There are homomorphisms of a group into itself that are one-to-one but not onto. Example :  $f:\Bbb Z \rightarrow \Bbb Z$ defined by $f(n)=2n$. $f$ is a homomorphism with kernel ${0}$. Therefore it is one-to-one. Image of the homomorphism is $2\Bbb Z$. Therefore it is not onto.   Is there an example of a homomorphism of a group onto itself that is not one-to-one?   It has to be an infinite group $G$ having a normal subgroup $N$ such that $G/N≈G$. Index of this normal subgroup $N$ in $G$ must be infinite.
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Let $G$ be any group such that  $$G\cong G/H$$ where $H$ is a normal subgroup of $G$.  If $G$ is finite, then $H$ is the trivial subgroup $\{e\}$. Does the result still hold when $G$ is infinite ? In what kind of group could I search for a counterexample ?
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I am trying to make a picture in my head so I can understand and remember the rules.  So if $f \circ g$ is onto, it is onto because the function $f$ maps every element from a set $B$ to a set $C$ (thus $f$ is onto) and if $f \circ g$ is one-to-one then every element from set $A$ is mapping an element of set $B$ (and thus is one-to-one).  If both $f$ and $g$ is onto then $f \circ g$ is onto and if both $f$ and $g$ is one-to-one then $f \circ g$ is one-to-one and if both $f$ and $g$ are bijective then $f \circ g$ is bijective?  If $f \circ g$ is bijective, we can't say anything, but that $f$ is onto and that $g$ is one-to-one?  If $f$ is onto and $g$ is one-to-one, nothing can be said?  If $g$ is one-to-one and $g$ is onto, nothing can be said?
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I came across an intuitive statement on factor groups that I realized I did not know how to prove.   Let $G$ be a group, and let $H \unlhd G, K \unlhd G$ such that $H \cong K$. I want to show that $G/H \cong G/K$.   I'm not sure how exactly to construct such an isomorphism from an isomorphism $\varphi: H \to K$. Is that the correct approach? By extending such an isomorphism to an automorphism on $\varphi^{\uparrow}:G \to G$, then we could apply the First Isomorphism Theorem to the composition of the natural homomorphism $\mu: G \to G/K$ and that automorphism (for $\ker \mu\varphi^{\uparrow} = H$, if such a $\varphi^\uparrow$ exists).  Edit: Hamou points out this is false in the infinite case: can it still be salvaged in the finite case? Are there restrictions we can put on $G, H,$ and $K$ such that this is true?
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Which groups $G$ have precisely two automorphisms, i.e., precisely one non-trivial automorphism?  Examples: $G= C_3, \mathbf{Z},\ldots$.  I think $G$ has to be abelian. In fact, we have $ \vert G\vert \geq 3$. Therefore, if $G$ is not abelian, we have at least two non-trivial inner automorphisms.  If we can show that $G$ is cyclic the above examples are all of them.
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My professor proved that the kernel is a normal subgroup under a homomorphism f by saying. Let $h\in H$  by applying f to it  $f(ghg^{-1})=f(g)f(h)f(g)^{-1}=e $  My question is how is this a proof that the kernel is normal. What is the reason we apply f to it ? By applying f dont we just prove that $f(ghg^{-1})$ is normal ?
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Let $F_a$ be the group which is freely generated by $a$ elements. How to show that there is a homomorphism from $F_a$ onto $F_b$ if and only if $b\le a$?  I was thinking one possibility is if $F_a$ generated freely by $\{x_1, \dots , x_a\}$ and $F_b$ generated freely by $\{y_1, \dots , y_b\}$ then consider map which says $\phi(x_i)=y_i$ for $1\le i \le b$ and $y_1$ otherwise. I do not know if this works or even how to proceed. How to show this fact?
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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?
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Let $G,H$ be topological groups and let $\rho: G \rightarrow H$ be a homomorphism of topological groups.  I understand this to mean that $\rho$ preserves group structure and is a map between the topologies.    If we say $\rho$ is continuous I understand that this means the map between the two topologies must be a continuous map i.e. the preimage of open sets will be open etc.  However I have come across the following claim, which in my eyes using the definitions above seems trivial. It goes as follows:  Let $G,H$ be topological groups and let $\rho: G \rightarrow H$ be a homomorphism of topological groups. Then $\rho$ is continuous if and only if $\rho^{-1} (V)$ is open for each $V$ in a basis of neighbourhoods $\mathcal{V}$  of the identity $1_h$.   Maybe I have not fully grasped one of the definitions.
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let $G$ be   a multiplicative group of non-zero complex analysis.consider the group homomorphism $\phi:G\rightarrow G$ defined by $\phi(z)=z^4$.  1.Identify kernel of $\phi=H$.  2.Identify $G/H$  My try:  let $z\in \ker \phi$  then $\phi(z)=1\implies z^4=1$ let $z=re^{i\theta}\implies r^4\cos 4\theta =1;r^4\sin 4\theta =0$   then $r=1$ and $\tan 4\theta=0\implies 4\theta=0\implies \theta=\frac{n\pi}{2}$  Is it correct?  I cant proceed in the 2nd problem.  Any hints in this regard.
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If I have two groups, $G$ and $H$ and two injective homomorphisms $\phi:G \to H$ and $\psi: H \to G$, then by the first isomorphism theorem applied to $\phi$, we have that $G \cong \mathrm{Im} (\phi)$, a subgroup of $H$. Similarly, $H$ is isomorphic to a subgroup of $G$. For finite groups, this guarantees that $G \cong H$ but does this hold for infinite groups?  Weird things can happen for infinite groups, e.g $n\mathbb Z \subsetneq \mathbb Z$ but $n \mathbb Z \cong \mathbb Z$. I'm wondering if this kind of thing stops an isomorphism from occurring.  I think this question generalises to rings, modules, fields and so on. Can this be answered for all of these structures?
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This has always been one of my most favorite exercises from Group Theory, and I was surprised to see that this hasn't been asked before. To repeat:     Characterize the natural numbers $n$ such that there is a surjective group homomorphism from $S_n$ to $S_{n-1}$.   I have a solution which I will post in a couple days (if someone doesn't recreate it), but I am more interested in seeing how other people would approach this problem. I am very interested in alternate proofs of this characterization.
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I need to find all group homomorphisms of $S_n\to\mathbb{C}^*$ (for $n\ge2$), and also all group homomorphisms of $A_n\to\mathbb{C}^*$ (for $n\ge2$).  Someone on this site, two years ago, seems to have already asked the same question (). I understand what the questioner says there, but I don't understand it when he says:     Therefore there is exactly one $g:\mathbb{Z}/2\mathbb{Z}\to\mathbb{C}^*$, so all elements of $a\in\mathbb{C}^*$ give us $\operatorname{ord}(a)\mid g[\mathbb{Z}/2\mathbb{Z}]$, so $a$ can only be $-1$ or $1$. Now we know that $f:S_n\to\{-1,1\}$ &hellip;   What is he saying there? And how to go on from there? (I do not understand the answer given either &hellip;)  I hope that someone here can help me further. (And if I'm not allowed to ask this question because is has already been asked two years ago, please tell me what to do instead).  Thank you in advance!
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Can you kindly provide an example of a subgroup that is not normal?  I have been told many times that, for coset multiplication to be defined, the subgroup must be normal.  I have seen the proof and examples of quotient group multiplications.  Now, I am trying to find out where the process will break down if the subgroup is not normal.  Suppose that $x \in a_1H = a_2H, y \in b_1H = b_2H$, where $H$ is not normal. So $x = a_1h_1 = a_2h_2$ for some $h_1, h_2 \in H$, and $y = b_1h_1^* = b_2h_2^*$ for some $h_1, h_2 \in H$.  $xy = (a_1b_1)(h_1h_1^*) = (a_2b_2)(h_2h_2^*)$.  However, both $h_1h_1^*, h_2h_2^*$ are still elements of $H$.  So I don't see how $a_1b_1H \neq a_2b_2H$.
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Let $(e_1, ..., e_n)$ be the standard $\Bbb Z$-basis of $\Bbb Z^n$. Let $x_1,..., x_n$ be elements of an abelian group $G$. Prove that there exists a homomorphism $f$ $:$ $\Bbb Z^n$ $\rightarrow$ $G$ such that $f$$(e_i)$ $=$ $x_i$ for all $i$.   Can anyone please help me out here?
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I'm currently working on a proof, but didn't get the right idea, a hint would be appreciated:  Show that $\mathbb{Z}/n^2\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/n \times \mathbb{Z}/n$  So, lets assume there is an isomorphism between these groups, my first idea to was to construct a contradiction with the homomorphism property. But I don't know how, there should be an element in  $\mathbb{Z}/n \times \mathbb{Z}/n$ which is not in $\mathbb{Z}/n^2$ to reach this kind of contradiction. Of course there is $n \in \mathbb{Z}/n^2$ which is not in the other group but I don't know if this helps.  Thanks
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Question statement is as in the title:     $\varphi:G\to H$ is a Lie Group Homomorphism, then $\ker \varphi$ has Lie algebra $\ker D_e \varphi$   Since $\varphi$ is a Lie group homomorphism, in particular it is continuous. As $H$ must be Hausdorff, every point set is closed, and hence $\{e\}$ is closed in $H$, which implies that $\ker\varphi = \varphi^{-1}(e)$ is closed as well.  So $N=\ker\varphi$ is an embedded Lie subgroup of $G$. Therefore, the inclusion $j:N\to G$ is a Lie Group homomorphism, and induces a Lie algebra homomorphism $D_e j :T_e N \to T_e G$. So I only need to show that $D_e j$ is in fact an isomorphism between $T_e N$ and $\ker D_e \varphi$.  But I'm stuck at this point. I know that for $X\in T_e N$, we have, for $f\in C^{\infty}(H)$,  $$ 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
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I can't figure out why such an epimorphism cannot exist.  I think the fact that a bijection exists between the $\mathbb{Q}$ and $\mathbb{Z}$ throws me off a little. What is it about having a homomorphic surjection that causes problems?