JJR

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2nd year Math student

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Mahler aficionado


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Jan
9
 revised Bijection between permissible cycle types and conjugacy classes
deleted 68 characters in body
Jan
9
 asked Bijection between permissible cycle types and conjugacy classes
Dec
8
 accepted For $G$ a group, if $|G| = n$, $n$ composite, will $G$ have a proper subgroup?
Dec
4
 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?
Dec
4
 asked For $G$ a group, if $|G| = n$, $n$ composite, will $G$ have a proper subgroup?
Dec
4
 comment Is the conjugation map always an isomorphism?
Thanks for the helpful answer and comments!
Dec
3
 accepted Is the conjugation map always an isomorphism?
Dec
3
 asked Is the conjugation map always an isomorphism?
Nov
28
 accepted How to prove that $\mathrm{GL}_2(\Bbb Z_2)$ has only six subgroups
Nov
28
 comment How to prove that $\mathrm{GL}_2(\Bbb Z_2)$ has only six subgroups
Aha! I knew I had forgotten that we knew that subgroups of order 2 or 3 must be cyclic.
Nov
28
 answered What are the prerequisites for Stochastic Processes
Nov
28
 asked How to prove that $\mathrm{GL}_2(\Bbb Z_2)$ has only six subgroups
Nov
26
 accepted Cardinality of the Union is less than the cardinality of the Cartesian product
Nov
26
 comment Cardinality of the Union is less than the cardinality of the Cartesian product
Thanks Clive, appreciated
Nov
26
 revised Cardinality of the Union is less than the cardinality of the Cartesian product
deleted 1 characters in body
Nov
26
 asked Cardinality of the Union is less than the cardinality of the Cartesian product
Nov
25
 accepted Trouble with this multiplication table
Nov
25
 comment Trouble with this multiplication table
So 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?
Nov
25
 comment Trouble with this multiplication table
Note: I also know that since $df = a = fb$, this table is not abelian.
Nov
25
 asked Trouble with this multiplication table
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