I'm reading Serre's Trees and in Example 1.5.3 he claims that $PSL_2(\mathbb{Z})$ is isomorphic to $\mathbb{Z}_2 \ast \mathbb{Z}_3$.   $\mathbb{Z}_2 \ast \mathbb{Z}_3 = \langle x, y | x^2 = y^3 = 1 \rangle$. But does $PSL_2(\mathbb{Z})$ even have elements of order $2$ or $3$ to begin with?   Previously, I thought since all elements of $PSL_2(\mathbb{Z})$ is in the form of  $$ \begin{bmatrix} 1 &amp; n \\ 0 &amp; 1 \end{bmatrix}. $$  Hence $\mathbb{Z}$ injects in $PSL_2(\mathbb{Z})$.  Let $\phi: \begin{bmatrix} 1 &amp; n \\ 0 &amp; 1 \end{bmatrix} \mapsto n$.  $\phi$ is a homomorphism and in fact bijective. Hence $\mathbb{Z} \cong PSL_2(\mathbb{Z})$.  Thus $PSL_2(\mathbb{Z})$ cannot have any $2$-torsion or $3$-torsion.   I must be missing something. Any help is appeciated!
44559
Prove that $PSL(2,\mathbb{Z})=C_2 \star C_3$.  Now $C_2 \star C_3=\langle a,b\ |\ a^2, b^3 \rangle$ i.e. the free product.  But how do I show that presentation of $PSL(2,\mathbb{Z})$ is this?
73590
I am currently working to find all ring homomorphisms $\phi : \mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}(\sqrt{2})$.   My work so far: Since $\mathbb{Q}(\sqrt{2})$ is a field extension of $\mathbb{Q}$, a field, it is obviously a field as well. Therefore, its only ideals are $\{0\}$ and itself. Since every homomorphism's kernel is an ideal, we have that either $Ker(\phi)=\{0\}$ or $\mathbb{Q}(\sqrt{2})$.   In the latter case, we have the trivial homomorphism. In the former case, we have that the homomorphism is injective (and therefore bijective).   My question: I think it is true that in the former case $\phi(1)=1$, but am not sure why. If this is the case, can I prove inductively that $\phi(x)=x, \forall x\in \mathbb{Q}(\sqrt{2})$?  Thank you in advance for your help.
81342
I came over a question in ring theory which I am not being able to proceed upon:     When is the ring $m\mathbb{Z}$ isomorphic to the ring $n\mathbb{Z}$, where $m, n \in \mathbb{N}$?   I know that to show isomorphism, I need to show that it is onto and the kernel consists of only $\{0\}_{m\mathbb{Z}}$ but I am not being able to write things down properly. Will someone help? What is the relation between $m$ and $n$?
61678
Here are my thoughts so far:    $C_p = \langle x \mid x^p =e\rangle$ $C_q = \langle y \mid y^q =e\rangle$ $C_p \times C_q$ has elements of the form $(x^a,y^b)$ There are $p$ possible values for $x^a$ and $q$ possible values for $y^b$. So there are $pq$ possible elements in $C_p \times C_q$. $C_{pq} = \langle z\mid z^{pq} = e\rangle$ and there are $pq$ elements in this group. For an isomorphism from $C_p \times C_q$ to $C_{pq}$ we send $(e,e)$ to $e$ to ensure that the identities are mapped to each other. However, I'm not sure how to define an isomorphism for the other elements.
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.
32893
I've been given a homework problem that requires me to show that the rings $\mathbb{C}[x,y]/(y - x^2)$ and $\mathbb{C}[x,y]/(xy-1)$ are not isomorphic.    This is my attempt at a solution:  For $\mathbb{C}[x,y]/(y - x^2)$, we can parametrize in the following way: $x = t$ and $y = t^2$. Then this ring is isomorphic to $\mathbb{C}[t]$.  For $\mathbb{C}[x,y]/(xy-1)$, we can parametrize $x = t$ and $ y = 1/t$. Then this is isomorphic to $\mathbb{C}[t, 1/t]$.   But $\mathbb{C}[t, 1/t]$ is not isomorphic to $\mathbb{C}[t]$.  Am I on the right track? If not, any helpful hints?
120663
the answer should be $$|\operatorname {SL}_2(\mathbb Z/N\mathbb Z)|=N^{3}\prod_{p|N}(1-{1 /p^2})$$ But first how to prove $$|\operatorname {SL}_2(\mathbb Z/p^e\mathbb Z)|=p^{3e}(1-{1 /p^2})$$
27124
I'm struggling a fair amount with this exercise:     Find a subring of $M(2,\mathbb{Q})$ which is isomorphic to a) $\mathbb{Q}$ x $ \mathbb{Q}$ b) $\mathbb{Q}$ c) $\mathbb{Q}[x]$/$x^2$   Now I know a subring must be a subgroup, must contain the elements $0,1$ and must be closed under multiplication. As we are looking for isomorphisms then they must also be ring homomorphisms and must be bijective.  I tried to come up with a random subring and attempt to prove it is an isomorphism. E.g $\left\{\left.\begin{matrix}  a &amp; b \\ c &amp; 0  \end{matrix}\right|a,b,c \in \mathbb{Q}\right\}$ is a subgroup of $M(2,\mathbb{Q})$ and then try and show it's an isomorphism. However I'm stuck here as I don't actually know how to go about finding these specific isomorphisms and I completely lose what I'm doing. Any help would be great.
110292
I am trying to prove that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic. I was thinking of arguing the following:  Suppose there exists an isomorphism $\varphi: \mathbb{Z}[x]\rightarrow\mathbb{Q}[x]$. Because isomorphisms are by definition surjective, there exist $x, y \in\mathbb{Z}[x]$ such that $\varphi(x) = c \in \mathbb{Q}[x]$ and $\varphi(y) = d \in \mathbb{Q}[x]$ for any $c, d\in\mathbb{Q}[x]$. Because $\varphi$ is an isomorphism we must have $\varphi(x+y) = \varphi(x) + \varphi(y)$ for all $x, y \in \mathbb{Z}[x]$. Namely, because polynomial addition is defined componentwise, we must have that the constant term of $\varphi(a + b) = c_{0} + d_{0}$ (where $c_{0}, d_{0}$ are the constant terms of $c$ and $d$ respectively. I would then argue that because $\mathbb{Z}$ and $\mathbb{Q}$ are not isomorphic as additive groups, no such isomorphism $\varphi$ exists. Is this a valid proof?  I've seen proofs that argue that because $\varphi(1) = 1$ for any homomorphism we have $1 = \varphi(2(1/2)) = 2(\varphi(1/2))$ so $\varphi(1/2)$ must be contained in $\mathbb{Z}[x]^{\times}$. Then because $\mathbb{Z}[x]^{\times} = \mathbb{Z}^{\times} = \{\pm1\}$ we have $2 \times \pm1 \neq1$, a contradiction. Is this any different than arguing that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ each have a different number of units?
55480
Determine whether $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$ are isomorphic groups or not.  pf) Suppose that these are isomorphic. Note that $\mathbb{Z}\times \mathbb{Z}$ is a subgroup of $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times\left \{ 0 \right \}$ is a subgroup of $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$. Since $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times\left \{ 0 \right \}$ are isomorphic, $\mathbb{Z}\times \mathbb{Z}/\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/\mathbb{Z}\times \mathbb{Z}\times \left \{ 0 \right \}$ are isomorphic. But the first one is isomorphic to the trivial group and the second one is isomorphic to $\mathbb{Z}$. It is a contradiction.  Is my proof right? If not, is there another proof?
103428
Show there is a surjective homomorphism from $\mathbb{Z}\ast\mathbb{Z}$ onto $C_2\ast C_3$, where $\ast$ denotes the coproduct in the category $\mathsf{Grp}$.   Note the exercise in this book (Algebra Chapter 0, Aluffi) is meant, or at least hinted at, to be done using the universal property of the coproduct, ie: coproduct of $A$ and $B$ in $\mathsf{Grp}$ is initial in the category $\mathsf{Grp}^{A,B}$, instead of using the definition of the free product.  By property of the coproduct, for any object $A$ and two homomorphisms $f_1,f_2:\mathbb{Z}\to A$, and two homomorphisms $i_1,i_2:\mathbb{Z}\to\mathbb{Z}\ast\mathbb{Z}$, there exists a unique homomorphism $\sigma:\mathbb{Z}\ast\mathbb{Z}\to A$  such that the diagram commutes, ie: $f_1=\sigma i_1$ and $f_2=\sigma i_2$. Now let $A=C_2\ast C_3$. Now what?
120588
In general, I struggle working with quotients of polynomial rings, so I was hoping someone might be able to help with the following exercise:  Show that $\mathbb{Q}[x]/(x^2-2x)$ is isomorphic to $\mathbb{Q}[x]/(x^2-1)$, but the latter is not isomorphic to $\mathbb{Q}[x]/(x^2)$.  Mostly I want to understand what these quotients all &quot;look like.&quot; We did few examples in class and few homework problems with such quotients, and the ones we do work with are much simpler (typically involving $\mathbb{Z}_2$ or $\mathbb{Z}_3$ or $\mathbb{R}$ instead of $\mathbb{Q}$). Any help is appreciated, thank you!
52306
Prove or contradict: A finite group with exactly $2$ conjugacy classes always  isomorphic to $\mathbb{Z}_2$.   At first I was trying to work with familiar groups to contradict it(permutations, cyclic, dihedral and general linear) but I could not find any counter example.  So I was trying to prove it.Let $G$ be a group that has exactly $2$ conjugacy classes, obviously one of those classes is $e$, and the other class is the rest of the elements.  So I'll define $f:G \rightarrow \mathbb{Z}_2$ as follows - $f(e) = 0$, and $f(g) = 1$ for every other $g \in G$. It is easy to see that $f$ is homomorphism, and also that $\ker (f) = e$, and by the first isomorphism theorem, we get that $G \cong \mathbb{Z}_2$.  Is it correct?
115398
I need help understanding the following isomorphisms:  $$\mathbb Z[\sqrt{-5}]/(2,1+\sqrt{-5}) \cong \mathbb Z[X]/(X^2+5,2,1+X) = \mathbb Z[X]/(2,1+X) \cong \mathbb Z/2\mathbb Z$$  In general, I am also wondering whether the following are true, and if so briefly why:   $\mathbb Z[\sqrt{D}] \cong \mathbb Z[x]/(x^2 - D)$ for all squarefree integers $D$ $(R/I)/J \cong R/(I +J)$, ie why do we have things like $\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+5) $ where $\mathbb Z[x] / (2) \cong \mathbb Z_2[x]$?   Also for example for something like the following:  $$\,\mathbb Z[x]/(2,x^2+5)\simeq \mathbb Z_2[x]/(x^2+1)$$  Is it true because of the third isomorphism theorem? I'm guessing we set $R = \mathbb Z[x]$, $I = (2)$, and $J = (x^2 + 5)$, but please expand more on each step of the reduction.
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!
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.
92119
Why does $PSL(2,\mathbb C)\cong PGL(2,\mathbb C)$ but $PSL(2,\mathbb R) \not\cong PGL(2,\mathbb R)$?