I have difficulties with this task  Find a homomorphism $\Phi : (G, \circ) \to (G, \circ)$, that is surjective but not injective.  I tried to find such a homomorphism, by finding a surjective function, yet functions I tried were not homomorphisms. And then I end up with $Ker(\Phi)=\{e_G\}$. So I found an isomorphism.
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What is a (simple) example of a group $G$ with an endomorphism $G\rightarrow G$ That is surjective, but not injective?
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Let $A$ be a finitely generated abelian group and $f:A\rightarrow A$ a surjective homomorphism. How do I prove that $f$ is an isomorphism?  And if $f$ were injective instead of surjective would the statement still hold?
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As the title suggests. Let $G$ be a group, and suppose the function $\phi: G \to G$ with $\phi(g)=g^3$ for $g \in G$ is a homomorphism. Show that if $3 \nmid |G|$, $G$ must be abelian.  By considering $\ker(\phi)$ and Lagrange's Theorem, we have $\phi$ must be an isomorphism (right?), but I'm not really sure where to go after that.  This is a problem from Alperin and Bell, and it is not for homework.
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Given a group $G$ with $a\in G$ fixed, define $\phi: G \to G$ by $\phi(x) = axa^{-1},  x \in G$, I am wondering when $\phi$ is an isomorphism. I think that it is always an isomorphism because it is one to one, and the order of the domain ($|G|$) and the order of the range are equal. Am I wrong in thinking this?
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The Problem:  Let $X, Y, Z$ be sets and $f: X \to Y, g:Y \to Z$ be functions.  (a) Show that if $g \circ f$ is injective, then so is $f$.  (b) If $g \circ f$ is surjective, must $g$ be surjective?  Where I Am:  So, I really have trouble with these, for some reason. I can draw pictures and make sense of the problems, but writing down proofs is very difficult for me.   Basically, for (a), I ended up with some complicated statement involving an implication implying another implication and then tried to derive a contradiction. It just got so convoluted that I couldn't make sense of it anymore, and I know there's a quick, elegant way to show it.  For (b), I know that $g$ need not be surjective. Once again, though, proving it directly from definitions has given me a bit of a headache.  Any help here would be appreciated. Thanks in advance.  The Proofs!  Ok, I did it. Thanks for the help, everyone! Let me know if there's anything wrong with these proofs, or if they could bet any better.  (a) Suppose $f$ is not injective. Then $$ f(x_1)=f(x_2) \implies x_1 \ne x_2 \text{        }(*).$$ Let $f(x_1)=y_0=f(x_2)$ and let $g(y_0)=z_0$. Then $$ (g \circ f)(x_1) = (g \circ f)(x_2) = g(y_0) = z_0. $$ Since $g \circ f$ is injective, $$ (g \circ f)(x_1) = (g \circ f_2) \implies x_1 = x_2. $$ However, this contradicts $(*)$. Therefore, $f$ must be injective.  (b) Suppose $g$ is not surjective. Then  $$ \forall y \in Y, \exists z \in Z \text{ such that } g(y) \ne z \text{         }(**).$$  Since, $g \circ f$ is surjective,  $$ \forall z \in Z, \exists x \in X \text{ such that } g(f(x)) = z \text{       } (***). $$  Let $f(x) = y$. Then,  $$ g(f(x)) = g(y) = z. $$  Because of $(***)$, this is true for all $z \in Z$, which contradicts $(**)$. Therefore, $g$ must be surjective.
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Quoting " Let $\phi : \Bbb Z \rightarrow \Bbb Z$ be given by $\phi(n) = 7n$. Prove that $\phi$ is a group homomorphism. Find the kernel, and the image of $\phi$."   My understanding:  Part 1:   Given $n,m \in \Bbb Z$, let's check that $\phi(n+m) =\phi(n)+\phi(m)$  $$\phi(n+m) = 7(n+m) = 7n + 7m = \phi(n)+\phi(m)$$  Therefore as the group operation in $\phi$ is preserved, $\phi$ is a group homomorphism.  Part 2 (revised):   $\phi$ is one-to-one as:  $\space \space \phi (n) = \phi(m) \Rightarrow 7n =7m \Rightarrow n=m  $. We also know that $\phi(e)=e$ by the property of homomorphism. Therefore, no other object than $0$ in the domain can map to $0$ in the codomain. It follows that: $$\ker(\phi)=\{ x \in \Bbb Z : \phi(x)=e\}=\{e\}$$   Part 3 (revised):  I claim that the only possible image-elements of $\phi$ are the multiples of $7$ in $\mathbb Z$ denoted $7\mathbb Z$.   To prove this I have to show that Im$(\phi) \subset 7 \Bbb Z$ and that $ 7 \Bbb Z \subset $ Im$(\phi)$.  Proving Im$(\phi) \subset 7 \Bbb Z$ :      $\forall y \in $ Im$(\phi)$ such that $  y =\phi(x)$ where $x \in 7 \Bbb Z$. It follows that $y \in 7 \Bbb Z$, therefore Im$(\phi) \subset 7 \Bbb Z$   Proving $ 7 \Bbb Z \subset $ Im$(\phi)$:   $\forall y \in  7\Bbb Z$ such that $  y =7x$ where $x \in \phi(x)$. It follows that $y \in \phi(x)$, therefore $ 7 \Bbb Z \subset $ Im$(\phi)$.   Any input to my understanding is much appreciated.
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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!
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Let $H$ be the normal subgroup of $G$.   Is it true that $(H[G,G])/H$ is isomorphic to $[G/H,G/H]$?  If so, I want to make a surjective homomorphism $\phi\colon H[G,G]\to [G/H,G/H]$ with Kernel $H$ to prove it.  But, if I define $\phi(h[g_1,g_2])=[g_1/H,g_2/H]$ then I have some trouble.
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How do you show that every group homomorphism from $(\mathbb{Q}, +)$ to $(\mathbb{Q}, \times)$ is the trivial map? I am trying to use proof by contradiction by assuming there is an element $\frac{a}{b}$ such that some homomorphism $\phi$ has $\phi(\frac{a}{b}) = \frac{c}{d} \ne 1$. But I cannot seem to deduce any contradictions from here. Maybe using direct proof is a better approach? Any help is appreciated.
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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?
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This is related to this .  I just didn't want a prolonged discussion in the comments.  Let $\phi: G \to G'$ be a homomorphism.  Let $G$ be a finite group.  Let $K \leq G$ be the kernel of $\phi$.  Let $I \leq G'$ be the image of $\phi$.  Let $H' \leq G'$.  Find a formula relating the order of $\phi^{-1}(H')$ in terms of $H', I, K$.    Attempt at a solution: $|\phi^{-1}(H')|=|H'\cap I|\cdot |K \cap \phi^{-1}(H')|$.  My justification for this is as follows.  $\phi^{-1}(H')/(K \cap \phi^{-1}(H'))\cong H' \cap I$
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How many homomorphisms are there from $\Bbb Z_{20}$ onto $\Bbb Z_{8}$? How many are there to $\Bbb Z_{8}$?  I can see that there are no onto homomorphisms because $|\Bbb Z_{20}|/ |\Bbb Z_{8}|$ is not an integer, but for "to" I can't figure how to solve this.  I can see that the trivial homomorphism and the natural one $\phi(x) = x\mod8$ work, but I can't see how I would find others. I can also see that $|\phi(\Bbb Z_{20})| = 1, 2, $ or $4$ because it divides  $\Bbb Z_{20}$ and $\Bbb Z_{8}$, but from here I'm stuck too.
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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.
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Let $G$ and $H$ be groups and assume that there exist onto homomorphisms $$\phi :G\rightarrow H,\qquad\psi :H\rightarrow G.$$ If $H$ and $G$ are finite then it is clear that $H$ and $G$ are isomorphic. It is also valid for finitly generated abelian groups. However, in general this is not true even in the abelian case.     My question is: if we know that $G,H$ are finitly generated and such $\phi ,\psi$ exist then are $G$ and $H$ isomorphic?
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If $ \phi: R \to R'$ is a homomorphism of $R$ onto $R'$ and $R$ has a unit element,  $1$, show that $\phi(1)$ is the unit element of $R'$.  I am having trouble proving this. I think I just need a hint on how to start and I will be able to solve. But basically I have tried a few things like $\phi(1) = \phi(a)\phi(b)$ but I cannot assume it is a division ring. Also is it okay to assume from the wording of the question that $R'$ has a unit element.  Thanks in advance.
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Let me ask Prob.F6 in Pinter's algebra book.  'Let $p$ be a prime and $f$ be a homomorphism from $G$ to $H$ where $G$ and $H$ are groups.  If range of $f$ has an element of order $p$, then $G$ has an element of order $p$.'   If $f$ is injective, life gets easier. But without the injectivity of the homomorphism, i feel lost... Would somebody help please?
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So I believe I've finished the proof, but it all amounts to case-work and I really think (and hope) there's a better approach than this. Here's what I've done: Attempt: Within $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ identities must be mapped to identities thus what defines each isomorphism is the bijective function between its last three elements: \begin{matrix} (0,0) &amp; \mapsto &amp; (0,0) \\ (0,1) &amp; &amp; (0,1) \\(1,0) &amp; &amp; (1,0) \\(1,1) &amp; &amp; (1,1) \\ \end{matrix} If every bijective function between these last three elements constitutes a homomorphism (including the first unchanged mapping of identities), then $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3$ in that isomorphisms in $\text{Aut}_\mathsf{Grp}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})$ can be identified with permutations of three elements, which is exactly what $S_3$ is. I then, essentially go case-by-case showing every one of the six different bijections is a homomorphism. For two of those bijections (which are just compositions of other bijections I've already checked) I apply the following lemma:  If $\phi :G \to H$ and $\psi : H \to K$ are bijective homomorphisms, then $(\psi \circ \phi)$ is a homomorphism. As proof, assume the premise. We then have $\forall a,b \in G:$ $$\psi (\phi (ab))= \psi(\phi(a) \phi(b))=\psi (\phi(a))\psi (\phi(b))$$ $\implies (\psi \circ \phi)$ is a homomorphism.  So my questions are:  Is the proof that I have sufficient? Is there a better proof for this problem? (A hint towards it rather than complete give-away would be much more appreciated)  EDIT: I have indeed looked at , but the approach carried out there uses vector spaces which I have not been introduced to or am supposed to use to solve this problem.