| bio | website | |
|---|---|---|
| location | UK | |
| age | ||
| visits | member for | 2 years |
| seen | 2 hours ago | |
| stats | profile views | 833 |
I'd rather walk than take a taxi.
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Examples of non-cyclic group with a cyclic automorphism group @JackSchmidt Oh. Whoopse! |
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Examples of non-cyclic group with a cyclic automorphism group ...and if $G$ is the Klein $4$-group?... |
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Subgroups of $\mathbb{Z}^k$ of finite index $n$ Ah, it is obvious now! |
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Subgroups of $\mathbb{Z}^k$ of finite index $n$ What do you mean by $\mathbb{Z}^k/(n\mathbb{Z})^k$? |
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Subgroups of $\mathbb{Z}^k$ of finite index $n$ @DerekAllums: I have neither of these books! I presume the operation has to be addition, as if $a$ and $d$ have absolute value other than $1$ then the matrices are not invertible (in $GL_2(\mathbb{Z})$), but then this looks like $\mathbb{Z}^3$ rather than $\mathbb{Z}^2$... |
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Subgroups of $\mathbb{Z}^k$ of finite index $n$ Also, my personal approach would be to think about how $\mathbb{Z}^n$ can map onto some abelian group $A$ of order $m$. This gives a number of subgroups of index $m$. Do this for all abelian groups of order $m$. For example, if $\langle a, b; [a, b]\rangle\rightarrow K_4$ the Klein $4$-group then $a$ and $b$ must both have order two. This gives you a unique kernel, $N=\langle a^2, b^2\rangle$. |
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Subgroups of $\mathbb{Z}^k$ of finite index $n$ I think I am not getting something - how are you viewing $\mathbb{Z}^n$ as matrices? |
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reviewed | Close Cost and Marginal cost |
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Cost and Marginal cost I am uncomfortable with a) the overly-long list of things you want us to find, and b) your offer of money. The latter point makes me wonder if this is some sort of assessed question, while the former point perhaps demonstrates that you have not tried anything. Combined, it just seems like you want someone to do your thinking for you. So, I would recommend removing the offer of money and I also telling us what your thoughts are on, say, revenue and profit functions. In the mean time, I am voting to close. Sorry. (Although I will vote to re-open once these things have been addressed.) |
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1d |
reviewed | Reviewed Value of series, Partialsum? |
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Value of series, Partialsum? Are you suggesting that we go $\frac{2n+1}{n^2(n+1)^2}=\frac{2(n+1)-1}{n^2(n+1)^2}=\frac{2-1}{n^2(n+1)}=\frac{1}{n^2(n+1)}$?...! |
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reviewed | No Action Needed $\lim_{x\to 0} \frac{\sin(t\sqrt{x^2-k^2})}{\sqrt{x^2-k^2}}=?$ |
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When does the isomorphism $G\simeq ker(\phi)\times im(\phi)$? hold? (The key word being split.) |
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When does the isomorphism $G\simeq ker(\phi)\times im(\phi)$? hold? @TobiasKildetoft Ah, right, gotcha! |
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When does the isomorphism $G\simeq ker(\phi)\times im(\phi)$? hold? @TobiasKildetoft: I do not think that even that is enough! My point is that in infinite groups the image can contain an isomorphic copy of itself which can be normal in the whole group without the kernel itself being normal. I do not think generation should be too difficult to get around. |
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When does the isomorphism $G\simeq ker(\phi)\times im(\phi)$? hold? (Counter-example: Take the group $\langle a, t; a^t=a^{-1}\rangle$, then this is not a cross-product. Take the map which kills $a$ in the presentation (and so maps onto $\mathbb{Z}$). Then note that $\langle t^2\rangle$ is normal and intersects trivially with the kernel of this map.) |
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When does the isomorphism $G\simeq ker(\phi)\times im(\phi)$? hold? @TobiasKildetoft The conclusion does not necessarily follow if $G$ is infinite... |
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2d |
awarded | Convention |
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Geometric question? @ZettaSuro: I think it is better if you explain to the OP why you think the "thank you so much" should be removed and ask them to remove it themselves, rather than removing it yourself. |
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reviewed | Reject suggested edit on Geometric question? |

