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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 6 months |
| seen | yesterday | |
| stats | profile views | 36 |
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May 13 |
comment |
Associativity with one operation or two (or more) operations @Martin It's been a while since I asked this question, but I've recently seen this 'associativity' described as the 'morphism property' - is this something you recognise? |
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May 13 |
accepted | Are algebraic numbers analogous to group elements with finite order? |
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May 12 |
accepted | $P \cap N$ is a Sylow $p$-subgroup of $N$, where $N$ is normal in $G$ and $P$ is a Sylow $p$-subgroup of $G$? |
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May 11 |
comment |
$P \cap N$ is a Sylow $p$-subgroup of $N$, where $N$ is normal in $G$ and $P$ is a Sylow $p$-subgroup of $G$? You could be right there, I don't recall that it does give a definition in the case where $p$ does not divide the order of $G$. |
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May 11 |
comment |
$P \cap N$ is a Sylow $p$-subgroup of $N$, where $N$ is normal in $G$ and $P$ is a Sylow $p$-subgroup of $G$? Nice to know I'm not seeing things! Is it really a non-standard definition though? For example Wikipedia has a page on the Sylow theorems which also states $n>0$. |
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May 11 |
comment |
$P \cap N$ is a Sylow $p$-subgroup of $N$, where $N$ is normal in $G$ and $P$ is a Sylow $p$-subgroup of $G$? Yes this is my point, all sources I have looked at (including that book) define a Sylow p-subgroup of a finite group with order $p^{n}m$ where $n>0$ and $m$ is not divisible by $p$. So by this definition $\{1\}$ is not a Sylow p-subgroup. |
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May 11 |
asked | $P \cap N$ is a Sylow $p$-subgroup of $N$, where $N$ is normal in $G$ and $P$ is a Sylow $p$-subgroup of $G$? |
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May 7 |
awarded | Caucus |
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May 5 |
comment |
Are algebraic numbers analogous to group elements with finite order? Thanks for that @Jack |
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May 5 |
comment |
Are algebraic numbers analogous to group elements with finite order? Unfortunately this answer is way above my head! |
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May 5 |
revised |
Are algebraic numbers analogous to group elements with finite order? deleted 11 characters in body |
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May 3 |
accepted | Intersection of conjugate subgroups is normal |
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May 3 |
comment |
Intersection of conjugate subgroups is normal @julien If you write your element-free argument as an answer I would be happy to accept. |
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May 3 |
revised |
Are algebraic numbers analogous to group elements with finite order? edited tags |
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May 3 |
comment |
Are algebraic numbers analogous to group elements with finite order? @Easy I don't think I mean to consider algebraic numbers as a group, as this again restricts us to one operation. I just meant is there a connection (maybe just superficial?) between elements of finite order in groups (e.g. $g^n = 1$ for some $g \in G$, $n \in \mathbb{Z}^+$) and algebraic numbers e.g. $\alpha \in {\mathbb{Q}}(\alpha)$ such that there is a finite sum ${\alpha}^n + a_{n-1}{\alpha}^{n-1} + \cdots + a_0 = 0$. |
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May 3 |
comment |
Are algebraic numbers analogous to group elements with finite order? @Easy Hi Easy, I had not come across the notion of 'torsion' yet, but I can read up about it now that you've mentioned it! (Having looked on Wikipedia it seems this is more advanced than I have covered so far.) |
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May 3 |
comment |
Are algebraic numbers analogous to group elements with finite order? @Glen O I'm not quite sure, maybe there is a more appropriate word, I just thought they are sort of similar ideas. There might not be any link of course, I am relatively new to algebra! |
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May 3 |
asked | Are algebraic numbers analogous to group elements with finite order? |
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May 2 |
comment |
Why can't we just say 1 instead of “unity”? A unital ring (with invertible elements) is a ring with unity anyway so at least there is no ambiguity for the term 'unital ring', whatever you mean by 'unit'. |
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May 2 |
accepted | Way to show $x^n + y^n = z^n$ factorises as $(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n$ |
