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### 73 Visitor Messages

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

As far as I'm concerned, you are welcome to post such information in any thread no matter how old it is.
4. Hey, way to go!
5. 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)
6. Done. Let me know if you need to edit more things in the future.
7. 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...
8. You should be able to edit it now. Let me know if there is a problem.
9. Hm, I'm interested in sending you a private message. Can you open up one space ?
10. 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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#### Basic Information

Date of Birth
January 12, 2000 (18)
Biography:
I know a thing or two about topology. Find number theory interesting but don't know much about it.
Location:
West Bengal, India
Interests:
Number Theory
Country Flag:
India

#### Signature

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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##### Visitor Messages
Total Messages
73
Most Recent Message
June 6th, 2018 01:03
##### General Information
Last Activity
June 6th, 2018 01:06
Last Visit
June 6th, 2018 at 01:06
Last Post
January 1st, 2017 at 22:39
Join Date
March 22nd, 2013
Referrer
MarkFL
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2
Referred Members
neelmodi, zuby

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