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Nov
17
comment 1-Cocycle of an Algebra
Co-cycle are a general concept which arise in the context of homological algebra. It would be helpful if you could also add the reference to where are you studying about this co-cycles.
Nov
16
revised $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
deleted 1 characters in body
Nov
16
answered Products, Naturality and Functors
Nov
15
comment Permutation representation argument validity
No suggestion or correction, your proof is correct.
Nov
14
revised Why is it worth spending time on type theory?
made a correction
Nov
14
comment Why is it worth spending time on type theory?
@ZhenLin So HoTT book aims to create an intuitive approach to intensional type theory? Ok, thanks for the correction, I'm gonna edit.
Nov
14
comment Why is it worth spending time on type theory?
@ZhenLin when I say new I was referring to the naive type theory, I don't know of any reference before HoTT book which treat type theory from a non formal point of view.
Nov
14
revised Why is it worth spending time on type theory?
added 1002 characters in body
Nov
14
answered Why is it worth spending time on type theory?
Nov
14
revised $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
added 183 characters in body
Nov
14
comment $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
@AndreasCaranti no, I'm not assuming that the group have finite order. Apparently I've assumed that the elements have finite order. I'm gonna edit the answer and add this detail. Thanks for pointing out.
Nov
14
comment $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
@some1.new4u if you tell me exactly what are the instrument you could use, I could try to come up with some simpler solution :)
Nov
14
comment $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
@some1.new4u it's possible, anyway till now I haven't come up with nothing simpler. Btw could you tell me which book are you referring to?
Nov
14
comment $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
@vadim123 thanks for pointing out my mistake :)
Nov
14
comment $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
@some1.new4u I was confused, now I've edited :)
Nov
14
revised $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
corrected answer
Nov
14
comment $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
@vadim123 My bad I've misread the question. I'm gonna edit soon.
Nov
14
answered $G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$
Nov
13
comment Without using Sylow: Group of order 28 has a normal subgroup of order 7
@azimut Ah, ok now I get it. Well I'd rather be complete :). Anyway thanks.
Nov
13
comment Without using Sylow: Group of order 28 has a normal subgroup of order 7
@azimut what's more complicated? I've just proved that a subgroup which is the only one of its order is characteristic and that a characteristic subgroup is normal. It's really that bad?