For groups $G$, $H$, and $K$, assume there exists a left-split short exact sequence: $$ 1 \rightarrow K \xrightarrow{\varphi} G \xrightarrow{\psi} H \rightarrow 1$$ Then $\varphi$ is an injective homomorphsim, $\psi$ is a surjective homomorphism, and ${\rm Im}(\varphi) = \ker(\psi)$. Furthermore, there exists a homomorphism $\pi: G \rightarrow K$ such that $\pi \circ \varphi = id_K$. How can I show that these assumptions imply that $H \triangleleft G, K \triangleleft G, G = HK,$ and $H \cap K = \{ 1 \}$?
122534
Here is the problem I am trying to prove: For groups $G,H,K,$ show that the following conditions are equivalent.  $G \cong K \times H.$ There exists a split short exact sequence: $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1.$ There exists a left-split short exact sequence: $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1.$ $H \triangleleft G, K \triangleleft G, G = HK $ and $H \cap K = \{1\}.$  My questions are: Here is the proof of $1 \implies 2.$ Assume that  $G \cong K \times H.$ We want to show that there exists a split short exact sequence: $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1.$ But we know that a short exact sequence is split if it is both left- and right-split. Also, know that a short exact sequence $1 \rightarrow K \rightarrow   G \rightarrow H \rightarrow 1$ is left-split if there is a retraction $r: G \rightarrow K$ s.t. $$r\circ i = id_K \quad \quad (1) $$ Where $i: K \rightarrow G$ and it is injective. And also, we know that a short exact sequence $1 \rightarrow K \rightarrow   G \rightarrow H \rightarrow 1$ is a right-split if there is a section $s: H \rightarrow G$ s.t. $$p\circ s = id_H \quad \quad (2)$$ Where $p: G \rightarrow H$ and it is onto. So, we need to find the functions $i, p, r$ and $s$ that satisfies $1.$ and $2.$ above. So, since $G \cong K \times H,$ we can say that we have $1 \rightarrow K \rightarrow K \times H   \rightarrow H \rightarrow 1,$ with $i:  K \rightarrow K \times H$ the embedding $i(k) = (k,1)$ and  $p: K \times H \rightarrow H$ the projection $p(k,h) = h.$ If we define $r:  K \times H \rightarrow K $ with $r(k,h) = k$ then we have $(r\circ i) (k) = r((k,1)) = k$ as required. Also, if we define $s: H \rightarrow K \times H $ with $s(h) = (1,h)$ then we have $(p\circ s)(h) = p ((1,h)) = h $ as required. 1- Is it correct? and is the importance of $G \cong K \times H$ is just to create the split short exact sequence? 2- Is the statement of my problem completely correct? 3- How can I prove $1 \implies 4$? 4- Is the right sequence of implications for proving the problem is proving $1 \implies 2 \implies 3 \implies 1$ and then proving $1 \Leftrightarrow 4$? 5- Is $2 \implies 3 \implies 1$ in my problem means $2 \implies 1$? or $2 \implies 1$ is a different thing
60139
I had an exam this morning, one of the questions asked about the truth of the statement      There is a space $X$ such that $S^1$ is homeomorphic to $X\times X$.   I said that this was false and this was my reasoning... was I correct? (Note, an earlier part of the question asked me to prove $\pi_1(Y\times Z)=\pi_1(Y)\times\pi_1(Z)$, I took this as a hint.)  Suppose for the sake of contradiction that such an $X$ exists. Then $\Bbb{Z}\cong G\times G$ where $G\cong\pi_1(X)$. Let $\Phi$ be the isomorphism.  Let $\Phi(1)=(g_1,g_2)\in G\times G$. Then at least one of $g_1,g_2$ is nonzero, since $\Phi(0)=(0,0)$ (the identity on $G\times G$, and $\Phi$ is injective). Without loss of generality $g_1\ne 0$. Then which element $n\in\Bbb{Z}$ has $\Phi(n)=(kg_1,0)$? Since $$kg_1=\underbrace{g_1+\dots +g_1}_k\in G,$$ and $\Phi$ is surjective, such an $n$ should exist. But this is not possible since $\Phi(n)=(kg_1,0)$ implies (since $\Phi$ is a homomorphism) that $kg_1=ng_1$ and $ng_2=0$, and since $k$ was arbitrary and this need not hold if $k\ne 0$.
19840
I recently had my second abstract algebra exam returned to me and I missed this particular question (unfortunately I left the problem blank, just a bad test day). I've been trying to figure out the problem though and could use some help.  a) Does there exist a surjective homomphism $\phi: \mathbb{Z} \rightarrow S_3$? Explain.  What I'm thinking: Let $\phi: \mathbb{Z} \rightarrow S_3$ be a homomorphism.$\mathbb{Z}$ is a cyclic group and thus $\phi(\mathbb{Z})$ is cyclic. However, $S_3$ is not cyclic. Thus $\phi(\mathbb{Z}) \neq S_3$ and $\phi$ cannot be surjective.  b) Assume that there exist homomorphisms $\phi: \mathbb{Z} \rightarrow S_3$ where Ker$(\phi) \neq \mathbb{Z}$. What are the possible kernels? Explain.  What I'm thinking: Since $1$ generates $\mathbb{Z}$, if $1\in$ Ker$(\phi)$, then $\mathbb{Z}=$Ker$(\phi)$. So could Ker$(\phi)$ = $n\mathbb{Z}$ for $n \in \mathbb{Z},  \; n\neq1$?  c)List at least two non-trivial homomorphisms $\phi: \mathbb{Z} \rightarrow S_3$ whose kernals are not $\mathbb{Z}$.  $\phi_1(n) = (1\; 2)^n$  $\phi_2(n) = (1\; 3)^n$  Any input is much appreciated!
107577
Question statement is as in the title:     $\varphi:G\to H$ is a Lie Group Homomorphism, then $\ker \varphi$ has Lie algebra $\ker D_e \varphi$   Since $\varphi$ is a Lie group homomorphism, in particular it is continuous. As $H$ must be Hausdorff, every point set is closed, and hence $\{e\}$ is closed in $H$, which implies that $\ker\varphi = \varphi^{-1}(e)$ is closed as well.  So $N=\ker\varphi$ is an embedded Lie subgroup of $G$. Therefore, the inclusion $j:N\to G$ is a Lie Group homomorphism, and induces a Lie algebra homomorphism $D_e j :T_e N \to T_e G$. So I only need to show that $D_e j$ is in fact an isomorphism between $T_e N$ and $\ker D_e \varphi$.  But I'm stuck at this point. I know that for $X\in T_e N$, we have, for $f\in C^{\infty}(H)$,  $$ D_e\varphi (D_e j(X))(f)=D_e j (X)(f\circ \varphi)=X(f\circ \varphi\circ j)=0$$  So $D_ej(T_e N) \subset \ker D_e \varphi$.  At this point, I can either show inverse inclusion: that is, for every $X\in \ker D_e \varphi$, we can find $Y\in T_e N$ such that $D_e j (Y)=X$, or for all $f\in C^{\infty}(G)$, $Y(f\circ j)=X(f)$  Or I can somehow show that dimensions of the two spaces are the same.   But I can't proceed from here.  It'd be great if someone could help me
4386
I have to show that, given $k|n,$ and  $\langle{r^k}\rangle$ is a normal subgroup of the Dihedral group $D_{2n}$ then $D_{2n}/\langle{r^k}\rangle\cong D_{2k}$  Given this, I know that I need to show that a particular homomorphism is a bijection.  The natural one I was thinking was $$\phi:D_{2k}\rightarrow D_{2n}/\langle{r^k}\rangle$$ $$r^is^j\mapsto r^is^j\langle{r^k}\rangle$$ First I have to show that this is a homomorphism, which is easy, i think, since $$\phi(r^{i_1}s^{j_1}r^{i_2}s^{j_2})=r^{i_1}s^{j_1}r^{i_2}s^{j_2}\langle{r^k}\rangle=r^{i_1}s^{j_1}\langle{r^k}\rangle r^{i_2}s^{j_2}\langle{r^k}\rangle=\phi(r^{i_1}s^{j_1})\phi(r^{i_2}s^{j_2})$$ Now that I have the homomorphism, I need to show bijectivity.  So given $\phi(r^{i_1}s^{j_1})=\phi(r^{i_2}s^{j_2})$ $$\phi(r^{i_1}s^{j_1})=\phi(r^{i_2}s^{j_2})\Rightarrow r^{i_1}s^{j_1}\langle{r^k}\rangle=r^{i_2}s^{j_2}\langle{r^k}\rangle$$ and it's here I'm getting stuck...drawing a blank
92230
Show that if $G$ is a group with two normal subgroups $H$ and $K$ such that $G=HK$ and $H\cap K=\{e\}$ then the map $(h,k)\rightarrow hk$ is an isomorphism of groups from $H\times K$ to $G$.  I am not sure whether or not I should be using the homomorphism theorem or not. (If $j:G\rightarrow H$ is a homomorphism then $G/\text{Ker}(j)$ is isomorphic to $\text{Im}(j)$)  Thanks
90216
Suppose that $H$ and $K$ are normal subgroups of a group $G$ such that $HK=G$. I need to show that  $G/(H\cap K)\cong (G/H)\times (G/K)$.  So from the second isomorphism theorem we have that:  $HK/H\cong H/(H\cap K)$ which gives that $G/H\cong H/(H\cap K)$  $HK/K\cong K/(H\cap K)$ which gives that $G/K\cong K/(H\cap K)$  So we must now show that:  $H/(H\cap K)\times K/(H\cap K)\cong G/(H\cap K)$  So consider the map $\theta:H/(H\cap K)\times K/(H\cap K)\rightarrow G/(H\cap K)$ such that $\theta(h(H\cap K)\times k(H\cap K))=hk(H\cap K)$  Now this is a homomorphism and is surjective. To check injectiviy we see that if:  $\theta(h_1(H\cap K)\times k_1(H\cap K))=\theta(h_2(H\cap K)\times k_2(H\cap K))$ then we have that:  $(h_1k_1(H\cap K))=(h_2k_2(H\cap K))$ so that $(h_1k_1)(h_2k_2)^{-1}(H\cap K)=(H\cap K)$ which gives that $h_1k_1(H\cap K)=h_2k_2(H\cap K)$ So we are done.  Is the above correct?  Thanks for any help
103896
Let $R$ be a commutative ring with unit and $M$ a right $R$-module. Consider the assignment $\varphi: \operatorname{Hom}_{R} (R,M) \rightarrow M$, $f \mapsto \varphi(f) = f(1)$. Show that $\varphi$ is an isomorphism.  My approach: We have to prove $\varphi$ is a morphism, then $\varphi$ is a injection, surjection, hence $\varphi$ is an isomorphism. $\forall f,g \in \operatorname{Hom}_{R} (R,M)$ $\varphi(fg) = (fg)(1) = f(1)g(1) = \varphi(f) \varphi(g)$ $\Rightarrow \varphi$ is a morphism $\forall f,g \in \operatorname{Hom}_{R} (R,M)$ $\varphi(f) = \varphi(g) \Rightarrow f(1) = g(1) \Rightarrow f = g$ $\Rightarrow \varphi$ is injective $\Rightarrow \varphi$ is a homomorphism  How can I prove $\varphi$ is surjective to conclude $\varphi$ is an isomorphism? Sorry for my poor English.
103432
Just starting to properly learn the chapter on homomorphisms, and it will be great if I can understand the following question I have:  How could I actually show that when there are two groups $G$ and $H$ (finite), and $|G|$ and $|H|$ are coprime, then then homomorphism $\Phi:G \rightarrow H$ must be trivial.   My idea: Since $Im(\phi)$ and $Ker(\phi)$ are respective subgroups of the groups $G$ and $H$, then by Lagrange's Theorem $|Ker(\phi)|||G|$ and $|Im(\phi)|||H|$. I think if we just show that the |$Im(\phi)$| is $1$, then it implies that the $|Ker(\phi)|$ is equal to the order of $G$. Hence $\phi$ must be trivial. But if this is correct, I cant seem to show $|Im(\phi)|=1$. However, is this the correct way?   Can someone please help me show the homomorphism $\Phi$ is trivial if my idea is completely off?
40824
Those 5 Statements are all equivalent to the Statement that $H$ is a normal subgroup of $G$:  $(1) \forall g \in G, h\in H$ we have $ghg^{-1} \in H$  $(2) \forall g \in G, gHg^{-1} \subseteq H$  $(3) \forall g \in G, gH = Hg$  $(4)$ Every right coset of $H$ is a left coset  $(5) H$ is the kernel of a homomorphism of $G$ to some other Group  In class we used the Definition 3 can someone explain me why 3 and 1 are equivalent?  Edit: It would be convenient to show the equivalence of all 5 statements
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!
25878
Does there exist any surjective group homomorphism from $(\mathbb R^* , .)$ (the multiplicative group of non-zero real numbers) onto $(\mathbb Q^* , .)$ (the multiplicative group of non-zero rational numbers)?
32086
Let $G$ be any group, and $H \leq G$ a subgroup.  Suppose that for each $x \in G$, there exists a $y \in G$ such that $xH \subseteq Hy$. In other words, every left coset of $H$ is contained inside some right coset of $H$.  Question: what can we say about $H$? In particular, does this imply that $H$ is normal?  I know that if $G$ is finite, then $H$ must be normal, because then $xH \subseteq Hy$ implies $xH = Hy$, since $|xH| = |Hy|$.
112299
Prove or disprove: for any comm. ring $R $,   For three $R $-modules $M $, $N $ and $P $, $M \oplus P \simeq N \oplus P $ implies $M \simeq N $. Let $f:M\to N $ and $g:N\to M $ be two $R $-module homomorphisms such that $gf=1_M $ (where $1_M = \operatorname{id}_M$ means the identity map on $M$). Then, $N\simeq\operatorname{Im}(f)\oplus\ker(g)$.   If $R $ is a PID and the modules are f.g., 1 is true, but I have no idea on general cases.
50532
Consider the following multiple choice problem:  Let $H$ be the set of all group homomorphsims $\phi:{\bf Z}_3\to{\bf Z}_6$. How many functions does $H$ contain? A.1  B.2 C.3 D.4 E.6  Since $1$ generates ${\bf Z}_3$, one can analyze $\phi(1)$ case by case, which may be rather time consuming, for me, at least. Since this is a multiple choice problem, is there any quick way to solve it?
115145
Does there exist a surjective homomorphism from $(\mathbb R,+)$ to $(\mathbb Q,+)$ ?  ( I know that there 'is' a   'surjection' , but I don't know whether any surjective homomrophism from $\mathbb R$ to $\mathbb Q$ exist or not . Please help . Thanks in advance )
96281
According to this ,      A group homomorphism is a map $f:G \rightarrow H$ between two groups   such that the operation is preserved: $f(g_{1}g_{2}) = f(g_{1})f(g_{2})$ for all $g_{1}, g_{2} \in G$, where the product on the left-hand side is in $G$ and on the right-hand side is $H$.   Question  Let G be a group and let $Aut(G)$ be the group of automorphisms of G.  (a) For any $g \in G$ define $\phi_{g} : G \rightarrow G$ by $\phi_{g}(x)=g^{-1}xg$.  Check that $\phi_{g}(x)$ is an automorphism.  (b) Consider the map  $\Phi : G \rightarrow Aut(G)$  $g \mapsto \phi_{g}$  Check that $\Phi$ is a homomorphism.  My Attempt  I did the first part.  Now, I need to do the second part of this problem.  Here is my attempt for the second problem.  Observe that if we evaluate $\Phi$ at $gh$ with arbitrary $x$, then:  $\Phi(gh) = \phi_{gh}(x) = (gh)^{-1}x(gh) = h^{-1}(g^{-1}xg)h = h^{-1}\phi_{g}(x)h = \phi_{h}(\phi_{g}(x))$.  The problem is that I'm stuck with this part.  I think that the operation of that $\Phi$ would be the composition of two functions.  Another thought: It would not make sense to say that:  $\Phi(gh) = \Phi(g)\Phi(h)$  because,  $\Phi (gh) = h^{-1}g^{-1}xgh$  But  $\Phi(g) = g^{-1}xg$  $\Phi(h) = h^{-1}xg$  $\Phi(g)\Phi(h) = g^{-1}xgh^{-1}xh$  The problem is that G is not assumed to be abelian (Well, if it is, then the equality holds.).  Then, we can't take the product of $\Phi$'s to match with $\Phi_{gh}$.  If I'm right, the solution to the second part of this problem is unlike the one to the first part of this problem.  Any comments or suggestions?