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Algebra II

Post Icon Posted: Submitted by richardhp on 4 June 2008 - 11:37am.

Joined: 2007-10-01
Posts: 178

Does anyone know where I can get answers to the ma242 04/05 paper? doesn't seem to be on mathstuff anywhere.

‹ Week of hell Algebra 2: Colourings ›
Post Icon Posted: 5 June 2008 - 8:14am

Joined: 2007-10-01
Posts: 178

never mind, can someone at least give me a hint to this interesting problem:

G simple and H proper subgroup, prove |G| divides |G:H|!
Post Icon Posted: 5 June 2008 - 10:19am

Joined: 2006-10-10
Posts: 520

Hmm... Well, a sufficient condition of that is the order of G being less than or equal to the number of cosets of G in H

Since G is simple, then all proper subgroups are non-normal, right? Dunno if that helps with calculations.
Post Icon Posted: 5 June 2008 - 10:34am

Joined: 2006-10-05
Posts: 697

I don't know what simple is (I'm on p7 of revision guide) but you can easily prove |G|=|G:H||G/H|$ \in\mathbb{R} $ for finite groups, then divide through. If not already assumed, use the simpleness to prove it's finite.

I have yet to find something non-trivial in the revision guide. I was worried at 20 pages long but the first half being blindingly easy makes sense.

Edit: Unrelated but is divisibility defined anywhere for an infinite number? Maybe defined w.r.t a sequence (or series) that tends to infinity. Something like if( there exist N s.t. $ \forall n \geq N $, d|$ x_n $ then d|$ (lim_{n\rightarrow\infty}x_n) $?
Post Icon Posted: 5 June 2008 - 10:49am

Joined: 2006-10-10
Posts: 520

A simple group means that its only normal subgroups are itself and {e}. I presume you have to use that fact somewhere in the proof?

And, as a guess, I'd imagine that the sensible way of defining division at infinity would be to let every number divide it? Not sure... I mean, this is one of the problems of thinking of infinity as a number, because there are multiple ways of constructing it. Infinity could be thought of as 2*2*2*2*2*.... in which case 3 doesnt define inf, or as 2*2*2*2*....*3*3*3*3*....*5*5*5*5*....
Post Icon Posted: 5 June 2008 - 10:55am

Joined: 2007-10-01
Posts: 178

the group G is finite, and you have to show |G| divides |G:H|! i.e. the factorial of the number of co-sets.
Post Icon Posted: 5 June 2008 - 12:16pm

Joined: 2007-10-03
Posts: 397

How about this :
-- Warning : Spoilers --
Let $ |G:H|=n $, you have a homomorphism $ \phi : G \rightarrow \mathrm{Sym}(n) $ given by the action (as a permutation) of the elements $ g $ on the set of cosets of $ H $ given by $ \sigma_{g_1}(g_2H) = g_1g_2H $
Suppose $ H \not=G $ as the result doesn't hold in that case.
If $ H={e} $ then the result is obviously true.
Otherwise, as $ G $ is simple, H is normal so no two elements of $ G $ can be mapped to the same element of $ \mathrm{Sym}(n) $ (if they were, you'd be able to quotient out as you would have a non trivial kernel, contradicting the fact that G is simple) (so $ \phi $ is injective), and because the image of $ \phi $ forms a subgoup of $ \mathrm{Sym}(n) $, by Lagrange this image must divide the order of $ \mathrm{Sym}(n) $, which is n!.
Therefore $ |G| $ divides $ |G:H|! $
-- Warning : End of spoilers--

Spoiler warnings courtesy of Colin.
Post Icon Posted: 5 June 2008 - 12:18pm

Joined: 2007-10-17
Posts: 109

Hmm... Well, a sufficient condition of that is the order of G being less than or equal to the number of cosets of G in H

I thought that was impossible, by Lagrange's. Alex gives what's probably meant to be Lagrange's Theorem, but I'm not sure what he's written is correct. As for the problem, I'm thinking.

Edit: Ah yes, I was trying to come up with something like that. Sam, you should bear in mind that hints are sometimes better than answers.
Post Icon Posted: 5 June 2008 - 12:23pm

Joined: 2007-10-03
Posts: 397

I edited. Any better ?
Post Icon Posted: 5 June 2008 - 12:29pm

Joined: 2006-10-10
Posts: 520

LMFBAO!!!!!
Post Icon Posted: 5 June 2008 - 2:04pm

Joined: 2006-11-02
Posts: 1082

Unrelated but is divisibility defined anywhere for an infinite number? Maybe defined w.r.t a sequence (or series) that tends to infinity. Something like if( there exist N s.t. $ \forall n \geq N $, d|$ x_n $ then d|$ (lim_{n\rightarrow\infty}x_n) $?

What is the actual context of your question (i.e. why would you need such a concept?)? If you're talking about the integers and divisibility is the normal concept then no because there are no "infinite numbers" and limits of integer sequences tend to either not exist or be trivial (i.e. a tail of the sequence is going to be $ (\dotsc, \ell,\ell,\ell,\ell,\ell,\dotsc $) for the limit $ \ell \in \mathsbb{Z} $.

Alex gives what's probably meant to be Lagrange's Theorem, but I'm not sure what he's written is correct. As for the problem, I'm thinking.

Well, since $ |G/H| = |G:H| $ it implies that all groups with a subgroup $ H $ such that $ |H| \geq 2 $ have square order, which obviously isn't true (not too sure what the $ \in \mathbb{R} $ is doing there). He probably meant to say that $ |G:H| = |G|/|H| $.

I edited. Any better ?

No, you should make the text hidden until you click on "Warning : Spoilers". :P
Post Icon Posted: 5 June 2008 - 2:11pm

Joined: 2006-10-10
Posts: 520

(...,nyuuki,nyuuki,nyuuki,nyuuki,nyuuki,...)
Post Icon Posted: 5 June 2008 - 7:16pm

Joined: 2006-10-05
Posts: 697

Yeah, I may have got that confused along the way. I learnt it 1 day ago tho.
Post Icon Posted: 7 June 2008 - 6:38pm

Joined: 2007-10-01
Posts: 178

One question states:

The symmetric group $ S_6\, $ has exactly 12 distinct conjugacy classes, true or false. The answer says true listing the classes as (in cycle-type notation):

$ 6, 5, 4, 4+2, 3^2, 3, 3+2^2, 3+2, 2^3, 2^2, 2, 1 $

I am confused however as to why $ 3+2^2\, $ is a cycle type since this would have 7 elements and therefore not be in $ S_6 $ or am I missing the point somehow?
Post Icon Posted: 7 June 2008 - 7:19pm

Joined: 2006-10-10
Posts: 520

Don't worry about that, I'm pretty certain it's a mistake. There are a few crossing outs on that paper which make me think whoever wrote them was using answers from another paper as inspiration. So he/she probably got confused and worked in S7 for a bit. There's only 11 cycle types in S6
Post Icon Posted: 7 June 2008 - 7:31pm

Joined: 2007-10-03
Posts: 397

Yes :

$ \begin{array}{rcl} 6 & (123456) & 120 \\ 5+1 & (12345) & 144 \\ 4+2 & (1234)(56) & 90 \\ 4+1+1 & (1234) & 90 \\ 3+3 & (123)(456) & 40 \\ 3+2+1 & (123)(45) & 120 \\ 3+1+1+1 & (123) & 40 \\ 2+2+2 & (12)(34)(56) & 15 \\ 2+2+1+1 & (12)(34) & 45 \\ 2+1+1+1+1 & (12) & 15 \\ 1+1+1+1+1+1 & () & 1 \\ \end{array} $

$ 120+144+90+90+40+120+40+15+45+15+1=720=6! $
Post Icon Posted: 7 June 2008 - 7:55pm

Joined: 2007-10-01
Posts: 178

that's a relief

EDIT: Another Question!!!

06 paper last part of question 2, it asks:

Let n = 118 and m = 2006. How many of the homomorphisms from Cn to Cm are injective?
(Hint: compute how many homomorphisms have kernel C118 C2 or C59)

In the answers it basically says there is one with kernel C118, 59 with kernel C2 and 2 with kernel C59 which I do not understand but am willing to accept. Then as if by magic they deduce that the number of injective homomorphisms is 118 - 1 - 2 - 59 = 58, is this something we ought to know?
Post Icon Posted: 7 June 2008 - 9:30pm

Joined: 2006-10-05
Posts: 697

What I just put is wrong because of a basic mathematical error. n divides m.

Piss. F***ing basic devision.
Post Icon Posted: 7 June 2008 - 9:51pm

Joined: 2006-11-02
Posts: 1082

Corrected :p

The kernel needs to be a normal subgroup of the domain of your homomorphism, so the only choices for $ C_{118} $ are $ C_{118} $, $ C_{59} $, $ C_2 $ and $ \{1\} $ (since $ 118 = 2\cdot 59 $ and 59 is prime). Also, since $ 2006 = 2\cdot 17 \cdot 59 $, the possible images will be isomorphic to one of $ C_{118}, C_{56}, C_{2} $ or $ \{1\} $, which you can easily check with the 1st IT. Homomorphisms from a cyclic group are uniquely determined by the image of a generator $ g $ and your image will be generated by $ \varphi(g) $ and, since $ \varphi(g^n) = \varphi(g)^n $, the order of $ \varphi(g) $ divides the order of $ g $. Supposing $ b $ is a generator of $ C_{2006} $ and $ a := b^{17} $ (which has order $ 118 $), you've got the following choices:

  • The one with kernel $ C_{118} $ is the one which maps $ g $ to $ {1} $,
  • The one with kernel $ C_{59} $ are those which map $ g $ to $ a^{59} $ (the only element of order 2),
  • The 58 with kernel $ C_{2} $ are those which map $ g $ to one of $ a^{2k} $ for $ 1 \leqslant k \leqslant 58 $ (the elements of order 59),
  • The rest have kernel $ {1} $ and are hence injective.

There are in 118 homomorphisms in all because you need to map $ g $ to one of the $ b^{17k} = a^k $ for $ 1 \leqslant k \leqslant 118 $ (these are all the elements of $ C_{2006} $ of order 118 or a divisor of it and they're all distinct). So all you need to do now is subtract: 118 - 1 - 1 - 58 = 58.

Edit: There's an easier way of finding the answer, considering that the injective homs are those which map $ g $ to an element of order $ 118 $ (because all the powers $ \varphi(g) $ will be distinct). These are exactly the $ a^k $ for which $ k $ is coprime to $ 118 $ and there are $ \phi(118) = 58 $ of these (where $ \phi $ is Euler's totient function)...
Post Icon Posted: 7 June 2008 - 10:14pm

Joined: 2006-10-10
Posts: 520

So you actually needed to know the prime decomposition of 2006? When I did this paper I couldn't do it without a calculator.

2008 = 2^3*251

Probably worth remembering this for Tuesday
Post Icon Posted: 7 June 2008 - 10:16pm

Joined: 2007-10-03
Posts: 397

Reminder : 118 - 1 - 2 - 59 = 56

The correction to Cosmin's proof goes :

  • One element of order 1 that is $ a^{118} $
  • One element of order 2, that is $ a^{59} = a^{-59} $
  • 58 elements of order 59, that is $ a^{2k} \text{ for } 1 \le k \le 58 $

Hence we get 118 - 1 - 1 - 58 = 58, which is the expected answer.
Post Icon Posted: 7 June 2008 - 10:39pm

Joined: 2006-10-05
Posts: 697

The correction to Sam's correction goes

  • One element of order $ 1 $ that is $ 1=a^{118} $
Post Icon Posted: 7 June 2008 - 10:43pm

Joined: 2006-10-10
Posts: 520

The meta-correction to Alex's correction about Sam's correction;

Sam wasn't wrong, so your correction is actually an annotation
Post Icon Posted: 7 June 2008 - 10:43pm

Joined: 2007-10-03
Posts: 397

The correction to Alex's correction of Sam's correction goes

  • One element of order $ 1 $ that is $ 1 = a^{118} = a^{2 725 644 149 258} $
Post Icon Posted: 7 June 2008 - 11:02pm

Joined: 2006-10-10
Posts: 520

http://objection.mrdictionary.net/go.php?n=2615512
Post Icon Posted: 7 June 2008 - 11:16pm

Joined: 2006-10-05
Posts: 697

Awesome. (Is actually not a rickroll, I very rarely click on any link from this website because 11.3% of them are rickrolls, but I'm actually glad I did this one)
Post Icon Posted: 7 June 2008 - 11:19pm

Joined: 2007-10-03
Posts: 397

Is 11.3% the proportion of posts on this website that were written by Cosmin ?
Post Icon Posted: 7 June 2008 - 11:21pm

Joined: 2006-10-05
Posts: 697

In fairness I made that number up. But I think I'll work out an approximation of how many cosmin has written, i think it will be much less.
Post Icon Posted: 7 June 2008 - 11:25pm

Joined: 2006-11-02
Posts: 1082

The correction to Alex's post goes:

In fairness I made that number up. But I think I'll work out an approximation of how many cosmin has written, i think it will be much less.
Post Icon Posted: 7 June 2008 - 11:27pm

Joined: 2007-10-03
Posts: 397

The correction to Cosmin's correction to Alex's post goes :

In fairness I made that number up. But I think I'll work out an approximation of how many cosmins were written, i think it will be much less.
Post Icon Posted: 7 June 2008 - 11:32pm

Joined: 2006-10-05
Posts: 697

Cosmin accounts for 19.7% of the posts on the forums, not including the groups i'm not a part of. Of this moment of course.

Edit: Sam: I make the no. of posts as 4606 in total, could you make me a nice pie chart? Since I haven't got excel or word or anything that could do one.
Post Icon Posted: 8 June 2008 - 12:04am

Joined: 2007-10-03
Posts: 397

Post Icon Posted: 8 June 2008 - 12:09am

Joined: 2006-11-02
Posts: 1082

The problem with that is that everything you do on the website counts as a post and so the percentages are false (for instance if you buy a hoodie I think it shows up as one post, same for a lot of website stuff like front page news and events, etc.). For some the percentage is going to be close but for others it isn't. I guess Gary might be able to do a good pie chart with his data. :P
Post Icon Posted: 9 June 2008 - 2:33pm

Joined: 2006-08-31
Posts: 696

That and every thread goes very off-topic and ends up being lots of one word or one sentence replies that have little or no content.

No?
Post Icon Posted: 9 June 2008 - 4:26pm

Joined: 2006-11-02
Posts: 1082

No.
Post Icon Posted: 9 June 2008 - 4:31pm

Joined: 2006-10-05
Posts: 697

If it doesn't go massively off topic tho the thread will die. That and cosmin would get a new hobby.
Post Icon Posted: 9 June 2008 - 4:40pm

Joined: 2006-11-02
Posts: 1082

I could point out that your post made it go off topic. :P
Post Icon Posted: 9 June 2008 - 7:23pm

Joined: 2006-08-31
Posts: 696

I'd start a thread to complain, but then it would easily be the longest on the forums. Besides, it was already off-topic, and your comments have further pushed it from original topic too.
Post Icon Posted: 9 June 2008 - 7:55pm

Joined: 2007-02-14
Posts: 105

Surely the second post in the thread was the one that made it go off-topic?
Post Icon Posted: 9 June 2008 - 7:58pm

Joined: 2006-11-02
Posts: 1082

Besides, it was already off-topic, and your comments have further pushed it from original topic too.

I was talking to Alex when I said that, refering to his correction of Sam's post.
Post Icon Posted: 9 June 2008 - 8:02pm

Joined: 2006-10-05
Posts: 697

I attempted to unpublish mine and everyone else's digression, but 2 exec members who are very guessable conspired to revive them. I tried to correct my mistake but

$$Cos(min(Maths,Society))\rightarrow\; Anarchy$$

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