8,041 reputation
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bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 6 months
seen Aug 11 at 15:14

Postgraduate student at the University of Bath, UK, studying geometry and representation theory. Currently thinking about cluster algebras and related objects.


Apr
10
answered Isomorphisms between groups
Apr
9
comment What are the consequences of presentation of an algebra by generators and relations?
A presentation of such an algebra as the path algebra of a quiver with admissible relations is a very useful one if you want to study its representation theory - this is a huge subject, but the book "Elements of the Representation Theory of Associative Algebras" by Assem, Simson and Skowroński covers lots of it.
Apr
7
revised Prove the order of an element divides the order of the group using cosets
added 9 characters in body
Apr
4
revised Prove the order of an element divides the order of the group using cosets
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Apr
4
answered Prove the order of an element divides the order of the group using cosets
Apr
4
comment Prove the order of an element divides the order of the group using cosets
That proof is valid - but I think you're just reproving Lagrange's theorem now. It may help to know what your statement of Lagrange's theorem is exactly, because there are different versions, some of which will be more immediately helpful than others.
Apr
4
comment Prove the order of an element divides the order of the group using cosets
If $H$ is the subgroup generated by $a$, what is the order (i.e. number of elements) of $H$?
Apr
4
comment What is Betti number of a group?
Said another way - in the textbook's language, the Betti number is not the number of generators, but the number of factors isomorphic to $\mathbb{Z}$ (and I agree that this is the better definition). As $\mathbb{Z}$ is not finite, no finite abelian group can have a factor isomorphic to $\mathbb{Z}$.
Apr
4
comment What is Betti number of a group?
Yes - this is why the number of generators comment on Wikipedia is wrong! The finite abelian group $\mathbb{Z}_5$ is generated by one object, but has Betti number zero (its free part is $0$). I think the confusion on Wikipedia is because they are talking about groups arising from topological spaces, and if you put enough restrictions on the space, you will only see groups isomorphic to $\mathbb{Z}^n$, so then the Betti number really is the number of generators - but this is not the case even for general topological spaces.
Apr
4
comment What is Betti number of a group?
One additional comment; I think the "number of generators" comment on Wikipedia is slightly wrong, but at the very least it means the minimal number of generators in a generating set, not the number of elements which are generators. So the "answer" for $\mathbb{Z}_6$ would be $1$, because it is generated by $1$ (or by $5$ - but one element is sufficient). Something like $\mathbb{Z}^2$ isn't generated by a single element; but it can be generated by two elements (e.g. $(1,0)$ and $(0,1)$), so the Betti number is $2$.
Apr
4
comment What is Betti number of a group?
True - but the fact the OP says "Betti number" rather than "Betti numbers" suggests that this is not the point. I don't know if one of the cohomological Betti numbers will also agree with the rank.
Apr
4
revised What is Betti number of a group?
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Apr
4
comment What is Betti number of a group?
That page is about the Betti number for topological spaces; the $k$-th Betti number of the space is the rank of its $k$-th homology group, which deserves to be called the Betti number of the group! To calculate the homology group of a group, you need to put a topology on it; for finite abelian groups, the most obvious one is the discrete one, in which case the $0$-th Betti number is the number of elements, and all the others are $0$ - this probably isn't what you wanted!
Apr
4
answered What is Betti number of a group?
Apr
3
comment What does $s\in \{(u_0,u_1)\in \Bbb R \times\Bbb R^3\}$ mean?
Saying subset is a little dangerous; elements of Cartesian products $A\times B$ are ordered pairs, whereas subsets (presumably of $A\cup B$ in this context) are unordered. This only really matters if you have $s\in A\times A$; the elements $(a_1,a_2)$ and $(a_2,a_1)$ of $A\times A$ are different, but the $2$-element subsets $\{a_1,a_2\}$ and $\{a_2,a_1\}$ of $A$ are the same.
Apr
2
revised For which values does the Gamma function yield an integer result?
edited title
Mar
19
comment Why is $*$ defined only for homotopy classes, and not individual paths between points?
Small typo: you mean "if $a=b$ and $c=d$ then $a*c=b*d$". I'm assuming you mean loops and not paths, and $*$ is the operation that says "do one loop and then the other", but maybe this is wrong...if you do mean this, then the operation is well-defined on paths, essentially automatically, because two paths are "equal" only when they are exactly the same. The more interesting statement is that it is well-defined on homotopy classes, because now some quite different looking paths are considered equal.
Mar
14
comment Finding remainder when $32^{32^{32}}$ is divided by $7$
A quick hint - you can use Fermat's little theorem to reduce the exponent modulo $6(=7-1)$ (because $a^{p-1}\equiv1\bmod{p}$ for any prime $p$).
Mar
11
comment Proving that $\dim(\mathrm{span}({I_n,A,A^2,…})) \leq n$
@clueless I decided to make a bold edit and replace the "diag" by "dim" (and $<$ by $\leq$ in the title), as it seems you agree that this is what it should be. It is better if anybody finds this page later if the question makes sense and agrees with the answer! However, if you don't agree with this edit, please roll it back (or if you don't know how, leave a comment and I'll do it myself.)
Mar
11
revised Proving that $\dim(\mathrm{span}({I_n,A,A^2,…})) \leq n$
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