@PeterTamaroff undergrad is before you get your Bachelor's degree (the first college degree usually) and graduate is when you are in graduate school after that.
@PeterTamaroff i'm a masters student now so not much of anything really. I have a paper in number theory though so i would call myself a number theorist
I do think it works, even though I guess if you stop to think about it carefully you realize it's amusingly convoluted. If you want you could say eg "not nearly as well-known" or look for other equivalents. At any rate, it should be "the proof is X" rather than "the proof are X." :)
In the place where I wrote are, I meant something like: locally barelled generalized inner product spaces are much less known than topological vector spaces.
is the same as putting $(1,0)$ in the basis of eigenvectors. Then applying your diagonal matrix to this. And then changing back into the standard basis.
@anon: for the first time in my life, I read the 2nd part of the question and missed its 1st part. Usually whenever it happens, I miss the last part of OP
but the problem when I calculate all the 20 automorphisms, I obtain that there are 1 of order 1, 5 of order 2, 11 of order 4, and 3 of order 5, and I dont know if I made a mistake or what? I just dont know of any groups with this elements :(
it seems to me that the automorphism that sends $\sqrt[5]{2} \to \eta\sqrt[5]{2}$ and leaves $\eta$ fixed is an automorphism of order 5, if we call this $\sigma$ then $\sigma^k$ must also have order 5, yes?
@Ilya It’s nice to get questions off the Unanswered list. I’ve been known to write up that kind of short answer after a while if no one else has done it.
I am learning for an exam and don't understand a proof in my learning materials
is this a good question for your front page, or do you dislike such questions?
I actually have the proof, only it makes a claim which to me sounds wrong. I suspect that it is my reasoning which is wrong and not the reasoning of the prof who wrote the script, but I can't find my mistake.
@rumtscho That’s a part of graph theory about which I know virtually nothing, or I’d take a stab at it myself. It sounds like a perfectly reasonable question to me, if presented properly. Be sure to make it self-contained, with at least a link to the Wikipedia page on the G-H algorithm.
the underlying set of a semi-direct product is the cartesian product, but the multiplication is not "coordinate-wise"
your best bet is to just give the group with a presentation, the automorphism that fixes the real 5th root of 2 and squares a primitive 5th root of unity is one generator, and the automorphism that fixes the 5th roots of unity and cyclically permutes the roots is another generator.
that is, you want the subgroup of $S_5$ generated by (1 2 3 4 5) and (2 3 5 4).
it will be a non-trivial relation because $\sigma$ and $\tau$ do not commute, your group is non-abelian
well the automorphism of order 5 sends the roots to each other like this: 1-->2-->3-->4-->5
the automorphism of order 4 leaves the real root unchanged. it sends $\eta \to \eta^2$ which sends root 2-->root 3, it sends root 3-->root 5 (the root that has $\eta^4$), and being a 4-cycle, has to send root 5-->root 4
@BrianMScott thank you - I am not very fluent in the Tex library on your site, and remembering how it works seemed like too much effort when all the symbols I needed were available in Unicode :)
@rumtscho No harm done. (I think that all I actually did was to enclose the mathematical expressions in single dollar signs and replace the membership symbol, the $\le$, and the $\ge$ by \in, \le, and \ge, respectively.)
@rumtscho I’m rather curious myself, as I see no basis for one. But upvotes and downvotes aren’t always reasonable or predictable. Give it a day or so to see what happens; there may not be anyone around who can answer it anyway, though I may take a look if it remains unanswered after I’ve had some sleep.
