I just showed that $D_4$ is solvable since it is isomorphic to $\mathbb{Z}_2 x \mathbb{Z}_2$ so since $\mathbb{Z}_2$ is solvable so is $D_4$.  Does something similar extend to $D_5$ or is there a better way of showing both?
68335
Is the dihedral group $D_n$ nilpotent? solvable?  I'm trying to solve this problem but I've been trying to apply a couple of theorems but have been unsuccessful so far. Can anyone help me?
46731
I am stuck in the following problem:  Show that every group of order 4 is an extension of $\mathbb{Z}_{2}$ by $\mathbb{Z}_{2}$. Which of the exact sequences splits?  Ok, i have to consider two cases for a group of order 4: $\mathbb{Z}_{4}$ and $\mathbb{Z}_{2}\times \mathbb{Z}_{2}$, so i get two exact sequences:  $0\rightarrow \mathbb{Z}_{2}\overset{\psi }{\rightarrow}\mathbb{Z}_{4}\overset{\varphi }{\rightarrow}\mathbb{Z}_{2}\rightarrow 0$  and $0\rightarrow \mathbb{Z}_{2}\overset{\tilde{\psi} }{\rightarrow}\mathbb{Z}_{2}\times \mathbb{Z}_{2}\overset{\tilde{\varphi} }{\rightarrow}\mathbb{Z}_{2}\rightarrow 0$  For both cases i checked the conditions of exact sequence and proved that these are exact sequences. Now i have a problem how to show if they split or not.  We say that an exact sequence splits if for the first sequence there exists a group homomorphism $\pi :\mathbb{Z}_{2}\rightarrow \mathbb{Z}_{4}$ with $\varphi \circ \pi =id_{\mathbb{Z}_{2}}$ and for the second $\tilde{\varphi} \circ \tilde{\pi} =id_{\mathbb{Z}_{2}}$, right? But how to show if these group homomorphisms exist or not?  Can anybody help me, please? I would appreciate any hints and comments. Thank you in advance!
93809
I need some help with the following task: Show that  for all $n\geq2$ is $D_{2n}/Z(D_{2n})\cong D_n$ $D_n$ is the dihedrial group with $2n$ elements and looks like $D_n=\{id,r,...,r^{n-1},s,sr,...,sr^{n-1}\}$ Sadly I have no idea how $D_{2n}/Z(D_n)$  looks like but I guess I need to know that to show that it is a homomorphism and bijectiv. Thanks for helping me
6343
I'm asked to show that $G=GL_2(\Bbb Z/2\Bbb Z)$ is isomorphic to $S_3$. I have few ideas but I don't manage to put them all together in order to obtain a satisying answer. I first tried using Cayley's theorem ($G$ is isomorphic to a subgroup of $S_6$), and I also noticed that $\operatorname{Card}(G)=\operatorname{Card}(S3)=6$ &amp; that they're both non-abelian group.  Is this enough to say that considering $S_3$ is a subgroup of $S_6$ with the same cardinality than $G$, it has to be isomorphic to it ?  Could anyone give me some elements to get a more rigorous proof or lead me to an other path to show this statement ?  Thanks in advance
36498
I recently read the well known theorem that for a group $G$ and $H$ a normal subgroup of $G$, then $G$ is solvable if and only if $H$ and $G/H$ are solvable. In my book, only the fact that $G$ is solvable implies $H$ is solvable was proven. I was able to show that if $H$ and $G/H$ are solvable, then so is $G$, but I can't quite show that $G$ is solvable implies $G/H$ is solvable.   My idea was this. Since $G$ is solvable, there exists a normal abelian tower $$ G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}. $$ I let $K_i=G_i/(H\cap G_i)$, in hopes of getting a sequence $$ G/H\supset K_1\supset\cdots\supset K_r=\{e\}. $$ My hunch is that the above is also a normal abelian tower. However, I'm having trouble verifying that $K_{i+1}\unlhd K_i$ and that $K_i/K_{i+1}$ is abelian.  Writing $H_i=H\cap G_i$, I take $gH_{i+1}\in K_{i+1}$ for some $g\in G_{i+1}$. If $g&#39;H_i\in K_i$, then I want to show $g&#39;H_igH_{i+1}g&#39;^{-1}H_i$ is still in $K_{i+1}$, but manipulating the cosets threw me off. I also tried to use either the second or third isomorphism theorems to show that $K_i/K_{i+1}$ is abelian, but I'm not clear on how to apply it exactly. I'd be grateful to see how this result comes through. Thank you.
30512
Find the derived subgroup of $A_4$.   Since it is $A_4$, for a permutation $\sigma$ to be in $A_4$, $\sigma$ must have a cycle structure of $2$ cycles. Therefore, $\sigma=(ab)(cd)$.  The commutator of such elements, would be obviously another permutation of $2$-cycle (of length $2$).  Does it mean that $A'_4=A_4$?
42042
Let $D$ be an integer such that it is not a perfect square in   $\mathbb{Z}$.       If $D \equiv 1$ mod $4$, the set $$T:= \left\{\left( \begin{matrix} a &amp; b  \\ \frac{(D-1)b}{4} &amp; a+b \\ \end{matrix} \right) : a, b \in  \mathbb{Z} \right\}$$ is a subring of $M_2(\mathbb{Z})$ which is isomorphic   to the ring $$O := \left\{a+b\frac{1+\sqrt{D}}{2}: a, b \in \mathbb{Z} \right\}.$$   My problem is to find an isomorphism. However, I find this relation : $$\det \left( \begin{matrix} a &amp; b  \\ \frac{(D-1)b}{4} &amp; a+b \\ \end{matrix} \right)$$ $$= a(a+b)-\frac{b^2(D-1)}{4}$$ $$= \left(a+b\frac{1+\sqrt{D}}{2}\right)\left(a+b\frac{1-\sqrt{D}}{2}\right)$$  and $D \equiv 1$ mod $4$ $\iff$ $\left(\frac{1+\sqrt{D}}{2}\right)\left(\frac{1-\sqrt{D}}{2}\right) = -k$ for some $k \in \mathbb{Z}$  Does someone could give me a little hint to help me complete this problem?
22095
Let $n \geq 1$ be an odd integer. Show that $D_{2n}\cong \mathbb{Z}_2 \times D_n$. I define a map $$\phi:D_{2n} \rightarrow \mathbb{Z}_2 \times D_n$$ by $R \mapsto (0,r^{\frac{n+1}{2}})$ and $M \mapsto(1,m)$. Then I am stuck at showing the map is bijective. By the way, do we need to show that $(0,r^{\frac{n+1}{2}})(1,m)(0,r^{\frac{n+1}{2}})^{-1}=(1,m)$ ? If we can prove this, then what can we conclude?
120173
Prove that $D_{12}\cong S_3 \times C_2$.   I really dont know how I should start this question. My gut feeling says in some way I have to consider normal subgroups of $D_{12}$ but I cannot see how this will lead necessarily to a unique solutions.  No full solutions please hints only (partly because I cannot give any more of an attempt and I dont want this downvoted)  Ideally I would like a non geometric solution so that similar techniques can be used for general groups
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!
9711
The question  got me curious and thinking about the following:   For which $n$ and $k$ are there nontrivial homomorphisms from the symmetric group $S_n$ to the cyclic group $\mathbb{Z}/k\mathbb{Z}$ of order $k$? I'm thinking it might be an open problem. Is this a problem with a known solution? If so, how can we prove it?   Thanks!
65300
Let $SL_2(\mathbb{Z})$ denote the group (under usual matrix multiplication) of $2\times2$ matrices with integer entries and determinant $1$. Let $H$ be the subgroup of $SL_2(\mathbb{Z})$ consisting of those matrices such that the diagonal entries are all equivalent to $1 \pmod 3$ and the off-diagonal entries are all divisible by $3$.   What is the index of $H$ in $SL_2(\mathbb{Z})$? There are a total of $3^4=81$ different equivalence classes of matrices in $SL_2(\mathbb{Z})$ modulo $3$ (each of the entries can have $0,1,2$ as remainders). Now, the given condition implies only one of the possible $81$ combinations modulo $3$. How do we proceed? any hints? Thanks beforehand.
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?
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!
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?
62882
Let $G$ be a solvable group and $f: G \rightarrow H$ an isomorphism. My question is if $H$ is a solvable group.   Let $1=G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_n = G$ be a normal series of $G$ whose factors are abelian groups. Clearly, $1=f(G_0) \trianglelefteq f(G_1) \trianglelefteq ... \trianglelefteq f(G_n) = H$ is a normal series of $H$ but are the quotients abelian?  Is there a weaker condition for invariance?
62678
Let G be a group of order 80  a) Show that G has a normal Sylow 5-subgroup or a normal Sylow 2-subgroup b) Show that G is solvable  Approach: $80=5*2^4$  $Syl_5(G) \equiv  1mod \text{ } 5$ and $Syl_5(G) | 16$, so the only possibilities are 1 and 16    $Syl_2(G) \equiv 1 mod \text{ }2$ and $Syl_2(G)|5$, so the only possibilities are 1 and 5  How do we discard one of the possibilities to imply there exists a normal sylow p subgroup in G?