Show $\otimes$ and $*$ are the same operation on $\pi_1(G, x_0)$ where $(f\otimes g)(s) = f(s) \cdot g(s)$ where $\cdot$ is the group operation on the topological group $G. $  This is a question from the Munkres text which is really frustrating me. It's number 7 on page 335 for anyone with the book. I've already proved the first two parts.  There is a hint and that is to compute $(f*$ $e_x)$ $\otimes$ $(e_x$ $*g)$.   When I do this it is pretty easy to see that this is the same as $f \otimes g$ but am at a loss as to how this is the same as $f*g$.
63937
Let G be a topological group and let $s_1$ and $s_2$ be loops in G (both loops are based at the identity e of G). Is it true that the loop $s_1s_2$ (where the multiplication is the one of the group structure of G) is equal, in $\pi_1(G,e)$, to the loop $s_1*s_2$ where this product is given by first going around $s_1$ and then $s_2$ (i.e., do we have $[s_1s_2] = [s_1*s_2]$)? If yes, what is the proof for this?
71428
This is an exercise from textbook Analysis I by Amann/Escher.  Let $G$ be a group and $X$ a nonempty set. Then $G$ acts (from the left) on $X$ if there is a function $$G \times X \rightarrow X, \quad(g, x) \mapsto g \cdot x$$ such that: GA1: $e \cdot x=x$ for all $x \in X$. GA2: $g \cdot(h \cdot x)=(g h) \cdot x$ for all $g, h \in G$ and $x \in X$. Let $\mathrm{S}_{m}$ be the set of all permutations of $\{1,2,\cdots,m\}$. Then $(\mathrm{S}_{m},\circ)$ is a group where $\circ$ is function composition operation. Show that $$\mathrm{S}_{m} \times \mathbb{N}^{m} \rightarrow \mathbb{N}^{m}, \quad(\sigma, \alpha) \mapsto \sigma \cdot \alpha := \left(\alpha_{\sigma(1)}, \ldots, \alpha_{\sigma(m)}\right)$$ defines an action of $\mathrm{S}_{m}$ on $\mathbb{N}^{m}$.  I am stuck at proving the GA2, i.e. $\pi \cdot (\sigma \cdot \alpha) = (\pi \circ \sigma) \cdot \alpha$.  My attempt: Clearly, $\operatorname{id}$ is the identity element of $\mathrm{S}_{m}$ and $\sigma \cdot \alpha = \alpha \circ \sigma$. Because $\operatorname{id} \cdot \alpha = \left(\alpha_{\operatorname{id}(1)}, \ldots, \alpha_{\operatorname{id}(m)}\right) = \left(\alpha_1,\cdots,\alpha_m\right) = \alpha$, GA1 follows. For $\pi,\sigma \in \mathrm{S}_{m}$ and $\alpha \in \mathbb{N}^{m}$, we have $$\pi \cdot (\sigma \cdot \alpha) = (\alpha \circ \sigma) \circ \pi = \alpha \circ (\sigma \circ \pi) = (\sigma \circ \pi) \cdot \alpha$$  It is clear that $\sigma \circ \pi = \pi \circ \sigma$ if and only if $(\mathrm{S}_{m},\circ)$ is Abelian, which is not true in general. Please help me figure out where I am wrong!
23648
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$\in \mathbb{R}$.
61948
I am trying to show that if for a group $G$ we have $Z(G)=1$, then $Z(\operatorname{Aut}(G))=1$.  I have tried everything I can think of, but to no avail.  The first part of the problem was to show that $\operatorname{Inn}(G) \triangleleft  \operatorname{Aut}(G)$ - does this have any connection?  I couldn't find one.  I know that $\operatorname{Inn}(G) \cong G/Z(G)$, so in this case $\operatorname{Inn}(G) \cong G$ and $Z(\operatorname{Inn}(G))=1$.  Is there any way to get from this to $Z(\operatorname{Aut}(G))=1$?  How do we go about showing this?  Thanks.
25940
I've seen other posts () about the topic of $\operatorname{Inn}(G) \simeq G/Z(G)$, but what I want to ask is a detail. When we ensure that $\operatorname{Inn}(G) \simeq G/Z(G)$, we want to say that there exists some isomorphism $F$ such that:  $$F: G/Z(G) \rightarrow \operatorname{Inn}(G), \quad F(gz) = \tau_g$$  where $g \in G, z \in Z(G)$ and $\tau_g(h) = ghg^{-1}$ for all $h \in G$  But is this $F$ really an isomorphism? I'm not pretty sure due to there is not ONE $\tau_g$ for ONE $gz$, what we have is a total of $|Z(G)| = \text{order  of }Z(G)$ elements of the form $gz$ for just ONE $\tau_g$ due to $\tau_g = \tau_{gz}$ if $z \in Z(G)$
97142
Please give me a hint for this problem.  Let $R$ be a ring, and $I$ a right ideal of $R$. Show if $M$ is a left $R$-module, then $$ f  \   \colon (R/I) \otimes_{R} M \to  M/IM$$  defined by $ f((r+I) \otimes m) = rm + IM$ is an isomorphism of commutative groups.   And show that if $L$ is a left ideal, then  $$ (R/I) \otimes _{R} (R/L) \cong R/(I+L)$$ as commutative groups.
1529
Problem statement: I need to find out if $\mathrm{Aut}(D_4)$ is isomorphic to $D_4$ and explain my answer. I already know that it is isomorphic, so now all I need to do is to prove it.   I assume that first we need to look where it sends $s$ and $t$ and it must send them to elements that satisfy the same relations() And then need to show that those two are conjugation by some element in $D_4$(?)  Any help is appreciated. I was hoping for a duplicate post, but couldn't find it.  Thank you!
39294
Let $R$ be a commutative unital ring and let $M$ be an $R$-module and let $S$ be a multiplicative subset of $R$.  Today I proved both of the following: $$ S^{-1} R\otimes_R S^{-1}M \cong S^{-1} M$$ and $$ S^{-1} R \otimes M \cong S^{-1} M$$  Now I'm slightly confused.      Either my proofs are wrong or $C \otimes A \cong C \otimes B$ does not imply $A \cong B$. But I can't come up with an example. Can someone give me an example? (Or tell me that my proofs are wrong.)
113060
How many homomorphism are there from $(\mathbb Q^+ , \,\cdot\,)$ to $(\mathbb Z_n , +)$?   Could anyone give a hint how to proceed? I couldn't even start
40291
It is well known that the tensor product of $R$-modules over some ring $R$ does not, in general, distribute over infinite direct products, an obvious example being $\mathbb Z_p \otimes_\mathbb Z \mathbb Q \neq 0$. I also know that a sufficient condition for the tensor product to distribute is .  What other sufficient conditions are there? For instance, it would seem intuitive that $\mathbb Q \otimes_\mathbb Z \prod_\mathbb N \mathbb Z \cong \prod_\mathbb N \mathbb Q$. But is it?
113212
True or false. (Prove or give a counterexample.) Let $G$ be a group. Then $|g| = |\phi(g)|$, for all homomorphisms $\phi: G \to G$ and all $g \in G$. Solution. False. $\phi: \mathbb Z_{10} \to \mathbb Z_{12}$ defined by $\phi(x)=0$ for all $x \in \mathbb Z_{10}$ is a counterexample. This function is a homomorphism because $\phi(x+y) = 0 = 0+0 = \phi(x) + \phi(y)$ for all $x, y \in \mathbb Z$ (this function is discussed in problem 3 in assignment 7 as the function sending $[1]$ to $[0]$). The order of $g=1$ is infinity. The order of $\phi(1)=0$ is one.  The problem is that he said $G \to G$ is equivalent to $\mathbb Z_{10} \to \mathbb Z_{12}$, which is think is not right. Explain please? EDIT: Please look at this test and give me your honest opinion,
93857
The following is a question that comes from a counterexample to a problem about exterior algebras of modules and the notes mentions somehow this all relates to the fact that symplectic geometry is done over even dimension.    Let $F$ be a free $R$-module of rank $ n\geq 3$ where $R$ is a commutative ring with identity.  Let $T(F) = \oplus_{k=0}^{\infty} T^{k}(F)$ where $T^k(F) = F \otimes F \otimes \ldots \otimes F$ is tensor product of $k$ modules.  Let $\wedge F $ denote the exterior algebra of the $R$-module $F$, that is the quotient of the tensor algebra $T(F)$ by the ideal $A(F)$ generated by all $x \otimes x$ for $x \in F$.    Suppose $1 \leq k \leq n-1$.       How do we show $ \wedge ^k F \cong \wedge^{n-k} F$?
50009
Considering  $$0\to \Bbb Z \to \Bbb Z\to \Bbb Z/(2)\to 0$$ We want to show that the tensor product is not left exact by tensoring with $\Bbb Z/(2)$, which apparently gives us what we want.  But I did this, and I get  $$0\to \Bbb Z/(2)\overset{\sim}{\to}\Bbb Z/(2)\overset{0}{\to} \Bbb Z/(2)$$   Where I can't see why this isn't exact? Is there a reason why I can't have the zero map on the right?
46016
Ok, I'm reading some thesis of some former students, and come up with this proof, but it doesn't really look good to me. So I guess it should be wrong somewhere. So, here it goes:     Let $R$ be a unitary commutative ring, and $I$ be an ideal, and $N$ is an $R$-module. I'll now show that $$I \otimes_R N \cong IN\,.$$   First, since $1 \in R$, and $I$ is an ideal of $R$, we must have that $I = RI$. So then $$I \otimes_R N = RI \otimes_R N = R \otimes_R IN \cong IN\,.$$  But this must be so wrong, since if this proof is true, I could prove that any $R$-module is flat. So how come this proof is wrong?
4907
Let $\sim$ be the equivalence relation on $S_1 \times I$ given by $(x,t)\sim(y,s)$ if and only if $xt=ys$ where $I=[0,1]$  How do I prove that $(S_1 \times I)/\sim$ is homeomorphic to the unit disk $D^2$ with the induced topology?
52803
I came across the symbol $\otimes$ as below and I would like to know what this symbol $\otimes$ means:     $\text{.... the projection operator P is given by: }$   $$P = I_nd - \nabla G^T(\nabla G \nabla G^T)^{-1} \nabla G= I_{nd} - I_d \otimes uu^T,$$ where $I_x$ denotes the identity matrix of size $x\times x$ and $\mathbf{u}$ is the unit vector , $\mathbf{u} = (1,1,1,\dots,1)/\sqrt{n})$ in $\mathbb{R}^n$
115866
Let be $\mathbb{R}$ the real numbers and $\mathbb{Z}$ the integers. Let $G = \mathbb{R}/\mathbb{Z}$. Determine   $O_\pi(G)$ - the normal maximal $\pi$-group of $G$, $T(G)$ - torsion subgroup.