@MattN The kernel of the homomorphism $x \mapsto e^{ix}$ is $2\pi \mathbb{Z}$. Now the kernel of the homomorphism $x \mapsto \varphi(e^{i x})$ is a closed subgroup of $\mathbb{R}$ and it contains $2\pi \mathbb{Z}$. Moreover $x \mapsto \varphi(e^{ix})$ is of the form $x \mapsto e^{\lambda i x}$ for some $\lambda \in \mathbb{R}$ because you know what the homomorphisms $\mathbb{R} \to S^1$ are. What possibilities for $\lambda$ do you have?
@MattN There's nothing complicated about this! Your proof goes wrong because you claim something that you don't justify in the middle of it (just before "We claim"). I'm providing the reason for it.
@KannappanSampath Have you heard of this theorem in group theory? For a group $G=\{g_1,\ldots,g_n\}$ of order $n<\infty,$ the set $\{g_{\pi(1)}\cdots g_{\pi(n)}\,|\,\pi\in S_n\}$ is a full coset of $G'$ in $G.$
@BrianMScott I'm not discounting the fact that a good teacher is hard to replace, but self-study does have its own advantages, no? and together with the internet could be interesting....
@tb I think I argue that it has to have the form $z \mapsto z^r$ for all $z$ before the claim and after the claim I argue why $r$ has to be an integer.
@MattN No, there's no argument: you said that $z$ depends on $r$ and $x$. Where does the $r$ come from? The way you actually wrote it, $r$ depends on $z$ and $x$, why doesn't it depend on $z$?
@KannappanSampath Um, no. For some $g\in G$ we have $gG'=\{g_{\pi(1)}\cdots g_{\pi(n)}\,|\,\pi\in S_n\}.$ Note that there are no commas there. It's a set of products. It's the set of all possible products of all elements in the group.
@MattN My youngest brother wanted to be a vet. He became one and practised for a few years before going back to get a PhD in animal pathology; he now teaches at Michigan State.
@DavidWheeler Do you mean a group with three elements? But there's only one and it's abelian, so it should trivialize in this case... (Although I'm not sure why. Let me think.)
@KannappanSampath The abelian case is trivial: the commutator subgroup is trivial, so the cosets are just singletons. But when a group is abelian the product of all elements of the group is the same regardless of their order in the product. So indeed, the set in question is a singleton.
What is true is that some modern languages are more conservative than others in ways that are apparent even on casual acquaintance; Icelandic and Lithuanian are the classic examples.
But both of these have actually changed considerably even over the last few hundred years, if not in such obvious ways as many other languages.
One of our graduate assistants used to use potato, though only in speaking. Mind you, his handwriting was so bad that some of his letters might as well have been potatoes.
@Skullpatrol I’ve not really looked closely at it: it wants more concentration than I have right now. I don’t dispute that it has a long literary history, but so does (for example) French: it’s just that we call the 2000-year-old state of that language Latin.
Two friends of mine from the uni bought a Lenovo tablet, which is pretty awesome.
I am really tired of the Apple way of doing things. If I could have waited an extra year before switching cellphones I would probably have bought the Galaxy S2 android.
@N3buchadnezzar No, Apple were genius and signed each of the cellular companies in Israel a contract to sell an incredible amount of iPhones within a year. They did that with every company and each company thought they had an exclusive contract.
So they got to a point they were literally giving iPhones away just so they could fulfill the contract, because it would have cost them more otherwise.
You know. The Chaz, I just ran into a bumped thread in which you said that my comment was funny but you still plan on running for "funniest comment of the year 2011".
Hah. I don't know if either of those comments represents our best work from last year. I might have to go searching tonight to find my best. You'll do the same?
