Basic abstract algebra, pt.9

[Foamykhan]

I just love that infected mushroom is accompanying the vid :D

[1990marwan]

thats good :)))

[cmonington]

3 out of 2 people have problems with fractions..haha that's funny!

[UndeadTheta]

wow this blew my mind :) thank you!

[Ezzyman17]

Thank you, these videos really motivates me. Keep it coming.

[369franklin]

shouldnt it be....

there are three types of mathematicians

those that can count and those that can not.

[VeritySeeker]

@369franklin

lol, yes...

[369franklin]

@VeritySeeker

ha i thout it was a bit obvious for you to state!.. i ejoyed the fields arranged by purity too!! cant member what video it was in but found it v amusing!. the vidoes are really good! thankyou for taken your time to do them im sure they have helped loads of people as they helped confirm alot for me :)

[wintroller]

@369franklin

I don't get it...

[luayabd]

thank you very much

[danielkim116]

thank you!

[magestaff567]

Thanks for these videos. Very enlightening. If you went to my school I would buy you a beer or whatever is appropriate to thank you.

[luvu4evabby]

hey thanks for doin this!

[iyottube013]

hi, could you explain to get the KERNEL of a group? I am always confused with the definition.

and also the IMAGE. thanks..

please give examples..

[VeritySeeker]

Hi, a group does not have a kernel or an image. But a homomorphism does. A homomorphism is explained in the videos, and it is a function between two groups, say f: G -> H.

The kernel of this function is all elements in G that goes to the identity in H. The image are all elements in H that is hit by some element from G through f.

[VeritySeeker]

Take the homomorphism f: Z --> Z, where Z is the integers and f(a) = 2*a. The kernel is all elements in Z such that f(a)=0, and in this case it is only a=0. The image is all even numbers.

[firstevidentenigma]

@VeritySeeker ker f can be a singleton set. ie {e1}.

[VeritySeeker]

@firstevidentenigma Yes, absolutely. Did I write something else somewhere? If I did, it was not correct.

[firstevidentenigma]

@VeritySeeker No you didn't, but it seems as if it needed emphasis, is all.

[FSMGauss]

at 1:44 you say that H is a subgroup in G if for all a and b, ab^-1 is in H.

is this just stated as obvious, or is the proof coming later, or is it complicated?

thanks

[scruffydodge]

thanks for all the videos they have been really helpful and enligtening :D

A question... In this episode you say that the elements of a subgroup can be expressed as g to the power of some integer i. This suggests that the first non-zero elements of subgroup 2 (in angled brackets) 2^0, 2^1, 2^2, ... = 1,2,4, ...

But we know this group contains the set of all even integers 0,2,4,6, etc which doesn't match. It Seems that the elements of a subgroup g would be expressed by g*i (* = multiplication)?

[VeritySeeker]

Hi, yes elements can be represented by g*i in an ADDITIVE cyclic group. However, to make general definitions we fix a notation, and always use multiplication notation (even if we work with an additive group). So in the specific example of integers g^i actually means g*i.

It is just a matter of notation. I tried to explain this at the beginning of the video, but it might not have been clear.

[VeritySeeker]

In other words, perhaps more clearly:

g^i = g*g*g*...*g,

no matter what * should be. Even if * is addition...

[scruffydodge]

That cleared it up thanks

[tyebillion]

Look at the integers 1, 2, 3, ..., p-1 modulo p under multiplication, what kind of group is that ?

[VeritySeeker]

These groups are a "subset" of the groups of congruence classes relatively prime to modulo n. They are indeed groups. Such a group has order n-1 if and only if n is prime (which is the class of groups you are talking about).

[notoriousvcm]

I've got to say that there is also two types of Professors, those who can teach, and those who cant. Unfortunately I got the later kind. Thanks to you, however, I'm getting some of this stuff pretty darn good.

[flyfish89]

I learned more abstract in one night watching these videos than in 8 week in class. You have a gift my friend...keep them comming!

[VeritySeeker]

I am very happy if you got something out of the videos. Thanks a lot.

[chirho23]

How about examples of nonabelian groups?

[VeritySeeker]

There will be som examples of groups which are not abelian. I thought i would stay in the abelian world a little while before doing that. Thanks for commenting.

[dividenot]

So if I were to classify myself as a mathematitian who can't count, would i also be one of the 3 out of 2 who have problems with fractions?

Nice video. Thanks alot ;)

[VeritySeeker]

Good question ;).

I think you can be a master with fractions even if you cannot count, but I am not sure ;).

[joeseth05]

5 out of 4 people have problems with subsets.

[joeseth05]

hey, I know that background song, I got it on CD!

[vee3vee3]

Thanks for this helpful series.

When is pt 10 coming?

[vbcool83]

This series was awesome! Keep up the good work! Eagerly waiting for the next video in the series - when is it coming?

[vbcool83]

Oh, By the way, it would be really great if you can create a similar series on algebraic number theory...

[science2490]

thank you for this video! cant wait for the number theory ones.

did you study math at university?

[kautilya33]

Will it be too much if I ask for a tutorial on Lattice Theory. Remember that 5 out of 2 mathematicians have trouble with lattices:-)

[asdfmosin]

Just subscribed. Cool vids. Explainations are cool and all. But the best thing I like is that you chose dark black back ground with formal fond for text. Makes the vids look cool. I'm glad you didn't try out different color back grounds that sometime make power point presentations disastrous to the eye.

[spell4hell]

Thank you very much for your good work.

I like your videos about algebra... specially the nice cat pictures that appear from time to time :-))

[britishguy444]

Hey Verityseeker good job on the vid. I look forward to the number theory ones.

[vecter]

Glad to have you back :)

[VeritySeeker]

Thank you vecter :).

[fourty2]

can you do some proofs with permutation groups?

[VeritySeeker]

I will look at permutation groups very soon. Thanks for commenting.