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    Welcome to the team of math helpers
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    Haha, that would be Mark. I like the way sounds, don't you?
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    Hey Balarka,

    As far as I'm concerned, you are welcome to post such information in any thread no matter how old it is.
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    Hey, way to go!
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    Hey Balarka, thanks, glad you are enjoying the challenge! As for the origin of this particular problem, I'm not sure, I found it in https://www.math.muni.cz/~bulik/vyuka/pen-20070711.pdf (page 67, no solutions) and thought it was interesting. It only says "Japan 1990" so I'm guessing it's from a contest of some sort, yes (the second part of the challenge is a custom addition of mine).

    Your proof appears to be correct although I will need some time to look at it in more detail. My solution approaches the problem in a similar way as yours, but proceeds somewhat differently, though I have not fully checked mine either (I will post it in a week as per the forum rules, as well as a reference solution if I can find one)
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    Done. Let me know if you need to edit more things in the future.
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    Hey Balarka,

    topsquark was purposely given a wrong number and he wanted me to spot for the error, which it turns out, I was wrong. I guess I just am not good at math. Hehehe...
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    You should be able to edit it now. Let me know if there is a problem.
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    Hm, I'm interested in sending you a private message. Can you open up one space ?
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    I haven't had time to play with it, but it *is* interesting. The failure of right-exactness is what interests me: I suspect this is akin to why in groups left-split implies right-split, but not vice versa. So I would be tempted to investigate what a sufficient condition on $G$ would be so that the map: $Aut_N(G) \to Aut_N(H)$ is onto.

    I feel the center of $G$ will be relevant, here, so I would look at the two extremes: $G$ abelian, and $G$ with trivial center. I think outer automorphisms will be troublesome. On the other hand if $H$ is an extension of $N$, and $G$ is an extension of $H$, this might be fruitful (again, mimicking the field construction).

    For "concrete categories", for an object $A$ we can always form the group $Aut(A)$ of all isomorphisms $A \to A$. I can't say right off-hand if we can do so for ANY category, but this might be true (if our category is a poset, the automorphism groups are all trivial). So "some" generalization is possible, I'm not sure "how wide".
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About mathbalarka

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Date of Birth
January 12, 2000 (19)
About mathbalarka
I know a thing or two about topology. Find number theory interesting but don't know much about it.
West Bengal, India
Number Theory
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Some of my notes on number theoretic topics are on a few primality tests and on Hardy & Littlewood's result (incomplete). The other articles are on quintics : about a brief description of Kiepert algorithm and a short introduction on quintic-solving algorithms, respectively. I have also written up an introductory thread on Riemann Hypothesis


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Name: Graduate POTW Award (Jul-Dec 2013)
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Math Help Boards