# JJR

less info
reputation
17
bio website location age 20 member for 10 months seen 10 hours ago profile views 190

2nd year Math student

Avid tennis player

 Jan9 revised Bijection between permissible cycle types and conjugacy classesdeleted 68 characters in body Jan9 asked Bijection between permissible cycle types and conjugacy classes Dec8 accepted For $G$ a group, if $|G| = n$, $n$ composite, will $G$ have a proper subgroup? Dec4 comment For $G$ a group, if $|G| = n$, $n$ composite, will $G$ have a proper subgroup?How can I see this is true without Cauchy's Theorem. For example, how would I know that a group of even order must have an element of order 2? Dec4 asked For $G$ a group, if $|G| = n$, $n$ composite, will $G$ have a proper subgroup? Dec4 comment Is the conjugation map always an isomorphism?Thanks for the helpful answer and comments! Dec3 accepted Is the conjugation map always an isomorphism? Dec3 asked Is the conjugation map always an isomorphism? Nov28 accepted How to prove that $\mathrm{GL}_2(\Bbb Z_2)$ has only six subgroups Nov28 comment How to prove that $\mathrm{GL}_2(\Bbb Z_2)$ has only six subgroupsAha! I knew I had forgotten that we knew that subgroups of order 2 or 3 must be cyclic. Nov28 answered What are the prerequisites for Stochastic Processes Nov28 asked How to prove that $\mathrm{GL}_2(\Bbb Z_2)$ has only six subgroups Nov26 accepted Cardinality of the Union is less than the cardinality of the Cartesian product Nov26 comment Cardinality of the Union is less than the cardinality of the Cartesian productThanks Clive, appreciated Nov26 revised Cardinality of the Union is less than the cardinality of the Cartesian productdeleted 1 characters in body Nov26 asked Cardinality of the Union is less than the cardinality of the Cartesian product Nov25 accepted Trouble with this multiplication table Nov25 comment Trouble with this multiplication tableSo it is $S_3$. Now to go through and identify each element with the cycle. Is there a better way to do this? I feel like it should be possible in some other manner? Nov25 comment Trouble with this multiplication tableNote: I also know that since $df = a = fb$, this table is not abelian. Nov25 asked Trouble with this multiplication table