"Let $x,y$ be elements of the group G. Prove the following is an equivalence relation: $x$ is related to $y$ iff there exists some $g$ in $G$ such that $y = g \cdot x \cdot g^{-1}$"  Now I understand that I must prove that the relation is reflexive,symmetric and transitive. But without knowledge of the set or whether the group binary operation is commutative I'm getting very confused on how to do it. I know that I just need to use the group properties of associativity, closure invertibility and identity, but I'm unsure how.  Any help appreciated.
This is a question from the free Harvard online abstract algebra .  I'm posting my solutions here to get some feedback on them.  For a fuller explanation, see  post. This problem is from assignment 5.  a) Prove that the relation $x$ conjugate to $y$ in a group $G$ is an equivalence relation on $G$. b) Describe the elements $a$ whose conjugacy class (= equivalence class) consists of the element $a$ alone.  a) Let $G$ be a group and $R$ be a relation on $G$ defined by $a\sim b$ if $a$ is conjugate to $b$.  Then $a\sim b$ if there is a $g\in G$ such that $a = gbg^{-1}$.  Let $a$ be an element of $G$.  Then $a=eae^{-1}$.  So $a\sim a$ and $R$ is reflexive.  Let $a$ and $b$ be elements of $G$ such that $a\sim b$.  Then there is a $g\in G$ such that $a=gbg^{-1}$.  Then $b=g^{-1}ag$.  Since $g^{-1}\in G$, $b\sim a$.  Hence, $R$ is symmetric.  Let $a,b$, and $c$ be elements of $G$ such that $a\sim b$ and $b\sim c$. Then there are elements $g,g^\prime\in G$ such that $a=gbg^{-1}$ and $b=g^\prime cg^{\prime -1}$.  Then $a=g(g^\prime c g^{\prime -1})g^{-1}=(gg^\prime)c(g^{\prime -1}g^{-1})=(gg^\prime)c(gg^\prime)^{-1}$. Since $gg^\prime\in G$, $a\sim c$.  Hence $R$ is transitive.  Therefore $R$ is an equivalence relation on $G$. b) Let $S$ be the set of elements of $G$ such that, for $s\in S$, $s=gsg^{-1}$ for any $g\in G$.  Then $sg=gs$.  So $S$ is the set of elements that commute with every element of $G$.  In other words, $S$ is the center of $G$. Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem. Thanks.
Complete classification of the groups for which converse of Lagrange's Theorem holds Let $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}$ . I would like to know whether the given sequence converges.   I see  that,  $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}= \sum_{k=1}^{n} \frac{1}{n+\frac{k}{n}}.$ When $n$ gets sufficiently large the contribution by the $ \frac{k}{n} $ term is diminishing and $ a_{n} &lt; \sum_{k=1}^{n} \frac{1}{n} = 1 $.   Thank you. Prove that the normalisation of $A=k[X,Y]/(Y^2-X^2-X^3)$ is $k[t]$ where $t=Y/X$ (Reid, Exercise 4.5) This is a problem about finding the normalisation of a quotient polynomial ring. So I have to find the integral closure of the ring in its field of fractions. The problem statement is as follows:     Let $A=k[X,Y]/(Y^2-X^2-X^3)$. Prove that the normalisation of $A$ is $k[t]$ where $t=Y/X$.   Can I do this by showing that the field of fractions $\text{Frac} A$ of $A$ is equal to $k[t]$, and subsequently showing that the field of fractions is normal? (This could be done by showing that $k[t]$ is a UFD?)  I am lost at calculating/determining $\text{Frac}A$, or similarly proving that $k[t]=\text{Frac}A$. Also, how do I show that it is normal?  I hope you can help! Ksplice is an amazing software (or whatever it is). It changes the way kernel updates work: Instead of rebooting after an update, it patches memory, so you don't need to do anything. Ramanujan Summation It seems that under the light of  the following is plausible:  $$1 + {2^{2n - 1}} + {3^{2n - 1}} +  \cdots  =  - \frac{{{B_{2n}}}}{{2n}}(\Re)$$  Alas, I can't really find any concrete definition of Ramanujan Summation. Could someone provide a small explanation? Making a WinForms TextBox behave like your browser's address bar When a C# WinForms textbox receives focus, I want it to behave like your browser's address bar.  To see what I mean, click in your web browser's address bar. You'll notice the following behavior:    Clicking in the textbox should select all the text if the textbox wasn't previously focused. Mouse down and drag in the textbox should select only the text I've highlighted with the mouse. If the textbox is already focused, clicking does not select all text. Focusing the textbox programmatically or via keyboard tabbing should select all text.   I want to do exactly this in WinForms.  FASTEST GUN ALERT: please read the following before answering! Thanks guys. :-)     Calling .SelectAll() during   the .Enter or .GotFocus events won't   work because if the user clicked the   textbox, the caret will be placed   where he clicked, thus deselecting all   text.      Calling .SelectAll() during the .Click event won't work because the user won't be able to select any text with the mouse; the .SelectAll() call will keep overwriting the user's text selection.      Calling BeginInvoke((Action)textbox.SelectAll) on focus/enter event enter doesn't work because it breaks rule #2 above, it will keep overriding the user's selection on focus. Is there a limit to the number of Power Armor sets you can keep? I've seen videos of some folks having several armor sets (4 is the most I've seen so far) standing about on their settlement. I'm wondering if having too many would cause some sets to disappear. They're not like guns/items that you can safely store in settlement crates, the suits kind of just... stand around. (I know there's the extra step of taking out the fusion cores, so that NPC's don't use them.)  Is there a limit to the number of Power Armor sets you can keep? Finding an unknown linear transformation given that $T(1,1)=(1,0,2)$ and $T(2,3) = (1,-1,4)$ Let $T\colon\mathbb{R}^2\to\mathbb{R}^3$ be the linear transformation such that $T(1,1)=(1,0,2)$ and $T(2,3) = (1,-1,4)$.       Does such a linear transformation exist?  So far I've worked out that it cannot exist, as the first entry for $T(1,1)$ is 1, whereas the first entry for $T(2,3)$ is 1, which can only be obtained through 3-2, or half of 2.  Is there any way of presenting this formally? When I create language plperl , I get error: ERROR:  could not load library "C:/Program Files/PostgreSQL/9.1/lib/plperl.dll": The specified module could not be found.  But in my computer,  "plperl.dll" file is exist in "C:/Program Files/PostgreSQL/9.1/lib/..." folder   ( I can not post illustrative image, this forum require >= 10 reputations)  And if I select * pg_pltemplate, I get:  -[ RECORD 4 ]-+------------------------- tmplname      | plperl tmpltrusted   | t tmpldbacreate | t tmplhandler   | plperl_call_handler tmplinline    | plperl_inline_handler tmplvalidator | plperl_validator tmpllibrary   | $libdir/plperl I need to find the closure of these sets on the ordered square: $$A = \{(1/n)\times 0 | n\in \mathbb{Z}_+\}$$ $$B = \{(1-1/n)\times \frac{1}{2}| n\in \mathbb{Z}_+\}$$ $$C = \{x\times 0 | 0&lt;x&lt;1\}$$ $$D = \{x\times \frac{1}{2}| 0&lt;x&lt;1\}$$ $$E = \{\frac{1}{2}\times y | 0&lt;y&lt;1\}$$ I know that the closure is just the set itself united with its limit points. For $A$, the set of limit points, I imagine, are $\{&lt;1,0&gt;\}$, that is, the $x$ coordinate is $1$ and the $y$ is $0$. But  is says the reverse. What am I getting wrong? Following the same reasoning, $B' = \{&lt;1,\frac{1}{2}&gt;\}$ $C' = \{&lt;0,0&gt;,&lt;0,1&gt;\}$ What am I doing wrong? What is the point of stacking favorite notifications? Contraction Map on Compact Normed Space has a Fixed Point I recently came across the paper . The author raised a few common misconceptions regarding hypothesis testing and p-values, about which I have two specific questions:   The p-value is technically $P({\rm observation}|H_{0})$, which, as pointed out by the paper, generally does not tell us anything about $P(H_{0}|{\rm observation})$, unless we happen to know the marginal distributions, which is rarely the case in "everyday" hypothesis testing. When we obtain a small p-value and "reject the null hypothesis," what exactly is the probabilistic statement that we are making, since we cannot say anything about $P(H_{0}|{\rm observation})$? The second question relates to a particular statement from page 6(652) of the paper:      Since the p-value, or range of p-values indicated by stars, is not set a priori, it is not the long-run probability of making a Type I error but is typically treated as such.   Can anyone help to explain what is meant by this statement? Jensen's Inequality implies convexity? Suppose $\phi$ is a real function on $\mathbb{R}$ such that $$\phi \Bigl(\int_0^1f(x)\,dx\Bigr)\leq \int_0^1 \phi(f)\,dx$$ for every real bounded measurable function $f$. Prove that $\phi$ is then convex.    This one is in Rudin's Real and Complex Analysis, page 74, and, to be quite honest, I am unsure what my first step to solve this problem should even be. To help think of a general methodology, I attempted this similar question, from page 71 of the same book:      Assume $\phi$ is a continuous real valued function on $(a,b)$ such that $$\phi\Bigl(\frac{x+y}{2}\Bigr)\leq\frac{\phi(x)+\phi(y)}{2}$$ for all $x, y\in (a,b).$ Prove that $\phi$ is then convex (the conclusion does not follow if continuity is omitted from the hypothesis).   I take it that the latter question is just a special case of the former (assuming $\phi$ is bounded on $(a,b)$), so I attempted to solve the latter problem first. My attempt was to note that \begin{align*} \phi\big(\lambda x + (1-\lambda)y\big) &amp;\leq \phi\Bigl(\frac{2\lambda x + 2(1-\lambda)y}{2}\Bigr) \\ &amp;\leq \frac{1}{2}\Bigl(\phi\bigl(2\lambda x\bigr)+\phi \bigl(2(1-\lambda)y\bigr)\Bigr) \\ &amp; \leq \phi\bigl(\lambda x\bigr)+\phi\bigl((1-\lambda)y\bigr) \end{align*} with the latter inequality justified simply by applying the hypothesis again twice to the previous expression.   From here, however, I am stuck; I cannot justify "popping out" the $\lambda$ and the $(1-\lambda)$ in the last inequality via some bound afforded by the hypothesis and/or the continuity of $\phi$. Automatically inserting "Section", "Subsection", etc I've recently decided that I just have to finally learn C/C++, and there is one thing I do not really understand about pointers or more precisely, their definition.  How about these examples:   int* test; int *test; int * test; int* test,test2; int *test,test2; int * test,test2;   Now, to my understanding, the first three cases are all doing the same: Test is not an int, but a pointer to one.  The second set of examples is a bit more tricky. In case 4, both test and test2 will be pointers to an int, whereas in case 5, only test is a pointer, whereas test2 is a "real" int. What about case 6? Same as case 5? Display the original question beneath the answer edit form Often when I edit one of my answers, it's because I want to better address the original question. The only problem is that I can't see the original question while I'm editing my answer!  I'd love to see the original question text beneath the "answer edit" form. This way I could scroll down slightly to read it if I wanted to, or just ignore it if not.  Current workaround is to click the 'return to answer' link and hope that you remember to launch it in a new tab/window, then tab back and forth.    Imported from this uservoice item: Do we know Snoke predates the Empire? Print to standard printer from Python? Following an online tutorial that uses an object named "icosahedron" from Add > Mesh > Regular Solids > Archimedean > Icosahedron. This object is not available in my version of Blender (2.74). Presenter asks me to include an add-on named "Add Mesh: Regular Solids", which again is not listed in my User Preferences dialog. What can I do about it?