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| bio | website | ms.uky.edu/~jack |
|---|---|---|
| location | Lexington, KY | |
| age | ||
| visits | member for | 2 years, 9 months |
| seen | 14 hours ago | |
| stats | profile views | 2,168 |
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18h |
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Examples of non-cyclic group with a cyclic automorphism group I would like a torsion-free example $G$ that does not embed in $\mathbb{Q}$. |
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19h |
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Examples of non-cyclic group with a cyclic automorphism group I think switching $2$ and $3$ in the prime factorization of $a/b$ is an automorphism. |
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19h |
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Examples of non-cyclic group with a cyclic automorphism group (@Dominic: It wouldn't hurt to put in a little more of the proof sketch in the finite case, as the automorphism group of a subgroup and/or quotient group can be much larger than the automorphism group of the original.) |
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19h |
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Examples of non-cyclic group with a cyclic automorphism group @user1729: automorphism group is GL(2,2) = S3, nonabelian |
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19h |
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Examples of non-cyclic group with a cyclic automorphism group There are uncountably many non-isomorphic but really-very-similar examples: for instance the "cube free numbers" consisting of all rational numbers whose denominator is not divisible by the cube of any prime. Again the endomorphism ring is the largest subring contained in it ($\mathbb{Z}$), and the automorphism group is the group of units ($\pm1$). |
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20h |
asked | Partial cycles in projective resolutions of square-free algebra |
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22h |
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Are the quotient groups in a composition sequence necessarily subgroups? Actually quasi-simples are only one kind of example. A6 N 3^6 (the non-split extension of A6 by a 3-dimensional GF(9) module; perfect of order 262440) is an example that contains no non-simple quasi-simple subquotients. I think quasi-simples are still good examples, but they aren't the whole story at all. |
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22h |
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a flatness criterion There is a superficial problem that not all free $R$-modules have finite rank, but noticing this is only superficial is part of the exercise. A method avoiding Tor but using direct limits is in Lam's Lectures on Modules and Rings, 4.12 page 125. It shows how to view this as the dual to Baer's criterion for injective modules. |
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23h |
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Are the quotient groups in a composition sequence necessarily subgroups? @TobiasKildetoft: Correct. The answer to your question is no, but I'm not 100% on whether $N$ has to have abelian composition factors. If both $G/N$ and $N$ are non-abelian simple, then $G \cong G/N \times N$. Let me know if you want the proof (not hard, but it does use the CFSG, at least Schreier's conjecture). |
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23h |
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Are the quotient groups in a composition sequence necessarily subgroups? If $N$ is required to be simple instead of $G/N$, then the quaternion group of order 8 is a nice counterexample. |
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23h |
answered | Are the quotient groups in a composition sequence necessarily subgroups? |
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1d |
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Big Greeks and commutation In Carter's Simple Groups books he explicitly mentions that it does not matter (for his purposes) what order is taken. This might be because this is important in group theory, but it might aso be because he did not believe there was a completely standard order. |
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1d |
answered | A group with six elements which are given partially by relations. |
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1d |
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A group with six elements which are given partially by relations. @Stefan: you are correct that a list of expressions for the group elements does not typically determine the group. One also needs the relations that simplify products of those group elements. Not only do you need to specify $x^3$, but also $y^2$ and $yx$. Even if $x^3=1$, you need not get $S_3$. |
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1d |
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Suppose that $G$ is nonabelian. Must $|\mathrm{Out}(G)| = |\mathrm{Aut}(G)|/|\mathrm{Inn}(G)|$? It might be a good idea to include the count of outer automorphisms $\{ \alpha \in \operatorname{Aut}(G) : \alpha \text{ is outer } \} = \operatorname{Aut}(G) - \operatorname{Inn}(G)$ has size $|\operatorname{Aut}(G)| - |\operatorname{Inn}(G)|$ |
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2d |
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Ulm and Frattini Subgroups That is exactly what you are doing, so presumably it is true. :-) I think your pq equality is the key. I'll think about why it is true. |
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2d |
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Ulm and Frattini Subgroups Are you sure it is true? It seems like $\Phi(A)=0$ is not very hard to happen, so I worry that $\Phi^\omega(A)$ might actually be pretty small. I didn't study torsion-free groups well enough, but it seems like if some exts don't vanish, it should be easy to get $U(A) = \Phi(A) \neq \Phi^2(A)$. |
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2d |
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Finite group is generated by a set of representatives of conjugacy classes. No, the union could be bigger than $H$, but it can never be exactly equal to $G$. For instance in the alternating group on 4 points, take $H$ to be of order 2. Then $\cup_{g \in G} gHg^{-1}$ is the Sylow 2-subgroup of $G$, but it is not all of $G$. The real reason why it is not equal to $G$ is given in Hagen's answer. I also explain it here using the same argument. |
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2d |
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Finite group is generated by a set of representatives of conjugacy classes. add proof with full details |
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2d |
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Finite group is generated by a set of representatives of conjugacy classes. it means the conjugacy class containing $x$. definition added to answer |
