I'm stuck on the following exercise from Herstein's "Topics in Algebra":  "show that ($n$ is prime) $\Leftrightarrow ([a][b]=[0]\Rightarrow [a]=[b]=[0]) $ in $J_n$".    for the rightward implication I have:  $[a][b]=[ab]=[0]\Rightarrow n|ab\Rightarrow n|a$ or $n|b$ (or both) $\Rightarrow [a]=[0]$ or $[b]=[0]$  and I'm stuck on the leftward implication.  Why do I get $[a]=[0]$ OR $[b]=[0]$ and not $[a]=[0]$ AND $[b]=[0]$ in the leftward implication? Where is it that I go wrong?   I'd also appreciate any comment/hint about how to prove the remaining left implication.
121709
Suppose $$ab \equiv 0 \pmod n$$ and that $a$ and $b$ are positive integers both less than $ n$. Does it follow that either $a | n$ or $b | n$?  If it does follow, give a proof.  If it doesn’t, then give an example.   For this question is this a suitable answer: No it doesn't. $3$ and $4$ are both less than $6$. $$3 \times 4 \equiv 0 \pmod {6}$$ but $4 \not \mid 6$.
58706
I'm trying to understand a proof of  which says that in a integral domain every prime element is irreducible:    I didn't understand why this implication $p=ab\implies p|a$ or $p|b$, is not the contrary $p=ab\implies a|p$ and $b|p$ ?  I'm a little confused  Thanks in advance
67989
$\mathbb Z_n$ is J-semisimple iff $n$ is square free.     A ring  $R$ is said to be $J$ semisimple if intersection of all maximal ideals of $R$ is $\{0\}$.   If $n$ is square free then $n=p_1p_2...p_k$ then $\mathbb Z_n\cong \mathbb Z_{p_1}\times\mathbb Z_{p_2}\times...\mathbb Z_{p_k}$ Since $\mathbb Z_{n}$  is a PID every prime ideal ideal is a maximal ideal and hence the maximal ideals will be $\mathbb Z_{p_i}$ for each $i$; but intersection of all $\mathbb Z_{p_i}$' s is $\{0\}$ since $p_i$ 's are distinct.  How to do the converse?
83953
Let $p$ be a prime number. Let $M$ denote the ring of $2 \times 2$ matrices over the field $F$ of $p$ elements. For $A \in M$, let $C(A)$ denote the set of those matrices $B \in M$ such that $AB = BA$. What are the possible values of the dimension of $C(A)$ over $F$ for $A \in M$?  What I know is this: Given a square $n \times n$ matrix $A$ over a field $F$, it is always true that $A$ commutes with any $B=p(A)$, where $p(x)$ is a polynomial with coefficients in $F$. Now, because of the theorem of Cayley-Hamilton, the matrix $A$ is root of its  characteristic polynomial (which has degree $n$), so that means that every power of $A$ larger than $n-1$ can be written as a linear combination of powers up to $n-1$. For, our case, it means that: $C(A)=B=p_0I+p_1A$, where $p_0,p_1 \in F$.  So, that means $ 2\le \dim(B)\le 4$ (cases where $p_0=0$ and $p_1=0$). Is this correct? But what about $p$ prime - what it has to do with anything?
20619
From problem 2.3.25 in Topics in Algebra, 2$\varepsilon$ by I. N. Herstein:     Let $G$ be the group of all $2 \times 2$ matrices $\left(\begin{array}{c c}a &amp; b \\ c &amp; d\end{array}\right)$ where $ad-bc \ne 0$ and $a,b,c,d$ are integers modulo 3, relative to matrix multiplication. Show that $o(G) = 48$.   I know that $o(G) \le 3^4 = 81$, since $a,b,c,d$ can each take one of 3 values (mod 3). I attempted to tighten this bound by finding the number of matrices such that $ad=bc$ (mod 3):   Suppose $ad=bc=0$ (mod 3). Then ($a = 0$ or $d = 0$) and ($b = 0$ or $c = 0$), leading to 36 possible values for $(a,b,c,d)$. Suppose $ad=bc=1$ (mod 3). Then ($a=d=1$ or $a=d=2$) and ($b=c=1$ or $b=c=2$), leading to 4 possible values for $(a,b,c,d)$. Suppose $ad=bc=2$ (mod 3). Then ($(a,d)=(1,2)$ or $(a,d)=(2,1)$) and ($(b,c)=(1,2)$ or $(b,c)=(2,1)$), leading to 4 possible values for $(a,b,c,d)$.   So, there are in total $36+4+4 = 44$ such $\left(\begin{array}{c c}a &amp; b \\ c &amp; d\end{array}\right)$ where $ad-bc=0$ (mod 3). That means there are at most $81-44 = 37$ such $\left(\begin{array}{c c}a &amp; b \\ c &amp; d\end{array}\right)$ where $ad-bc\ne 0$, i.e., $o(G) \le 37$. However, this contradicts the problem. Where did I go wrong? Can someone set me on the right path?
88756
I am struggling with the end of my proof.   So far I have used the fact that $S_n=\langle T\rangle$ where $T=\{(i\text{ }j)|1\leq i&lt;j\leq n\}$ to show that $S_n$ is generated by the set $\{(1\text{ }2),(1\text{ }3),\cdots,(1\text{ }n)\}$ of $n-1$ transpositions.   I am struggling with proving that $S_n$ is also generated by the set $\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$ of $n-1$ transpositions. Essentially I want to show that each $(1\text{ }i)$ is a product of these transpositions to reach this conclusion.   Here is what I have so far...    Consider any $(1\text{ }i)$ transposition in $S_n$. Notice by the composition of a permutation that \begin{eqnarray*} (1\text{ }i-1)(i-1\text{ }i)(1\text{ }i-1)&amp;=&amp;\begin{bmatrix}     1       &amp; i-1 &amp; i   \\     i-1       &amp; 1 &amp; i    \end{bmatrix} \begin{bmatrix}     1       &amp; i-1 &amp; i   \\    1       &amp; i &amp; i-1  \end{bmatrix} \begin{bmatrix}    1       &amp; i-1 &amp; i   \\     i-1       &amp; 1 &amp; i \end{bmatrix}\\ &amp;=&amp; \begin{bmatrix}     1       &amp; i-1 &amp; i   \\     i-1       &amp; 1 &amp; i    \end{bmatrix} \begin{bmatrix}    1       &amp; i-1 &amp; i   \\     i       &amp; 1 &amp; i-1 \end{bmatrix}\\ &amp;=&amp; \begin{bmatrix}    1       &amp; i-1 &amp; i   \\     i       &amp; i-1 &amp; 1 \end{bmatrix}\\ &amp;=&amp; (1\text{ }i) \end{eqnarray*}  Now we want to use this property, call it STAR, to show that $S_n$ is also generated by the set  $$Y=\{(1\text{ }2),(2\text{ }3),\cdots,(n-1\text{ }n)\}$$ of $n-1$ transpositions. Notice for each $(i\text{ }j)$ in $Y$ that $j-i=1$. Suppose $n=2$. Then $S_2=\{(1),(1\text{ }2)\}$, and we done because $2-1=1$. Suppose $n=3$. Then $S_3$ is generated by the set $\{(1\text{ }2),(1\text{ }3)\}$. By STAR,  $$ (1\text{ }3)=(1\text{ }2)(2\text{ }3)(1\text{ }2) $$ $$ (1\text{ }2)=(1\text{ }1)(1\text{ }2)(1\text{ }1) = (1\text{ }2) $$ Therefore $S_3$ is generated by $\{(1\text{ }2),(2\text{ }3)\}$ and for each $(i\text{ }j)$ we have $j-i=1$.    And here is where I am struggling. I do not know how to apply induction to this. I want to show that each side part will continue breaking down using the property I stated until all of $S_n$ is created. Here is why I think this: $$ (1\text{ }1)(1\text{ }2)(1\text{ }1)=(1\text{ }2) $$ $$ (1\text{ }2)(2\text{ }3)(1\text{ }2)=(1\text{ }3) $$ $$ \cdots $$ $$ (1\text{ }n-1)(n-1\text{ } n)(1\text{ } n-1)=(1\text{ }n) $$ Any input would be much appreciated. Best.
41628
I've been working on this problem listed in Herstein's Topics in Algebra (Chapter 2.3, problem 4):     If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, show that $G$ is abelian.   I managed to prove it, but I'm not very happy with my result (I think there's a neater way to prove this). Anyway, I'm just looking to see if there's a different approach to this.   My approach:  Let $j=i+1, k=i+2$ for some $i\in \mathbb{Z}$.  Then we have that $(ab)^i = a^ib^i$, $(ab)^j = a^jb^j$ and $(ab)^k = a^kb^k$.  If $(ab)^k = a^kb^k$, then $a^jb^jab =a^jab^jb$.   We cancel on the left and right and we have $b^ja = ab^j$, that is $b^iba = ab^j$.  Multiply both sides by $a^i$ on the left and we get $a^ib^iba = a^jb^j$, so $(ab)^iba = (ab)^j$. But that is $(ab)^iba = (ab)^iab$.   Cancelling on the left yields $ab=ba$, which holds for all $a,b \in G$, and therefore, $G$ is abelian.  Thanks!
71010
Let $G$ be the group  of all $2 \times 2$ matrices of the form $\begin{pmatrix} a &amp; b \\ 0 &amp; d\end{pmatrix}$ where $ad \neq 0$ under matrix multiplication. Let  $N=\left\{A \in G \; \colon \; A = \begin{pmatrix}1 &amp; b \\ 0 &amp; 1\end{pmatrix} \right\}$ be  a subset  of the group $G$. Prove that  $N$ is  a normal subgroup of $G$ and prove that  $G/N$ is abelian group.   Here is my attempt!  To prove $N$ is normal I consider  the group homomorphism $f \colon G \to \mathbb R^*$ given by $f(B) = \det(B)$ for all $B$ in $G$. And I see that $f(N)$ is all the singleton $\{1\}$ since $\{1\}$ as a subgroup of $\mathbb R^*$  is normal, it follows that $N$ is also normal. Is this proof helpful  here? Then how to prove that $G/N$ is Abelian? I know $G/N$ is a collection of left cosets.   Thank you.
69528
I have to do the following question:  Given a unital ring $R$ and a natural number $n$ $\geq 1$, identify (with proof) the prime ideals of the matrix ring $\mathbb{M}_{n}(R)$.  I understand the theory, but putting it all together is proving difficult.   I understand that the prime ideals of $\mathbb{M}_{n}(R)$ should be $\mathbb{M}_{n}(\mathbb{P})$.  Can anyone suggest any steps as to how to prove this?
42366
In $\mathbb{Z}\left [ x \right ]$, let $I = \left \{ f\left ( x \right ) \in \mathbb{Z}\left [ x \right ]: f\left ( 0 \right ) \text{ is an even integer} \right \}$. Is $I$ a Prime ideal of $\mathbb{Z}\left [ x \right ]$?  Let $f_{1}\left ( x \right ),f_{2}\left ( x \right ) \in \mathbb{Z}\left [ x \right ].$ Let $f_{1}\left ( x \right )f_{2}\left ( x \right ) \in I$. Clearly, I is a proper ideal of $\mathbb{Z}\left [ x \right ]$ since the elements in $\mathbb{Z}\left [ x \right ]$ evaluated at 0 can produce an odd integer but this cannot be true in I. Now, $f_{1}\left ( 0 \right )f_{2}\left ( 0 \right )$ is an even integer in I. Hence, either $f_{1}\left ( 0 \right )$ has an even constant integer or $f_{2}\left ( 0 \right )$ has an even constant integer. Thus, I is a prime ideal by definition. Is this correct? I would now like to show if I is a maximal ideal. Again, clearly, I is a proper ideal by the above argument. Let B be an ideal of $\mathbb{Z}\left [ x \right ]$ and I $\subseteq B\subseteq \mathbb{Z}\left [ x \right ]$. How should B be defined so that I would conclude either $B=I$ or $B=\mathbb{Z}[x]$?
515
The question is      Prove that $a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}$.   Hint:First prove $b\mid a \Leftrightarrow a\mathbb Z\le b\mathbb Z$ and then prove $a\mathbb Z+b\mathbb Z=(a,b)\mathbb Z$   I managed to prove $b\mid a\Leftrightarrow a\mathbb Z\le b\mathbb Z$. About the next claim: From the first statement ($b\mid a...$) it follows that since $(a,b)\mid a$ and $(a,b)\mid b, a\mathbb Z+b\mathbb Z\le(a,b)\mathbb Z$. My problem is proving the opposite side (a.k.a $a\mathbb Z+b\mathbb Z\ge(a,b)\mathbb Z$). How can I do so and how can I proceed from $(a,b)$ to $[a,b]$?
8795
${ \mathbb{N} }^{ \left\{ 0,1 \right\}  }$ and ${ \left\{ 0,1 \right\}  }^{ \mathbb{N} }$ to be more specific, and is there a countable subset in each one of them? How do I find them?
38871
I'm having trouble deriving the following fact from the basic properties of $+_{\mathbb{N}}$ and the definition:  Definition. $\forall n, m \in \mathbb{N}: (n \geq m) \leftrightarrow (\exists a \in \mathbb{N}): n = m + a$. Fact. $\forall a, b \in \mathbb{N}: (a \geq b ~\land~ b \geq a) \rightarrow a = b$.  I've tried several different strategies, but none works. I think the fact that 0 isn't the successor of any natural number is the key both in the base case and the inductive step, but I keep going in circles trying to make use of that (and other basic facts about addition on $\mathbb{N}$). I would show you my notes, but there is really nothing of substance in them. I've made almost no progress since yesterday.
79801
I have to do this exercise for my math study, and I'm having trouble with doing the second part of it.  Let $x, y, \epsilon \in \mathbb{R}$ and $\epsilon &gt; 0$. Prove: $|x - y| \leqslant \epsilon$ $\forall \epsilon &gt; 0 \Leftrightarrow x = y$  I think I have the left implication $\Leftarrow$:  Assume $x = y \Rightarrow x - y = 0 \Rightarrow|x - y| = 0 = |0| \Rightarrow|x - y| = 0 &lt; \epsilon \Rightarrow |x - y| \leqslant \epsilon$ $\forall \epsilon &gt; 0 \in \mathbb{R}$  Is this argument correct?  For the right implication, I do only have an idea of how to prove it. I think I have to assume first that $x &gt; y$ and then $x &lt; y$ and get a contradiction form both.   Could you please explain me the right implication and tell me if my left implication is correct?  Thanks in advance!
64148
I'm reading Beardon's Algebra and Geometry:     Show that $\mathbb{Z}$ with the operation $m*n=m+(-1)^m n$, is a group.   I've been thinking on how I should show associativity, I thought about doing this:  $$(m*n)*p$$  Then assuming:  $$m*n=m+(-1)^m n$$  I have:  $$(m+(-1)^m n)*p$$  Which results in:  $$[m+(-1)^m n]+(-1)^{[m+(-1)^m n]}p$$  Then:  $$m*(n*p)$$  Again, assuming that:  $$n*p=n+(-1)^n p$$  I have:  $$m*(n+(-1)^n p)$$  Then:  $$m+(-1)^m [n+(-1)^n p]$$  And show that:  $$[m+(-1)^m n]+(-1)^{[m+(-1)^m n]}p=m+(-1)^m [n+(-1)^n p]\tag{A} $$  But I've tested it with some values on Mathematica and found that when:  $$m = 1; n = 0; p = -1;$$  The equality does not hold. I'm stuck again. Can you please provide me some hints on how to do it? I'm not looking for a full answer.    When I was testing the equality on Mathematica, I felt that permutations of $m,n,p$ on the set $\{-1,0,1\}$ could be enough to prove it, for example:  $$\begin {array} { c | c | c } m,n,p &amp; m,n,p &amp; m,n,p \\ \hline   -1, -1, -1 &amp; - 1, -1, 0 &amp; - 1, -1, 1  \\   -1, 0, -1 &amp; - 1, 0, 0 &amp; - 1, 0, 1  \\   -1, 1, -1 &amp; - 1, 1, 0 &amp; - 1, 1, 1  \\   0, -1, -1 &amp; 0, -1, 0 &amp; 0, -1, 1  \\   0, 0, -1 &amp; 0, 0, 0 &amp; 0, 0, 1  \\   0, 1, -1 &amp; 0, 1, 0 &amp; 0, 1, 1  \\   1, -1, -1 &amp; 1, -1, 0 &amp; 1, -1, 1  \\   1, 0, -1 &amp; 1, 0, 0 &amp; 1, 0, 1  \\   1, 1, -1 &amp; 1, 1, 0 &amp; 1, 1, 1  \\ \end {array} $$  But I'm not sure if this is true. What do you think?
117839
Yes, I know that there are tons of solutions of this up here, but I, essentially, wanted to try it a different way and ah, well.  Let $|G| = p^2$ for some prime $p$.  Consider $x \in G$. So, $|x| = p, $ or $ p^2$. If it is the latter, then, we are done.  So, if $|x|= p$, then, let, $H = [h|  h=x^i, i \in Z] \Rightarrow |H|= p$.  As $p$ is the smallest prime dividing $|G|$, $H$ is a normal subgroup.  Now, consider some $y \in G$, but not in $H \Rightarrow y^p= e$. Then, its easy to show that $G = \langle x,y \rangle$.  Now, consider a homomorphism $f:G \rightarrow \operatorname{Aut}(H), g \to c_g $, wherein $c_g$ represents conjugacy by $g$. So, for example, $c_g(x) = gx g^{-1}$.  Then, it is clear that, $H \subset \operatorname{Ker}(f) $, as it is cyclic and therefore abelian. So, we need only look at $G/H$.  Basically, I am asking for a hint as to why $c_y(x) = yxy^{-1} = x = c_e(x)$, where $x$ and $y$ are the generators of $H$ and $G/H$, respectively?  Because, if I establish it for the generators, then it follows that it applies for for all elements, i.e, that $\operatorname{Im}(f)$ is trivial and therefore that $G$ is abelian.  Thank You!
105287
I am trying to prove $(a)$ is a maximal ideal in $\mathbb{Z}$,  if and only if $a$ is prime number.  Now I wrote: assume $a \in\mathbb{Z}$, while it's not prime number we can write as $a=xy$, for some integers $x$ and $y$. then $(a)\subset (x)$ and $(a)\subset (y)$  while if $(a)$ is maximal ideal, it's not exist $(k)$ such that $(a)\subset (k)\subset\mathbb{Z}$. so $(a)$ cannot be maximal ideal. we can say by contradictory if $a$ is prime, $(a)$ is maximal ideal.  is this correct?  and how to prove the other way.