This comes up in reading David Speyer's answer to  Given a finite group $G$ and two non-conjugate elements $x, y,$ how does one construct a unitary representation $\rho$ of $G$ such that $\rho(x)$ and $\rho(y)$ have different traces? (The same question makes sense for infinite groups, but it is far from clear that this is always possible in the infinite setting, even if you drop "unitary").
120864
Let $x$ and $y$ be elements that are not conjugate in $G$.  Then there is some irreducible character $\chi$ such that $\chi(x) \not = \chi(y)$.   Clearly the "irreducible" part isn't important, since any character can be written as the sum of irreducible characters, but I'm having trouble going beyond that.  I'd appreciate a good hint over a full answer, and I'd be most interested in a way to construct a group representation $\varphi:G \to GL(V)$ of $G$ such that the character of the representation takes different values on $x$ and $y$.
78203
Given a completely positive and trace preserving map $\Phi : \textrm{L}(\mathcal{H})\to\textrm{L}(\mathcal{G})$, it is clear by the Kraus representation theorem that there exist $A_k \in \text{L}(\mathcal{H}, \mathcal{G})$ such that $\Phi(\rho) = \sum_k A_k \rho A_k^\dagger$ for all density operators $\rho$ on $\mathcal{H}$. (I'll consider the special case $\mathcal{H} = \mathcal{G}$ for simplicity.)  If we use then the system+environment model to express this action as $\Phi(\rho)=\text{Tr}_{\mathcal{H}_E} (Y\rho Y^\dagger)$ for an isometry $Y$ from $\mathcal{H}$ to $\mathcal{H}\otimes\mathcal{H}_E$, where $\mathcal{H}_E$ is an ancilla modelling the environment, what is an explicit construction for a unitary $U$ that has the same action on inputs of the form $\rho\otimes\left|0\right&gt;\left&lt;0\right|_E$? That is, how can I construct an explicit dilation of the map to a unitary acting on a larger space? I understand that this is possible by Steinspring's dilation theorem, but actually constructing an explicit form for the dialated unitary I have had much less success with.
103890
Well be with you, gentlemen. I have the following problem from Aluffi's Algebra:  given a finite group $G$ with an unique element $f$ of order $2$, show that \begin{equation} \prod_{g\in G}g=f \end{equation} My reasoning is the following. Since $f$ is the unique element of order $2$, for each $g\in G$ such that $g\neq e$ and $g\neq f$, it must be that $g^{-1}\neq g$. So we can concoct a particular product coupling the elements with their inverses: \begin{equation} e\cdot f\cdot (g_1\cdot g_1^{-1})\cdot(g_2\cdot g_2^{-1})\cdots (g_n\cdot g_n^{-1}) \end{equation} which is really just $f$ in the end. My question is: how can I show that the order of the factors in this product really does not matter? I cannot see a way out.
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 $z \in Z(G)$
51026
I was thinking about the following: Say we got two groups $A$ and $B$, and we know that for any group C, there is a bijection  $$Hom(A,C) \to Hom(B,C).$$  Are $A$ and $B$ already isomorphic?  If the family of bijections was natural in $C$, this follows from Yoneda's lemma. But does it hold anyway without naturality, just from the cardinalities? I would suspect no but in simple cases (finitely generated). I'd like to hear any thoughts on how to come up with a proof or counterexample.  Also, does the answer change when we consider the contravariant case with an identification $Hom(C,A) \to Hom(C,B)$?
14245
Let $n&gt;2$ and $\tau\in A_n$ and let $C=\{g\tau g^{-1}|g\in S_n\}$ be the conjugacy class of $\tau$ in $S_n$.  Suppose that there is no odd element $x\in S_n$ such that $\tau x=x\tau$.  Prove that $C$ is a  union of two disjoint conjugacy classes in $A_n$.  I managed to observe that if $\sigma_1\in S_n$ an even element and $\sigma_2\in S_n$ an odd element, then:  $\sigma_1\tau\sigma_1^{-1}\neq\sigma_2\tau\sigma_2^{-1}$ otherwise we get that the odd element $x=\sigma_2^{-1}\sigma_1$ commutes with $\tau$.  So, I got one conjugacy class to be the conjugacy class of $\tau$ in $A_n$, say $Q$. It is now clear that it contains all the elements in $C$ of the form $g\tau g^{-1}$ for $g$ even. But I couldn't prove that $C\setminus Q$ can be generated by a conjugacy class in $A_n$ and moreover the sizes equivalence $|Q|=|C\setminus Q|$.
48711
Let $G$ be a non-trivial cyclic group with an unique generator. Show that $G$ has exactly two elements.   Let $g$ be a generator of $G$. My first idea to approach the problem was to show that if $|G|=n &gt; 2$ then there is an integer $k$ between 1 and $n$ such that $\langle g^k \rangle = G$ or, equivalently, $g \in \langle g^k \rangle$. My second idea was to analyse the problem in 2 separate cases. For the first one, if $n$ is prime, let $k$ be a positive integer then, by Lagrange's Theorem, $|g^k| \mid n$ and, therefore, $\langle g^k \rangle = n$. Hence, we must conclude that, if $n &gt; 2$ the generator cannot be unique. For the second case, where $n$ is composite, I was hoping to get to a contradiction, but could not advance any further. Is there a way to develop either of these ideas or is it necessary a new approach?
101271
I was going through my notes on the unitary irreducible representations of the Poincare group and the subsequent construction of one particle states and I stumbled across the following steps in the method of induced representations: ($\mathscr{H}$ is the Hilbert space and $\mathscr{H}_p$ is the subspace of one-particle states with momenta p. $\mathcal{S}_k$ is just the little group of the standard momentum $k$. $L(p)$ is the standard Lorentz boost; $L(p)k=p$)   (1) Let $|k,\sigma\rangle$ be a basis for $\mathscr{H}_k$ and let $\mathcal{D}$ be the (finite) unitary irrep of $\mathcal{S}_k$ induced by the unitary irrep $\mathcal{U}$. Then $\mathcal{D}$ acts on $\mathscr{H}_k$ by mixing the spin polarization states,  \begin{align} \mathcal{U}(h)|k,\sigma\rangle=\sum_{\sigma'}\mathcal{D}_{\sigma'\sigma}(h)|k,\sigma'\rangle,~~~ \forall h\in\mathcal{S}_k,~~~\mathcal{U}(h)\equiv\mathcal{U}(h,0).  \end{align} (2) Define the basis for $\mathscr{H}_p$ by \begin{align} |p,\sigma\rangle=\mathcal{U}(L(p))|k,\sigma\rangle. \end{align} (3) The action of Lorentz transformations on these states is then \begin{align} \mathcal{U}(\Lambda)|p,\sigma\rangle=\sum_{\sigma'}\mathcal{D}_{\sigma'\sigma}\big(h(\Lambda,p)\big)|\Lambda p,\sigma'\rangle  \end{align} where $h(\Lambda,p)=L^{-1}(\Lambda p)\Lambda L(p)\in \mathcal{S}_k$ is known as the Wigner rotation.  (4) Finally, we extend this action to an arbitrary state in the Hilbert space $|\Psi\rangle=\sum_{p,\sigma}\Psi_{\sigma}(p)|p,\sigma\rangle\in\mathscr{H}$ for some coefficients $\Psi_{\sigma}(p)$ by, \begin{align} \mathcal{U}(\Lambda)|\Psi\rangle&amp;=\sum_{p,\sigma}\big[\mathcal{U}(\Lambda)\cdot\Psi\big]_{\sigma}\big( p\big)| p,\sigma\rangle,\notag\\ \text{ where } \big[\mathcal{U}(\Lambda)\cdot\Psi\big]_{\sigma}\big(p\big)&amp;=\sum_{\sigma'}\mathcal{D}_{\sigma\sigma'}\big(h(\Lambda,\Lambda^{-1}p)\big)\Psi_{\sigma'}(\Lambda^{-1}p). \end{align}  These are my own notes which are largely influenced by Weinbergs construction, so they may be incorrect. I understand all of the steps except for one part of step 4; The part where I say $|\Psi\rangle=\sum_{p,\sigma}\Psi_{\sigma}(p)|p,\sigma\rangle\in\mathscr{H}$ for some coefficients $\Psi_{\sigma}(p)$. What are these coefficients? Are they the wavefunctions? Do they depend continuously on $p$, if so should the sum over $p$ be replaced by an integral? Should these coefficients be square-integrable; $\Psi_{\sigma}(p)\in L^2(\Sigma_m,\mu,\mathbb{C})$ where $\Sigma_m$ is the mass shell and $\mu$ a Lorentz invariant measure? Is this expansion of an arbitrary state even correct?
11401
Suppose $G$ is a finite group and I know for every $k \leq |G|$ that exactly $n_k$ elements in $G$ have order $k$. Do I know what the group is? Is there a counterexample where two groups $G$ and $H$ have the same number of elements for each order, but $G$ is not isomorphic to $H$? I suspect that there is, but I haven't thought of one.
104048
I will try and show that if our group $G$ is not cyclic then it is isomorphic to .  So suppose $G$ is not cyclic. Then its elements, excluding $e$, are of order $3$ or $2$. Supposing there is an element of order $3$ gives three distinct elements $g, g^2, g^3$. By hypothesis there exists  $h \in G$ such that $h$ is distinct from these three elements. So we can make $gh$ which must be equal to one of the elements previously mentioned. Setting this equal to the various elements:  \begin{align} &amp;gh=g \Rightarrow h=e &amp;&amp; gh=g^2 \Rightarrow h=g &amp; \\ &amp;gh=h \Rightarrow g=e, &amp;&amp; gh=g^3 \Rightarrow h=g^2 &amp; \end{align} Each of which is a contradiction. Therefore all the elements are of order two and we can get an isomorphism between $G$ and $H \times H$ where $ H$ is cyclic of order two. But this direct product is isomorphic to Klein's four-group so $G$ is isomorphic to $V$.  Is this correct?  Also it is just given in the book that $V$ is isomorphic to this product; how do I show this without checking that $f(gh)=f(g)f(h)$ for all combinations?
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?
77555
Let $\langle g\rangle$ be a cyclic group of order $n$. Suppose $1\leq q \leq n-1$, I want to show that $g^q$ generates $\langle g\rangle$ if and only if $\gcd(n,q)=1$.  Suppose $g^q$ generates $\langle g\rangle$, then   $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $\operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.  On the other hand, if $\gcd(q,n)=1$, then $\operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of  $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are distinct again and $g^q$ is a generator.  Is my proof correct at all and is there maybe a more elegant argument?
56924
In any finite group, number of elements not equal to their own inverses is even number  In my book they have paired elements with their inverses, being elements and inverses different from each other. How do i see this hint ? Thanks
118
This is most likely a lack of understanding of wording on my part. I was considerind the Klein 4-group as the set of four permutations: the identity permutation, and three other permutations of four elements, where each of those is made up of two transposes, (i.e., 1 $\rightarrow$ 2, 2 $\rightarrow$ 1 and 3 $\rightarrow$ 4, 4 $\rightarrow$ 3) taken over the three possible such combinations of four elements.  Here, then, is my question. I am assuming (?) that Aut(G) in this case is the set of permutations of the four elements of the Klein 4-group - or the three non-identity ones for the purpose of showing isomorphic to $S_3$. If this is the case, what does it mean to have a permutation of these three permutations that I mentioned above? As always, thanks for your help.
105524
If $G$ is a finite group with subgroups $H$ and $K$ such that $HK = G$ and $H\cap K = \{1\}$ we get that every element of $G$ can be written uniquely as $hk$ with $h\in H$ and $k\in K$. This then gives a map from $H\times K$ to itself sending $(h,k)$ to the unique $(h&#39;,k&#39;)$ such that $kh = h&#39;k&#39;$.  Conversely, given a map $\varphi$ from $H\times K$ to itself with suitable properties, one can define a group structure $G$ on $H\times K$ such that $H$ and $K$ are subgroups with $HK = G$ and $H\cap K = \{1\}$ and such that the multiplication is given via the map as $(h,k)(h&#39;,k&#39;) = (h\varphi_1(h&#39;,k),\varphi_2(h&#39;,k)k&#39;)$ where $\varphi_1$ and $\varphi_2$ are the projections on the first and second coordinate.  Has this construction been studied before? It seems like it would be a natural generalization of a semidirect product, but I have never seen it mentioned anywhere.
75242
Let $H$ and $K$ be subgroups of a finite group $G$.  A standard result is the formula $$\left|HK\right| = \frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$$ for the cardinality of the set $HK$. The proofs of this that I've seen always involve some arbitrary choice, such as the choice of a transversal for $\{hK \colon  h \in H\}$. Is there a way to show that these quantities are equal without making such an arbitrary choice?  For example, can they be shown to be the respective dimensions of a pair of canonically isomorphic modules?
54372
I am presented with the following task:     Show that if $U$ is the collection of all units in a ring $\langle R, +, \cdot\rangle$ with unity, then $\langle U, \cdot\rangle$ is a group.   I am still not confident with proofs in Abstract Algebra, and I'd like to receive some verification/slaughter/constructive critique on my reasoning here. I do not know if there are any mistakes, or how serious they are.     We know that there is an identity element, since the unity itself is a unit, and therefore in the group. It follows directly from the definition of a unit that there exists an inverse for every element in the group, and associativity follows directly from the Ring-axioms. As for closure, we need to show that the product of two units is also a unit. Let $a, b \in U$ be two elements with the property that $a\cdot b \ne e$, where $e$ is the identity element in $\langle U, \cdot \rangle$. We know that there exists elements $a', b' \in U$ with the property that $a' \cdot a = e = a' \cdot a$ and $b' \cdot b = e = b \cdot b'$. Let $a', b'$ be two such elements with the property that $a' \cdot b' \ne e$. We may now compute $(a\cdot b) \cdot (a' \cdot b')$, and from the Ring-axioms we have associativity under multiplication, so we may write the expression as $a \cdot (b \cdot a' \cdot b') = a \cdot (b \cdot b' \cdot a') = a \cdot (e \cdot a') = a \cdot a' = e$. Thus we have shown that the product of any two elements is a unity, which leads to closure. All the group-axioms are met and $\langle U, \cdot \rangle$ is a group.    I have to apologize if you find the argument needlessly long; I've been having some trouble with assuming things I am not supposed to assume (like what I'm going to prove), so I wanted to stay clear of that.