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The Kernel of a Group Homomorphism - Abstract Algebra

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The kernel of a group homomorphism measures how far off it is from being one-to-one (an injection). Suppose you have a group homomorphism f:G → H. The kernel is the set of all elements in G which map to the identity element in H. It is a subgroup in G and it depends on f. Different homomorphisms between G and H can give different kernels.
If f is an isomorphism, then the kernel will simply be the identity element.
You can also define a kernel for a homomorphism between other objects in abstract algebra: rings, fields, vector spaces, modules. We will cover these in separate videos.
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We recommend the following textbooks:
Dummit & Foote, Abstract Algebra 3rd Edition
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Milne, Algebra Course Notes (available free online)
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Teaching Assistant: Liliana de Castro
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Опубликовано:

5 сен 2024

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Комментарии : 271

@Socratica 2 года назад

Sign up to our email list to be notified when we release more Abstract Algebra content: snu.socratica.com/abstract-algebra

@senthilkumaranmahadevan4991 6 лет назад

Dear Socratica , I believe that your lecture series is just the most beautiful lecture series i have ever watched in abstract algebra. i am not afraid of abstract algebra any more thank you for such a beautiful series on math..... great work ....

@singh3549 3 года назад

true

@fatemekashkouie3662 2 года назад

Completely agree with that

@EssentialsOfMath 6 лет назад

Claim: The kernel of G is a subgroup of G. Proof: We have established so far that the kernel is a non empty set containing elements of G, combined with the operation of G, *. We know that the identity 1G is always in the kernel by definition. Also, we know * is associative. Therefore we need to show that the kernel is closed under *, and that all elements of the kernel have unique inverses. Consider two elements of the kernel of G, x and y. We know that f(x) = 1H and f(y) = 1H. Then f(x*y) = f(x) • f(y) = 1H • 1H = 1H. Thus x*y is in the kernel of G; the kernel is closed. Now consider an element z of the kernel. Since homomorphisms map inverses to inverses, we know that f(z-1) = f(z)-1. But f(z) = 1H, and the identity is it's own inverse, so f(z-1) = 1H, and z-1 is in the kernel. Thus the kernel of a group G with respect to a homomorphism f is a subgroup of G.

@ballaraviteja4122 5 лет назад

only explanation that i understood easily

@jasonlai4294 5 лет назад

Thanks for the solution, but, should the conclusion statement be that the kernel of a homomorphism f w.r.t. a group G is a subgroup of the group G? Since the "kernel" here isn't about the group but about the map f, stating the kernel of a group G might be somewhat misleading.

@mehulkumar3469 4 года назад

I notice everywhere you write kernel of G, but kernel is a property of homomorphism between two groups not the group, first correct your words, so I don't confuse.

@JMeaeavjaiad 4 года назад

You are proved kernel is a group but you missed to prove it is subgroup of G

@JMeaeavjaiad 4 года назад

To prove kernel is subgroup of G We know that the definition of subgroup of a group ab^-1 belongs to G since a,b belongs to H We also know 1 is the only element present in kernel it is identity element also Inverse of the Identity element is itself so we can claim kernel of a group G with respect to homomorphism f is subgroup of a group G

@petergartin5904 7 лет назад

I wish you were my abstract algebra prof.

@Socratica 7 лет назад

The best we can do is make more videos for you! Thanks for watching, and thank you for your kind comment! :)

@adiatarabi3786 3 года назад

^^^^

@WahranRai 3 года назад

the grass is always greener elsewhere

@chanfish2238 6 лет назад

University I spent 6 weeks to learn these = Here I use 20 min understand ... Thank You

@xigong3009 4 года назад

I am sorry but you cannot really understand something without doing some exercise problems.

@elangovan1592 4 года назад

Man....I feel the same way!!!!!!!

@littlefishbigmountain 2 месяца назад

@@xigong3009 So, so true. People talk about math as if it’s a matter of understanding, which it is, but sometimes it’s neglected to mention that math is also a skill practiced by doing. You can understand the concepts, but applying them gives a more direct experiential familiarity with the processes actually going on, the repetition of which provides a greater and greater subconscious intuition that is absolutely invaluable and unlocks new maths frontiers for you. Not doing practice problems is like watching archery on RU-vid and thinking you know enough to hit the bull’s-eye..

@anamaria-og6lo 8 лет назад

you are contributing to make a better world. thank you!

@Socratica 8 лет назад

Oh my goodness, what a lovely thing for you to say. Our viewers are just the nicest. Thank you so much for watching!! :)

@ModeZt 7 лет назад

Many years have passes since I learned this in the university.. It is a pleasure to recover that forgotten knowledge with such a wonderful teacher. Thank you!

@Socratica 7 лет назад

Isn't it wonderful that you can pick up where you left off? Hooray for lifelong learning!! Thank you for watching, and thank you for your kind comment! :)

@LastvanLichtenGlorie 5 лет назад

This video stopped me from giving up in Abstract Algebra when I was on the edge of giving up. I'm deeply in your debt. As soon as I have a decent salary I will be contributing.

@Socratica 5 лет назад

We're thrilled we could help, Russell. Your message really inspired us today - thank you so much for writing and letting us know. Thank you for watching! :D

@musicalBurr 7 лет назад

I like your challenge question at the end to show that the ker(f) is a subgroup of G. For anyone who is a little stuck (this is a common feeling among mathematicians - it's OK to feel that way you're in good company!) just write down everything you know again on a sheet of paper. So.... you have G,* and H,◊ and you have f: G -> H and you also know that f(x*y)=f(x)◊f(y). We also have our new definition for kernel which is ker(f) = { x in G | f(x)=1H} All you need to do to show that this set, ker(f), is a subgroup of G is show that it's 1) closed under * 2) Has an identity 3) Each element in ker(f) also has it's inverse in ker(f) and finally 4) It's associative. Just like we did back in the fourth video "Group or not group"! That's it. It's fun and not too tough - hope that helps anyone who's stuck.

@ThePharphis 5 лет назад

Is it necessary to check all of those? I thought for subgroups less effort is required since we're talking about subsets of something which is a group. For example associativity is a given, I think

@shreya2262 5 лет назад

associative property need not be proved for subgroups.

@jeetendragour5140 7 лет назад

I like the way of teaching her. It's so lucid and made the content easy to understand. Thank you.

@mazenabdelbadea784 4 года назад

It is 2020 and still watching this. Thank you, it really helped alot.

@bluetaylor7614 3 года назад

These are helping me get a better overview of Abstract Algebra. Thank you! Hope Socratica creates more Abstract Algebra videos as well as playlists on Topology and Analysis next.

@hardik1993ful 7 лет назад

I admire the presentation skill of the instructor. She presented it like a beautiful story.

@GelidGanef 8 лет назад

Yay! I've really been enjoying the python/programming videos, but I'd honestly forgotten why I subscribed to this channel? This is why. Your abstract algebra videos are phenomenal. Keep them coming!

@WilliamHuang1995 8 лет назад

+GelidGanef not to mention she makes math seem so much interesting than it is in college

@69Solo 8 лет назад

+William Huang True dat. If I had a maths teacher like her, I would be flying rockets to other planets, rather than my current part time job as a human resource. :-D

@Socratica 8 лет назад

+GelidGanef Thank you for the helpful feedback! Many more abstract algebra and python videos are in the works.

@kanikastudio2715 5 лет назад

True

@evanspaulmuwonge1511 Год назад

Lady Socratica; thank you so so so so so much. I have completely understood your video from the word Go to the word end. What a blessing to have u on you tube. What a blessing, what a blessing from the LORD that you lady exist in Abstract Algebra. Thank you so much,really much and really much. An amazing video. U have humbled my minds down to learn.

@Socratica 2 года назад

Socratica Friends! Do you want to grow as a student? We wrote a book for you! How to Be a Great Student ebook: amzn.to/2Lh3XSP Paperback: amzn.to/3t5jeH3 or read for free when you sign up for Kindle Unlimited: amzn.to/3atr8TJ

@pishposh54321 6 лет назад

I learned more in this video than i have in the past 2 months of my abstract algebra class

@Master1906 8 лет назад

This has helped me for one of my math modules. Explained succinctly and intuitively, can't ask for more! Thank you so much!

@nikhilallenki943 5 лет назад

I AM SO LUCKY TO HAVE YOU MADAM SO THANKFUL TO YOU FOR HELPING ME OUT IN WHAT I THOUGHT IS IMPOSSIBLE TO ME AND MAKING IT POSSIBLE TO ME

@someshbarthwal322 4 года назад

I first time in my life understand the meaning of kernel you guys are surely amazing, ❤❤❤❤

@jaimelima2420 4 года назад

I have watched many of these at this point. Besides being really a useful tool to learn a specific math topic which has a well deserved fame of being bit-cryptic and being able do it an efficient way, this innovative approach makes me think about how wrong the established approaches to transmit scientific knowledge is these days, being them on the 'math has to be dry and hard' or in 'math is fun' side. Learning should be a social experience, before becoming an individual one. IMHO this is the most important lesson I am taking from these classes.

@AnastasisKr 7 лет назад

Nice video, but you should mention the cokernel and draw an analogy with "onto" maps. I find the dual construction very enlightening when trying to intuit kernels.

@adityashekhar630 7 лет назад

awesome video. ur organization is doing a great job. your explanation is so clear. please make more videos on concepts of abstract algebra.

@Socratica 7 лет назад

Thanks! Many more Abstract Algebra videos are on the way. :)

@aadeshnikam2643 6 лет назад

Socratica we're waiting for it..

@joem8251 3 года назад

I doubt there is a better video on this subject, but please prove me wrong with a reply! This whole series is fantastic.

@pittdancer85 Год назад

I love that I was about to ask if the kernel is a subgroup of G, and then she said it was. I feel like I’m learning!

@cristopheririas1509 5 лет назад

I just want you to know I fell in love with your videos. although I am not a native English Speaker I completely got your explanation. Best Regards from Honduras!

@Nakameguro97 7 лет назад

omg, this kernel is totally consistent with the kernel in linear algebra (as it should be). Gotta luv it when terminology and concepts are consistent! Question is, should you learn linear algebra first or abstract algebra first?

@Socratica 7 лет назад

You can learn them in either order. However, if you learn Linear Algebra first, you'll be equipped with lots of examples for the ideas in abstract algebra. In fact, most abstract algebra textbooks assume you are familiar with matrices. So most people would probably find it easier to learn linear algebra first.

@nipun056 7 лет назад

This channel is absolutely incredible. Thanks so much for making these videos.

@2LightaManonFire 8 лет назад

The way that built up to ker(f) makes a lot more sense than the way i initially learned. Interesting mix of videos

@algerianinusa 5 лет назад

Hi Socratica, First of all, thank you so much for all these useful videos. Secondly, could you plz correct the negligible mistake at 3:35 f(x_2)=y -> f(x_2*x_2^(-1))=1_H

@imagaynig6501 5 лет назад

5 minute youtube video better for my understanding than 3 hrs of lectures. It's all good tho cuz my prof irl dumb handsome ;O

@Ivana9910 3 года назад

way better than my prof explains it. Well-planned and executed video that makes algebra much easier to understand when ideas are explained fully since I don't remember them all yet

@AshwaniMaurya-ph3vs 7 месяцев назад

I did my major in Physics. I would never have come this far in abstract algebra series. These lectures are tonic for my brain😅😅

@Socratica 7 месяцев назад

We're so glad you're exploring with us!! 💜🦉

@Riesig88 8 лет назад

when I saw this video uploaded I got so excited!!! keep up the AMAZING work with abstract algebra, you guys are the best!

@Socratica 8 лет назад

+Ilya Noskov Thank you! We're planning many more abstract algebra videos, and will be filming the next one this week!

@Riesig88 8 лет назад

+Socratica what are the topics you plan to cover? I think I am gonna be your patreon if there are more math videos!

@Socratica 8 лет назад

+Ilya Noskov For abstract algebra we're going to cover the most important structures: groups, rings, fields, vector spaces and modules. We're also going to begin making number theory videos in the next few weeks!

@asitisj 3 года назад

Clarity of your speech is helpful in seeing the terms and. relations apart .

@trinity-jaynehayward8509 3 года назад

Hi, This has been the most helpful thing during a pandemic when you can't go to uni! Thank you so much there is no way I could even attempt my coursework without you! :)

@afanoromolyricsofficial Год назад

I liked the way you teach with an authority. It makes the lecture more interesting!

@papaonn 5 лет назад

My 2cents : ( correct me if wrong, as a progress of learning humbly ). Definition of Subgroup S

@hunainghouri1768 3 года назад

This is the best explanation i have gone through till now. Thanks

@senahdongasso4580 4 года назад

When I watch this video it like , in French we say " une illumination" for me . Thank you very much

@MrCardeso 4 года назад

Beautifully presented! Thanks, Liliana and Socratica team!

@thairameher3754 Год назад

Thank you so much, I understood easily, I never forget about kernel.

@ethanjensen7967 3 года назад

This is great! Have you considered making a video about orbits and stabilizers?

@BedrockBlocker 2 года назад

I think the assumption x not equal 1 in 1:10 is not nececary. In fact, we can always choose x=1 and the proof still holds.

@josepher9071 4 года назад

Solid Snake voice: "Huh... Kernel. I'm trying to map to 1. But I'm dummy thicc And the elements of my group keep mapping to a non-identity"

@KKajice 3 года назад

underappreciated comment

@charliedexter3202 8 лет назад

You are doing a great job...finished all the abstract algebra vids in one sitting...Please upload more...thanks in advance :D

@Socratica 8 лет назад

Thanks, Charlie! More Abstract Algebra videos are on the way! We filmed several more just last week.

@shivamagarwal126 3 года назад

Watching this in 2020 and it is so elegantly explained. Thank you so much.

@devesh09 Год назад

This 4 min video takes my 1 hour to understand thoroughly not losing hope 😊

@bhumisworld2803 3 года назад

2021 !! And i found this videos what a great start of study with u..

@Maturetraders 3 года назад

Dear socratica, your teaching method is too much impressive but your lectures ends before it started so please add few examples more plzzzzzzzzzzzzzz😘love 💕 from INDIA 🇮🇳🇮🇳

@Socratica 3 года назад

We're planning on adding more short example resources on our website! Thanks for the encouragement. 💜🦉 You can sign up for our email list so you'll get notified when new stuff arrives! www.socratica.com/email-groups/abstract-algebra

@MIRZAADNANBAIG 5 лет назад

Interesting : Being a Lecturer, it was really very fruitful lecture for me. Thank you

@yajaveri6633 4 года назад

I took whole lot year while our lecture teaching. Only 5minn in socratica😎😍😘🥰

@Dilip_Ghosh_BJP 3 года назад

Super video. Short and *Concrete*

@spunpum 4 года назад

I love you. Thanks for these videos. they are very explanatory. Wish there were more math teachers in uni like you.

@bobzheng5251 5 лет назад

Great explanation! Have more confidence for the incoming midterm

@MdImran-hj2cd 6 лет назад

thank u madam for ur giving a good knowldege of mathematics ..i am very much impress to ur way teaching and understanding the concept of mathematics, i want to discuss the few general doubt of FUNDAMENTAL THEOREM OF HOMORPHISM OF GROUPS .i am grateful to u, if u prepare a video lecture on this topic plz maam...

@shuhaozhang7332 5 лет назад

Sol of challenge: (1) kernel is a homomorphism that contains all elements that map to identity of H, so it contains the identity of G (2)if x in kernel then f(x)=identityH, if y also in kernel then f(y)=identityH, so f(xy)=f(x)f(y)=identityH*identityH=identityH (x)if x in kernel then f(x)=identityH, so f(Identity G)=f(x&x^-1) = f(x)*f(x^-1)=Ih*f(x^-1)=f(x^-1) = Ih

@abidahaque853 5 лет назад

Great video! I watched a different one explaining isomorphisms/homomorphisms. So one way to prove a function is 1-1 is to say, Let f(x) = f(y)......x=y. Another way would be to say f(x)=identity iff x in Ker(f), or...?

@havock0701 7 лет назад

You need to go over theorems in the Algebra playlist! Like Sylows theorem. thanks

@akrishna1729 2 года назад

thank you so much for these - truly the simplest explanation of the subject, these videos have helped me so much !!

@lakhanpaul1458 7 лет назад

Thank for those free I was searching for Abstract Algebra professor And finally I got it 😊.Yes, I have solved the challengeThank You

@DeepObserver7689 2 года назад

Please make a video on Reproducing kernel operator with cyclic vector

@kusalweerasekara2305 8 лет назад

You are doing these videos quite interesting manner , We hope u will keep it up , I think u should cover whole content of this particular subject..

@malenaalmasi1774 4 года назад

I really love the clear notation.

@nehasusan7251 3 года назад

The very first line blew me away.

@amitmishra-fe6yi 3 года назад

Really your teaching style is so good ❤️❤️

@anadesign 7 лет назад

so sweet ,keep going, give ur self some times to learn c programing ,it will be amazing with algebra , belive me ,and it will not take long ,u can learn fundmentals in a week.algebra more difficult and complex than programming in beginner level

@sinisternightcore3489 4 года назад

Show that ker(f) is a subgroup of G: It is already shown that ker(f) is a subset of G and that it contains the identity 1_G. ker(f) is also associative because its group operation is the same as of G. To show ker(f) is closed, take any xa, xb ∈ ker(f). xa * xb = x f(xa * xb) = f(x) f(xa) ♢ f(xb) = f(x) 1H ♢ 1H = f(x) f(x) = 1H , therefore x ∈ ker(f). To show every element in ker(f) has an inverse, choose x1, x2 ∈ G such that x1, x2 → y as shown at 3:35 this yields: f(x1 * x2^-1) = 1H and by the same reasoning f(x2 * x1^-1) = 1H Call these: x1 * x2^-1 = xr ∈ ker(f) x2 * x1^-1 = xs ∈ ker(f) We can invert one of these step by step: x1 * x2^-1 = xr x1 * x2^-1 * xr^-1 = xr * xr^-1 x1 * x2^-1 * xr^-1 = 1G x1^-1 * x1 * x2^-1 * xr^-1 = x1^-1 * 1G x2^-1 * xr^-1 = x1^-1 xr^-1 = x2 * x1^-1 = xs This shows that xr is the inverse of xs.

@pras1293 6 лет назад

That was very nicely put.very nice explanation.THANKYOU SO MUCH .it was realy useful.

@sahilraja6631 4 года назад

Itna acha bolti hn dil krta h pora math ap smjhati Jaen mn sunt jaon nice

@charitylyngdoh8912 3 года назад

Great Jop👍👍.. Thank You Soooooo Much for making such a wonderful lectures🙏🙏🙏

@douggwyn9656 7 лет назад

Not a criticism, but around 3:35 some steps were skipped. Given x1 not equal to x2, we should show that x1 * x1~ and x2 * x1~ are distinct elements. As with previous uses of cancellation using inverse, it's not hard to do, but at the beginning level such details should be spelled out.

@anldemirel8429 3 года назад

Eyvallah bacım, teşekkür ederiz

@coldassassin6615 Год назад

this is a really helpful video, thankyou!

@bakkamydestination 2 года назад

Excellent classes

@OmarAhmed-ic4fw 3 года назад

This series is great but it needs to be completed by covering more topics.

@axeljebens2802 2 года назад

Awesome. This channel is exceptional!

@fazalhaq53 3 года назад

i am inspired from her way of expressing.. math as poems of Shakespeare

@aaronmtonga1425 5 лет назад

great videos the socratica team

@christopherellis2663 4 года назад

Are there any examples around which one might wrap his head? This is somewhat abstract.

@myworld-hl3gk 7 лет назад

love your videos. you make my interest in algebra. and im very thankful. more videos please...

@Socratica 7 лет назад

Thank you so much for watching! We're so glad you are finding our videos helpful. More on the way soon! :)

@KAREN-ye5pf 3 года назад

Mam, you make abstract algebra simple...thanks a lot

@RobElfrink 5 лет назад

At 1:16 why must the identity element be excluded? The proof will still work if x=1(G) I think. What do I oversee?

@papaonn 4 года назад

Thats to simply sepearate the case from identity to obtain a more rigourous proofs.

@jiansenxmu 6 лет назад

Note that we can donate to this program with some amount of Bitcoin:-)

@Vivek-lu4eq 8 месяцев назад

Please tell me the name of background music you play for this video.

@usmanaziz7087 4 года назад

Makes group theory easiest to understand 😍😍😍

@aayushbhattarai2282 2 года назад

Thank you for the clear explanation!

@amywallquist3405 4 года назад

do you have a video about ring homomorphisms and the kernel of those?

@yousify 4 года назад

Beautiful explanation!!

@fazalhaq53 3 года назад

@socratica tell me how we can prove that kernal is subgroup of G??

@cameronspalding9792 3 года назад

Not only is the the kernel of a homomorphism a subgroup but it is also a normal subgroup

@brunopinheiro5278 8 лет назад

socratica!!!!!!!!!, very good!!!!# socratica best channel of youtuber!!!

@lianajaafar9475 4 года назад

thank you very much this was very helpful

@saiganeshreddy7235 7 лет назад

Lets consider a case let {Xp |p=1,2...m }and{ Xq |q=1,2...n} Maps to y1 and y2 and the rest is 1-1 map ,how would u define a kernal in such case ? Do we require 2 kernals in that case ?

@dorveille1 3 года назад

I think the proof that group homomorphisms send identities to identities isn't quite right. It doesn't work if G = {e}, where e is the identity of G. In other words, G is the trivial group. The argument doesn't work because there is no x in G distinct from e. However, the assumption that x ≠ e is irrelevant to the argument. In particular, since the only element that is guaranteed to be in G is e, we might as well use that. In more detail, let f: G -> H be a group homomorphism. Then e=ee, so f(e) = f(ee) = f(e)f(e). Multiplying both sides by the inverse of f(e), which exists because H is a group, implies that f(e) is the identity in H.

@greeshmamathew3663 4 года назад

Let f be a non trivial homomorphism from ℤ10 to ℤ15. Then which of the following holds? A) Im f is of order 10. B) Ker ݂f is of order 5. C) Ker f is of order 2. D)f is a one to one map. What is the answer for this question?

@MuffinsAPlenty 4 года назад

Note that im f is a subgroup of Z15, so |im f| divides |Z15| by Lagrange's Theorem. Additionally, by the First Isomorphism Theorem, im f ≅ Z10/(ker f). By Lagrange's Theorem, |Z10/(ker f)| = |Z10|/|ker f|. This gives that |im f| divides |Z15| and |Z10|. But since f is non-trivial, |im f| is NOT 1. Can you figure it out from there?