Reputation
2,381
Top tag
Next privilege 2,500 Rep.
Create tag synonyms
Badges
1 10 36
Impact
~75k people reached

1d
comment Cubic Planar Graphs have $2^m-1$ Hamilton Cycles, contradicting Bosak…
Are you sure that $\Omega$ is always a group? Any group where every element has order two has order $2^n$ for some $n$.
Aug
1
comment Are these functions characters?
Thank you, but I am more interested in multiplicative characters so that I can study $f,g$ as characters of the same group.
Aug
1
revised Are these functions characters?
added 15 characters in body
Jul
31
revised Are these functions characters?
edited title
Jul
31
asked Are these functions characters?
Jul
29
comment Cubic Planar Graphs have $2^m-1$ Hamilton Cycles, contradicting Bosak…
I have constructed an injective homomorphism from a group with $H+1$ elements to one of $2^n$ elements. Thus $H+1$ must be a power of two.
Jul
27
accepted Prove that for $\alpha$ an ordinal, $\alpha \le \omega^\alpha$.
Jul
24
revised Cubic Planar Graphs have $2^m-1$ Hamilton Cycles, contradicting Bosak…
added 401 characters in body
Jul
24
answered Cubic Planar Graphs have $2^m-1$ Hamilton Cycles, contradicting Bosak…
Jul
23
revised Proving that if $p^2 = a^2 + 2b^2$ then also $p$ can be written in form $a^2 +2b^2$
added 1 character in body
Jul
23
comment Proving that if $p^2 = a^2 + 2b^2$ then also $p$ can be written in form $a^2 +2b^2$
With hindsight splitting into the three cases was unnecessary, you only need that $(p)$ splits into one or two prime ideals, but maybe it's instructional to leave it in.
Jul
23
answered Proving that if $p^2 = a^2 + 2b^2$ then also $p$ can be written in form $a^2 +2b^2$
Jul
11
comment For what functions is this theorem correct?
@joriki The cardinality. One of the conditions for the theorem to be true is surely that preimages are finite.
Jul
11
comment For what functions is this theorem correct?
@joriki Apologies, I meant preimage of an element in the image. I thought you were being sarcastic.
Jul
11
comment For what functions is this theorem correct?
@joriki How is that relevant?
Jul
9
comment For what functions is this theorem correct?
That is correct.
Jul
9
asked For what functions is this theorem correct?
Jun
23
comment Prove that for $\alpha$ an ordinal, $\alpha \le \omega^\alpha$.
Aha, thanks. I see.
Jun
23
comment Prove that for $\alpha$ an ordinal, $\alpha \le \omega^\alpha$.
My non-understanding lies elsewhere: why does the result follow from this?
Jun
23
asked Prove that for $\alpha$ an ordinal, $\alpha \le \omega^\alpha$.