Let $\cdot$ be an action of a group $G$ on a set $X$. For $g\in G$, define the map $\phi(g):X\to X$ by $\phi(g)(x)=g\cdot x$. I'm asked to show that $\phi(g)\in {\rm Sym}(X)$ and that $\phi :G\to{\rm Sym}(X)$ is a homomorphism.   I've shown that $\phi(g)\in{\rm Sym}(X)$, but I don't understand how to show $\phi$ is a homomorphism? I know it follows from the definition of an action, but what even is the function $\phi: G\to Sym(X)$ explicity?  The solution says that $\phi$ being a homomorphism follows immediately from the fact that $\phi (gh)(x)=g\cdot (h\cdot x)$, but aren't $\phi$ and $\phi (g)$ different functions?
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Checking out  and  (and some papers) was a bit difficult for understanding Group Action. The closest to understanding was this part:    I understand the definition of homomorphism (a structure-preserving map between two algebraic structures of the same type). And from what I understand the symmetric group is a group of functions (bijective functions) mapping elements of a set $X$ onto itself.  However I get confused at this line:     ...an action of $G$ on $X$ may be formally defined as a group homomorphism $\phi$ from $G$ to the symmetric group of $X$.   That seems to be the definition of group action, but I am having trouble understanding how to apply that.  What that says to me is it is a homomorphism (function) $f \colon X \rightarrow S$ where $S$ is a set of bijective functions on $X$. That itself seems weird, I thought a group action would result in a number or something related to $\{1, 2, 3\}$ (defined next).  Continuing, if we have a set $X = \{1, 2, 3\}$ and a group $(*,\{4, 5, 6\})$, then the symmetric group of $X$ would be something like $S = \{1 \rightarrow 2, 2 \rightarrow 3, \dots \}$, and $\phi$ would be something like $f : \{4, 5, 6\} \rightarrow S$. That doesn't make sense to me and I feel like I'm interpreting something wrong. I don't see how mapping to a set of functions $S$ will result in an action on the set.
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Good day. I'm not too sure how I should go about asking my question, so if it is in anyway confusing, feel free to edit it.   In my Group Theory course we defined and showed a few properties of the commutator subgroup $[G,G]$ of a group $G$. And then we stated and proved a lemma in lecture that I'm having some difficulty wrapping my head around. I'll state it here for clarity.  Lemma a) The commutator subgroup $[G,G]$ of $G$ is a normal subgroup of $G$. b) The quotient group $G / [G,G]$ is the largest abelian quotient of G in the sense that every homomorphism $\varphi : G \to A$ where $A$ is an an abelian group, induces a unique homomorphism $\bar \varphi : G/[G,G] \to A$ so that $\varphi =\bar \varphi \circ \pi$ where $\pi : G \to G/[G,G]$ is the quotient homomorphism.  Now, it's not the proof of this lemma that I'm having trouble with, it's the italicized text that is messing with me. I completely understand what it is that the italicized text is saying, I'm just not too sure why it is that the property gives us the "largeness" that is claimed. I can see that the above forces $G$ to "pass through" $G/[G,G]$ before it "reaches" $A$ (I apologize for the casual terminology), so I can sort of see why $G/[G,G]$ should be the largest quotient, in a non-rigorous sort of way.  Overall I'm very confused. Perhaps I'm missing something very simple. Any help or nudges in the right direction would be much appreciated.   As a side note I did try a search on Google and here on Stack Exchange and couldn't find anything that answered my question. So if I missed something obvious, then my apologies. Thank you in advance.
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The Gallian Abstract Algebra text has a number of exercises of the form 'Determine the number of homomorphisms between two groups $G$ and $H$'. It is pointed out that, in the case of a cyclic $G$, determining the image of a generator of $G$ under a homomorphism $\phi$ suffices to specify the entire mapping, since $\phi(g^n)=\phi(g)^n$ for any $g \in G$. Hence determining the number of homomorphisms simplifies to counting generators of $H$ and applying Lagrange's Theorem. Are there any other useful properties of homomorphisms that can be used as tricks for this type of problem or just general strategies when counting maps between $\textbf{non-cyclic groups}$? Two examples I see in the text exercises are:  (1): Determine all homomorphisms from $S_3$ to $G$, where $G$ is Abelian.  (2): Determine the number of homomorphisms from $Z_p\oplus Z_p\to Z_p$, where $p$ is prime ($Z_p$ being a subgroup of the additive integers).  Thanks in advance!
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I have trouble on this question: "Characterize the groups G for which the map $$\iota: G \to G$$ sending $$x \mapsto x^{-1}$$ for all $x \in G$, is an automorphism of $G$." From  post I have seen that $G$ must be abelian. Can anyone propose another characteristic of $G$ that I can derive from the given conditions besides abelian? Thank you so much!!
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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 $A, B:G\to \mathrm{Aut}(X)$ be two free group actions, where $X$ is a set and $G$ a group. A group action, $A$, is free if for all $g\in G\setminus{e}$, $A(g)x\neq x$ for all $x\in X$.Then there exists a function $f:X\to X$ such that for all $g\in G, x\in X$ $$f(A(g)x)= B(g)f(x).$$    I've been completely stumped for some time! How ought one think about such a question? I can't even think of a candidate $f$ or of some simple properties it ought satisfy! I've managed to dig up that the action being free is equivalent to $(g, x)\mapsto (A(g)x, x)$, $G\times X \to X\times X$ being an injection, but haven't been able to make use of this fact.
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We have a function $\varphi:G\rightarrow H$ is an isomorphism, show its inverse $\varphi^{-1}:H\rightarrow G$ is also an isomorphism  I am fine with showing it to be a homomorphism and surjective, this is my attempt at showing injectivity, I'm unsure if they would be sufficient  Assume: $\forall h\in H, \exists g\in G$ s.t. $\varphi(g)=h$ and $\forall g_1, g_2\in G, \varphi(g_1)=\varphi(g_2)\Rightarrow g_1=g_2$  Injectivity - Need to show $\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=g_2$  $\varphi^{-1}(g_1)=\varphi^{-1}(g_2)\Rightarrow g_1=\varphi(\varphi^{-1}(g_2))$ from our surjectivity assumption on $\varphi$ we know there exists such a $\varphi^{-1}(g_2)\in H$ and from our injectivity assumption on $\varphi$ we know that this solution is unique.   [Sufficient? What more could be added? How could I do this clearer?]
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Qusetion: Prove $\phi(G)$ is cyclic if $\phi$ : $G \to H$ is  group homomorphism and $G$ is cyclicDo I do something like this to prove this: Let $G$ be a cyclic group and $G$=$\langle g \rangle$Then $\phi$ : $\mathbb Z \to \langle g \rangle$ by $\phi(n)$= $g^{n}$, $n\in \mathbb Z$ I am not sure of my approach to this. Need some help
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The Question:  Find all primes $p$ such that $\phi(x)=x^{13}$ is a homomorphism $\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z$    My Thoughts:  So from what I understand  \begin{align} \ &amp; \phi:\Bbb Z/p\Bbb Z \rightarrow \Bbb Z/p\Bbb Z \quad\text{is a homomorphism} \\ \ \iff &amp; \phi(xy) \equiv \phi(x)\phi(y) \pmod p \quad \forall \; x,y\\ \ \iff &amp; (xy)^{13}\equiv x^{13}y^{13} \pmod p\quad \forall \; x,y \end{align}  But isn't $(xy)^{13}\equiv x^{13}y^{13} \pmod p\quad \forall \; x,y$ always true, regardless of $p$?  Or is there something I am not understanding?
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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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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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Let $\phi:G\to G'$ be a group homomorphism, and let $N'$ be a normal subgroup of $G'$. Show that $\phi^{-1}[N']$ is a normal subset of $G'$.  My attempt: $\phi^{-1}[N']=\{g\in G:\phi(n)\in N'\}$ $$g' n' (g')^{-1}\in N'$$ $$\Rightarrow\phi^{-1}(g' n' (g')^{-1})=\phi^{-1}(g')\phi^{-1}(n')\phi^{-1}(g'^{-1})$$ Thus $\phi^{-1}[N']$ is normal. Is this thinking correct? I feel like I am missing something.
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I was given a group and I have to prove   a) If $f\cdot g = f$ or $g\cdot f = f$ then $g = 1$.  Is it right to do it using just Identity axiom of group:  $f \cdot g = f = f \cdot1 \Longrightarrow g = 1 ?$  Thanks.
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Let $G$ be a group and $H$ be a subgroup of $G$ with finite index. I want to show that there exists a normal subgroup $N$ of $G$ with finite index and $N \subset H$. The hint for this exercise is to find a homomorphism $G \to S_n$ for $n := [G:H]$ with kernel contained in $H$.   The standard solution suggests to choose $\varphi$ as the homomorphism induced by left-multiplication $\varphi: G \to S(G/H) \cong S_n$. I'm not 100% sure if I understand this correctly. What exactly does $\varphi$ do? We take $g \in G$ and send it to a bijection $\varphi_g: G/H \to G/H, xH \mapsto gxH$? If so, how can I see that its kernel is contained in $H$? Also, the standard solution claims its image is isomorphic to $G/N$ and thus $N$ has a finite index in $G$, how can I see that the image is isomorphic to $G/N$?  Thanks in advance for any help.
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I want to find all group homomorphisms $\varphi: \mathbb{Z}/ m \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$, with $m$ and $n$ natural numbers.  Clearly $\varphi(0)=0$ since the identity in $\mathbb{Z}/m\mathbb{Z}$ must map to the identity in $\mathbb{Z}/n\mathbb{Z}$, and it follows that $m \varphi(1)=0$, so is $mx \equiv 0 \bmod n$?  This as far as I get. How can I find all the homomorphisms?
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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 am trying to prove a series of propositions:  Given any homomorphism p from $\mathbb{R}$[X] to $\mathbb{R}$[X], show that it is equal to $\phi_g$ for a unique g in $\mathbb{R}$[X], with $\phi_g$(f) = f(g(X)). I've expanded the expression $p(f) = p(\sum_{i=0}^n a_iX^i) =\sum_{i=0}^n p(a_i)p(X)^i$ but I'm not sure how to show $p(f) = \sum_{i=0}^n p(a_i)p(X)^i = \sum_{i=0}^n a_ig(X)^i = \phi_g(f)$.  Show that if h,g $\in \mathbb{R}$[X] are such that h(g(X)) = X, then g(X) = aX+b for a $\in$ R$^x$ and b$\in \mathbb{R}$.