphism group of $G$).   ($\mathbb Z_p$ is the integres modulo p, for example $\mathbb Z_2=\{0,1\},\ \  \mathbb Z_5=\{0,1,2,3,4\}$)  I thoght about:  Think about $(\mathbb Z_p)^n$ as vector space over $\mathbb Z_p$ of dimension n. Then to think about the automorphisms as linear transformations  (do I need to explain it more?). So $A(\mathbb Z_p)^n$ is equal to to the space of all invertible linear transformations of $(\mathbb Z_p)^n$ to itself.  Then we get that $A(G)\cong GL_n(\mathbb Z_p)$.   is that enough or I need to add something?
121238
Let G be $\underbrace{\mathbb Z_p\times\dots\times \mathbb Z_p}_{n \text{ times}}$. Find $A(G)$.   I know that $A(G)\cong GL_n(\mathbb Z_p)$.   I prove it by taking $\varphi$ from $A(G)$ and show that it's like matrix multiplication, than I took $\theta$ from $A(G)$ and I show that the composition of those two is a again matrix multiplication. Than I show that because we take matrix from $A(G)$ then there kernl is trivial so the matrix need to be not reversible.   I don't know if this is good way or not, I know it's good for $\mathbb Z_p\times\mathbb Z_p$ but I don't know if it's enough for $\mathbb Z_p\times \mathbb Z_p\times\dots\times\mathbb Z_p$ (n times).   If you have other wat please help me, there is a hint to think about vector space $(\mathbb Z_p)^n$ but I don't know it could help.
20705
In general, the group $\mathbb Z_{mn}$ under addition $\bmod mn$ is isomorphic to the direct product $\mathbb Z_m \times \mathbb Z_n$ if and only if $\gcd(m,n)=1.$ This is easy to see since $\mathbb Z_{mn}$ is generated by the element $(1,1)$, and if $\gcd(m,n)=1$, then this will only be zero if it is multiplied by itself $mn$ times (resulting in the cyclic group).   However I'm interested in constructing an isomorphism between the two groups as an explicit expression. I've been playing around a bit with some examples, one of which is the function $\phi : \mathbb Z_2 \times \mathbb Z_3 \to \mathbb Z_6$ where $\phi(a,b) = 3a + 4b \pmod 6$, which gives us the desired isomorphism. I'm interested here in the relationship between the integers $2$, $3$, and the coefficients $3, 4$ of $a$ and $b$. After playing around in Wolfram|Mathematica, I managed to obtain the following isomorphisms $\phi_{m,n} : \mathbb Z_m \times \mathbb Z_n \to \mathbb Z_{mn}$ for coprime $m,n$ in $\{1, 2, \dots, 10\}$.   First of all, the following results seem to hold in general.  $$\phi_{1, n}(a, b) = a + b \qquad \text{for all $n$}$$  $$\phi_{m, 1}(a, b) = a + b \qquad \text{for all $m$}$$  $$\phi_{2, n}(a, b) = na + (n + 1)b \qquad \text{for all $n$}$$ $$\phi_{m, n}(a, b) = K a + L b \iff \phi_{n,m}(a, b) = L a + K b \qquad \text{for all $m, n$}$$  Beyond this point, I could not spot a general pattern. The isomorphisms I obtained are:  $$\phi_{3, 4}(a, b) = 4a + 9b$$ $$\phi_{3, 5}(a, b) = 10a + 6b$$ $$\phi_{3, 7}(a, b) = 7a + 15b$$ $$\phi_{3, 8}(a, b) = 16a + 9b$$ $$\phi_{3, 10}(a, b) = 10a + 21b$$ $$\phi_{4, 5}(a, b) = 5a + 16b$$ $$\phi_{4, 7}(a, b) = 21a + 8b$$ $$\phi_{4, 9}(a, b) = 9a + 28b$$ $$\phi_{5, 6}(a, b) = 6a + 25b$$ $$\phi_{5, 7}(a, b) = 21a + 15b$$ $$\phi_{5, 8}(a, b) = 16a + 25b$$ $$\phi_{5, 9}(a, b) = 36a + 10b$$ $$\phi_{6, 7}(a, b) = 7a + 36b$$ $$\phi_{7, 8}(a, b) = 8a + 49b$$ $$\phi_{7, 9}(a, b) = 36a + 28b$$ $$\phi_{7, 10}(a, b) = 50a + 21b$$ $$\phi_{8, 9}(a, b) = 9a + 64b$$ $$\phi_{9, 10}(a, b) = 10a + 81b$$ If you want to try and generate a few yourself, here is the Mathematica function I implemented. I'm sure it can be optimised, but for the first 10 coprime integers it sufficed.  isomorphism[{m_, n_}] :=       Module[{x = Null},       For[i = 1, i &lt; m*n,           i++, (If[#[[2]] == Table[i, {i, 0, m*n - 1}],           x = ToString[#[[1]][[1]]] &lt;&gt; "a+" &lt;&gt; ToString[#[[1]][[2]]] &lt;&gt;          "b"; Break[]] &amp;) /@       Table[{{i, j},       Mod[i #[[1]] + j #[[2]], m*n] &amp; /@       Table[{Mod[i, m], Mod[i, n]}, {i, 0, m*n - 1}]}, {i, 1,      m*n}, {j, 1, m*n}][[i]]]; x];   I'd appreciate if anyone can help me find the pattern to determine $\phi_{m, n}$ in general.
53094
How to show that $\mathbb Z_{mn}$ is isomorphic to $\mathbb Z_m \times\mathbb Z_n$ when $m$ and $n$ are coprime? It is easy to show that the natural map from $\mathbb Z_{mn}$ to $\mathbb Z_m \times\mathbb Z_n$ is a ring homomorphism. How to show that it is bijective?
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!
13733
I have to compute Galois group of $\mathbb F_q/\mathbb F_p$ where $q=p^n$. I know already that $\mathbb F_q/\mathbb F_p$ is galois, so I don't need to prove it. Moreover, I know that \begin{align*}\sigma :\mathbb F_q&amp;\longrightarrow \mathbb F_q\\ x&amp;\longmapsto x^p\end{align*} is an automorphism that fix elements of $\mathbb F_p$ and thus $\sigma \in Gal(\mathbb F_q/\mathbb F_p)$.  Now, I know that if $k&lt;p^n$ then $\sigma ^k(x)=(x^{p})^k=x^{pk}\neq 1$ and thus $\sigma $ is of order $p^n$. Now if I can prove that $|Gal(\mathbb F_q/\mathbb F_p)|=p^n$, then $Gal(\mathbb F_q/\mathbb F_p)=\left&lt;\sigma \right&gt;$ and it would be finish, but I can't prove it since even $X^q-X$ split over $\mathbb F_q$, theis polynomial is not irreducible, and thus, I can't conclude. So how can I find an irreducible polynomial of degree $q$ that split over $\mathbb F_q$ ?
66503
Let $\mathbb Q$ be group of rational numbers under addition.  Then I want to find out the automorphism group of $\mathbb Q$. In a book it is given that Aut($\mathbb Q$) is isomorphic to $\mathbb Q^*$, the  multiplicative group of rational numbers. I stuck in this.
15293
In John Fraleigh's book, A First Course In Abstract Algebra Exercises 15.7 and 15.11, one shows that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (1,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \  \mbox{ and  } \ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,2)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$  One does this with the first isomorphism theorem. With the same idea I proved for example that  $$ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,3)\rangle \cong \mathbb{Z} \times \mathbb{Z}_1 \cong \mathbb{Z} \ \ \ \ \ \mbox{ and }\ \ \ \ \ (\mathbb{Z} \times \mathbb{Z})/ \langle (2,4)\rangle \cong \mathbb{Z} \times \mathbb{Z}_2 $$ So I conjectured that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_{k}$ where $k=\mathrm{gdc}(m,n)$. For the previous four cases, the homomorphism $\phi: \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z} \times\mathbb{Z}_k$ given by $$ \phi(x,y)=\left(\frac{nx-my}{k}, \ x \ \ (\mathrm{mod} \ k) \right) $$ is surjective with kernel =$\langle (m,n)\rangle$. However, this is not the case when $(m,n)=(4,6)$. So, I cannot use what I did for the four cases to prove the general case.     What I want to know is if my conjecture is true. If so, how can I give a general homomorphism? If it is not true, how can I find $k$, such that $(\mathbb{Z} \times \mathbb{Z})/ \langle (m,n)\rangle \cong \mathbb{Z} \times \mathbb{Z}_k$?   Thanks in advance for any help/hint/comment!
94962
I know an Automorphism is a group $G$ that is an isomorphism where $h:G \rightarrow G$, (G is being mapped to itself) and that $\text{Aut}(G)$ is the set of all Automorphisms in G. I was wondering how I would approach finding all functions in $\text{Aut}(\mathbb{Z}_n)$, let's say $\text{Aut}(\mathbb{Z}_4)$ for example. Is it as trivial as showing a bijective map from each element in $\mathbb{Z}_4$ to itself?
18163
In my recent study of Artinian modules I have been looking at Prüfer $p$-groups (which I will denote by $\mathbb Z_{p^\infty}$ from now on).   I am attempting to prove that $\mathbb Z_{p^\infty}$ is an Artinian $\mathbb{Z}$-module which is not finitely generated.      To do this I am trying to show that every proper non-trivial submodule of $\mathbb Z_{p^\infty}$ is equal to $\left(\frac{1}{p^n}\mathbb{Z_{}}\right)/\mathbb{Z}$ for some $n$.   I am using the following definition of $\mathbb Z_{p^\infty}$: $$\mathbb Z_{p^\infty}=\left\{x\in \mathbb{Q}/\mathbb{Z}\mid \exists n:p^nx=0 \right\}$$ I have read in several places, Atiyah &amp; MacDonald Example 6.3 and Wikipedia to name a few, that the subgroups of $\mathbb Z_{p^\infty}$ are of this form. This would demonstrate that $\mathbb Z_{p^\infty}$ is Artinian. However, I can't seem to find anywhere a proof or example showing why all the subgroups are of this form. I would like to understand this further, as it is not immediate to me that this would be the case, but my attempts to figure out why have so far been futile. I may be thinking too much and missing a simple explanation. I have looked at starting with an arbitrary subgroup $H$ such that $H\neq 0$ and $H\neq\mathbb Z_{p^\infty}$ and seeing if I can deduce anything from that. I got this far, but it doesn't seem to be going anywhere. I have also looked at Sylow's theorems, but I'm not sure if they will be any use to me as $|\mathbb Z_{p^\infty}|=\infty$ (unless I'm wrong about this!).  I would be extremely grateful if anybody could offer a few hints here as I am slowly losing my sanity trying to figure it out!  Thanks,  Andy.
90920
If $G$ is an abelien group, let $\hat{G}$ be the set of all   homomorphisms of $G$ into the group of nonzero complex numbers under   multiplication. If $\phi_1, \phi_1\in \hat{G}$ define $\phi_1\cdot  \phi_2$ by $(\phi_1\cdot \phi_2)(g)=\phi_1(g)\phi_2(g)$ for all $g\in  G$.      If $G$ is a finite abelian group prove that $o(G)=o(\hat{G})$ and $G$   is isomorphic to $\hat{G}$.   The case when $G$ is finite cyclic is easy because in this case $\hat{G} \cong G$. Let's consider the case when $G$ is abelian group of order $p^n$, $p$ - prime.   By Fundamental theorem on finite abelian groups we have the following: $$G=\langle a_1 \rangle_{p^{n_1}}\times \langle a_2 \rangle_{p^{n_2}} \times \dots \times \langle a_k \rangle_{p^{n_k}},$$  where $n_1\geq n_2\geq \dots \geq n_k&gt;0$ and $n_1+\dots+n_k=n$.   Let $A_i=\langle a_i \rangle_{p^{n_i}}$ and $A_i$ are cyclic groups then $A_i\cong \hat{A_i}$. So we have the following $$G=A_1\times \dots \times A_k\cong \hat{A_1}\times \dots \times \hat{A_k}.$$  Then I was trying to think about $\hat{G}$ and I think that $\hat{G}=\hat{A_1}\times \dots \times \hat{A_k}$ which follows that $G\cong \hat{G}$.   How to prove rigorously that $\hat{G}=\hat{A_1}\times \dots \times \hat{A_k}$? I am not able to prove it.
81719
In Section 9 of Chapter 6, Lang constructrs for the first time a non-abelian extension as a splitting field $K$ of $X^n - a$ when the ground field $k$ does not contain primitive $n$-th roots of unity $\zeta_n$. He constructs an injective homomorphism of the Galois group $G(K/k)$ into the matrices of type   $$ M=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix} \quad \hbox{with}\ b\in \mathbb{Z}/n\mathbb{Z} \quad \hbox{and} \quad d\in (\mathbb{Z}/n\mathbb{Z})^*,$$  where $(\mathbb{Z}/n\mathbb{Z})^*$ the multiplicative group of integers modulo n.  Afterwards, he states in Theorem 9.4 that if $n$ is an odd positive integer prime to the characterstic and $[k(\zeta_n):k]=\varphi(n)$ (i.e, the primitive roots of unity are not in the ground field), then the above homomorphism is, in fact, an isomorphism. Besides, the commutator group is $G(K/k(\zeta_n))$, so $k(\zeta_n)$ is the maximal abelian subextension of $K$.  Lang proves the theorem inductively over the number of prime factors in the decomposition of $n$ (i.e. if $n=9= 3\cdot 3$, then the proof "would take two steps" to complete). The induction start with $n$ prime is written clearly and I agree with it.   Now come the questions:   I do not understand Lang's argument to prove that the commutator subgroup is the subgroup $\begin{pmatrix}1 &amp; 0 \\ b &amp; 1\end{pmatrix}$, $b\in \mathbb{Z}/n\mathbb{Z}$. What is this projection on the diagonal and how does one see that the factor group of the commutator group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$?   By the way, direct computation of $x\, y\, x^{-1}\, y^{-1}$ for two arbitrary matrices $x=\begin{pmatrix} 1 &amp; 0 \\ b &amp; d \end{pmatrix}$, $y=\begin{pmatrix} 1 &amp; 0 \\ c &amp; e \end{pmatrix}$ yields a matrix $\begin{pmatrix} 1 &amp; 0 \\ b(1-e)+c(d-1) &amp; 1 \end{pmatrix}$. As $n$ is odd, $2 \in (\mathbb{Z}/n\mathbb{Z})^*$, so taking $e=1$ and $d=2$, we indeed reconstruct the above-mentioned group.   The second (inductive) part of the proof is a complete mystery to me with virtually every sentence being a Theorem which needs an extra proof. Can someone give me some hints?   2.a. Lang writes $n=p\,m$ and takes $\alpha$ to be the root of $X^n-a$. It is clear that $\beta=\alpha^p$ is a root of $X^m -a$. Now comes the argument "By induction we can apply the theorem to $X^m-a$". I agree that we can apply it here but the result is that $K=k(\beta,\zeta_m)$ is the splitting field of $X^m-a$. In the diagram below, there appears $k(\beta,\zeta_n)$ out of nowhere.   2.b. Now Lang proceeds to apply the theorem on $X^p-\beta$ over $k(\beta)$. Bergman's "Companion" reminds to check that $[k(\beta,\zeta_p):k(\beta)]=\varphi(p)$ before applying the theorem and gives a hint "$[k(\zeta_p):k]$ and $[k(\beta):k]$ are relatively prime". I disagree with this hint. Take $n=21= m p$ where $m=3$ and $p=7$. Then $[k(\zeta_p):k]=\varphi(p)=p-1=6$ but $[k(\beta):k]=m=3$ and both are not co-prime. Can we apply the theorem here at all? Or if so what is the argument? If Bergman's argument is valid, where is the fault in my logic?  2.c. Even assuming that we can apply the theorem in 2.b., then we can say something about $k(\alpha, \zeta_p)$ and not $k(\alpha,\zeta_n)$. How do we proceed to checking everything Lang states in the last section of the proof (basically, everything after the diagram)?  EDIT: I have found the essentially same question
119971
Prove that $\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}$ implies $\gcd(m,n)=1$.  This is the converse of the Chinese remainder theorem in abstract algebra. Any help would be appreciated.  Thanks!
36711
I'm searching for a way to generate the group $\mathrm{GL}(n,\mathbb Z)$. Does anyone have an idea?  The intention of my question is that I am searching for an easy proof of the existence of the epimorphism:  $\Phi:\mathrm{Aut}(F_n )\to \mathrm{Aut}(F_n/[F_n,F_n])=\mathrm{Aut}(\mathbb {Z}^n)=\mathrm{GL}(n,\mathbb {Z})$  I know that $\Phi$ is an canonical homomorphism since the commutator subgroup is characteristic in $F_n$. So I need some nice generators of $\mathrm{GL}(n,\mathbb {Z})$ to find their preimages in $\mathrm {Aut}(F_n)$ to prove the surjectivity of $\Phi$.  Thanks for help!
17898
Calculate (find the type of isomorphism) of $\mathbb Z_p/p\mathbb Z_p$ where p is prime.   I know that $\mathbb Z/p\mathbb Z$ isomorphic to $\mathbb Z_p$ by showing    $\phi : \mathbb Z \rightarrow \mathbb Z_p$  $\phi(x) = x  \bmod p$   $\ker \phi = p\mathbb Z$  and by isomorphism $1$ theorem.  So I get  $\mathbb Z_p/p\mathbb Z_p$ isomorphic to $(\mathbb Z/p\mathbb Z)/p(\mathbb Z/p\mathbb Z)$.   Not sure if I'm writing this in a correct way and if so then how to continue from here.    Any help and explanation will be appreciated.
55974
The question at hand:  The group $\mathrm{GL}_2(\mathbb F_p)$ of all $2\times 2$ invertible matrices with coefficients in the finite field $\mathbb F_p$ has order $(p^2 − 1)(p^2 − p)$. Let $\mathrm{SL}_2(\mathbb F_p)$ be the subgroup consisting of all matrices of determinant $1$, i.e. $\mathrm{SL}_2(\mathbb F_p)$ is the kernel of the group homomorphism $\det : \mathrm{GL}_2(\mathbb F_p) → \mathbb F_p$. Compute the order of $\mathrm{SL}_2(\mathbb F_p)$. Hint: $\mathrm{GL}_2(\mathbb F_p)$ is a disjoint union of fibers of $\det$.  I am feeling a little lost on this problem. I understand via my professor that $\mathrm{GL}_2(F_p)$ is the union of the fibres of det, that each fibre is a coset of $\mathrm{SL}_2(F_p)$, and that $\mathrm{GL}_2(F_p)$ is the order of $\mathrm{SL}_2(F_p)$ times the number of fibres. However, I am not sure how to show that the order is $(p^2-1)(p^2-p)$. Does anyone have any insight into this problem?
50076
I had this "almost bonus" question on the final in Group Theory recently: prove that there is no such group $G$ which would satisfy Aut$(G)\cong \mathbb{Z}_n$, where $n$ is an odd integer. I don't have much certainty if this proof is OK, and one's opinion would be appreciated.  Here's my attempt:     Suppose that there exists such a group $G$ that satisfies the above condition. Since $\mathbb{Z}_n$ is cyclic, Aut$(G)$ is also cyclic and $G$ is abelian, which implies that Inn$(G)\cong \{e\}$. Thus Aut$(G)=$Out$(G)$. (I don't think this fact is important here though). Now, since an automorphism sends a generator to a generator, and since each automorphism is completely determined by such mapping, |Aut($G$)| must be of factorial order. But an integer factorial is always even. However, $|\mathbb{Z}_n|$ is odd, which is a contradiction. Therefore, no such $G$ exists.   I'm afraid that my proof is not very nice, but at least a genuine attempt was made.
77054
I am trying to show the isomorphism  $$\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle.$$  I've tried to define $\psi:\mathbb Z^n \to \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle$ an epimorphism with $\ker(\psi)=\langle (a,...,a) \rangle$ so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.