Sep
15
answered Proof that S3 isomorphic to D3*
Sep
13
comment How is $\sqrt{3} * \sqrt{2} = \sqrt{6}$ mathematically possible?
If you identify real numbers with equivalence classes of Cauchy sequences (this is a standard construction of $\Bbb R$), multiplication of reals amounts to pointwise multiplication of their representative sequences. One checks this is well-defined, obeys all the original rules set forth for multiplication of integers and then rationals, and indeed is continuous when we view $\Bbb R$ as the metric completion of $\Bbb Q$.
Sep
12
comment Infinite primes of a number field
@Dune Different usage of terminology, sure. I more often than not have seen "places" to mean equivalence classes of absolute value. But I have seen "prime" used this way before too. Have you heard of the "prime at infinity" or seen $p\color{Red}{\le}\infty$ in subscripts of $\prod$s before? (BTW instead of calling your comment not correct I should said it's not relevant to OP's desires instead, sorry.)
Sep
12
comment Infinite primes of a number field
@Dune No, that's not correct. This question is about (classes of) absolute values arising from embeddings, not the ones arising from prime ideals of the ring of integers.
Sep
12
comment If a subgroup acts transitively on a set, then the index of the subgroup equals the index of the stabilizer?
Because multiplication of two cardinal numbers always returns the larger of them. If $\kappa<\alpha$ then $\alpha\kappa=\beta\kappa$ implies $\alpha=\beta\kappa=\max\{\beta,\kappa\}$, in which the max cannot be $\kappa$ (we'd have a contradiction), so it is $\beta$, hence $\alpha=\beta$. This proves the cancellation property. For any instance of $\kappa\ge\alpha$ we can exhibit $\beta$ for which $\alpha\kappa=\beta\kappa$ but $\alpha\ne\beta$.
Sep
12
comment If a subgroup acts transitively on a set, then the index of the subgroup equals the index of the stabilizer?
The only way it'd be valid to cancel it is if $|X|<[G:H]$ (strictly).
Sep
12
comment If a subgroup acts transitively on a set, then the index of the subgroup equals the index of the stabilizer?
Yes $[G:G_x]=[H:H_x]=|X|$ is always true, by orbit-stabilizer, but we're still stuck because we cannot divide by infinite cardinals (generally).
Sep
12
comment Showing that a generating function is equivalent to some fraction
@Snufsan on a technical level, there is a notion of convergence and divergence in the ring of formal power series - it's based on the $(x)$-adic topology. In this topology, $8^{100}$ and $8^{-100}$ are equally close to $0$, and $x^k\to0$ as $k\to\infty$.
Sep
10
awarded  Good Answer
Sep
10
awarded  Nice Answer
Sep
10
comment Requesting Hint: Prove $G \cong G \times H$ does not imply $H$ is trivial.
@DanDouglas in fact $\Bbb R\times\Bbb R\cong\Bbb R$ are isomorphic as vector spaces over $\Bbb Q$, not just as abelian groups. This is essentially because $\Bbb R$ has dimension $\frak c$ over $\Bbb Q$ and that $\frak c+c=c$. However we cannot write down such an isomorphism explicitly, because we invoke the axiom of choice to conclude it exists at all.
Sep
8
answered Basis for $\Bbb Z[x_1,\cdots,x_n]$ over $\Bbb Z[e_1,\cdots,e_n]$
Sep
8
comment Is anything known about $2\pi$ integer multiple arguments of the cosine integral?
The sine integral evaluated at integer multiples of $\pi$ are involved in the first, second, third, etc. ringing artifacts in Gibbs phenomenon. See one my earlier questions on it.
Sep
8
comment About Riemann's Hypothesis.
Language remark: such an $n$ would not be a "counterexample to RH," it would be a counterexample to an equivalent to the RH. A counterexample to RH would be a nontrivial zero with real part off the critical line.
Sep
6
comment inverse laplace transform of $s/(s^2+6s+13)$
Your partial fraction expansion is incorrect. Where are you getting $s+3$ from? Clearly the denominator has no rational roots, and partial fractions with radical expressions could turn out complicated - hence completing the square is the best thing to try.
Sep
5
comment Number of actions of $\mathbb Z$
There is no condition that $1\in\Bbb Z$ act trivially. While $1$ is the multiplicative identity, when we treat the integers $\Bbb Z$ as an additive group we do not see multiplication, we see addition, and $1$ is not the additive identity. Rather, $0$ is the additive identity, so $0$ is the thing that acts trivially.
Sep
4
comment Group's morphisms
The latter is abelian and the commutator subgroup of $S_n$ is $A_n$.
Sep
4
comment Proving a group, $G$, is a group action onto some set, $X$
If my understanding of your question is correct, your question should have been phrased something like "I have a function $G\times X\to X$, and I want to know what it means for this to define a group action, and how to test it in practice. Is my understanding of these three tests (closure, identity, composition) correct?" By the way, users are free to post homework questions, as long as they don't omit important details and explain all of their thoughts, ideas, work, etc. Indeed it would probably be better if you include the original problem(s) so we can help you better.
Sep
4
comment Proving a group, $G$, is a group action onto some set, $X$
Okay. Let's say you are starting with a function $G\times X\to X$, or equivalently a function $g:X\to X$ associated to each $g\in G$. Then closure is immediate, since the codomain is $X$. (In your reasoning I don't understand what $a$ is, or what $ax$ is supposed to mean.) For identity, you indeed show that $e(x)=x$ for all $x\in X$, where $e\in G$ is the identity. And for composition, you show $f(g(x))=(f\cdot g)(x)$, where $\cdot$ is the group operation in $G$ and $f,g\in G$.
Sep
4
comment Proving a group, $G$, is a group action onto some set, $X$
What does that mean? Do you have a function $G\times X\to X$ for some group $G$ and set $X$, and you want to know what it means to prove this function defines a group action?