The Kernel of a Group Homomorphism - Abstract Algebra

[@Socratica]

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

[@senthilkumaranmahadevan4991]

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]

true

[@fatemekashkouie3662]

Completely agree with that

[@EssentialsOfMath]

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]

only explanation that i understood easily

[@jasonlai4294]

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]

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]

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

[@JMeaeavjaiad]

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

[@chanfish2238]

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

[@xigong3009]

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

[@elangovan1592]

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

[@littlefishbigmountain]

@@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 KZbin and thinking you know enough to hit the bull’s-eye..

[@petergartin5904]

I wish you were my abstract algebra prof.

[@Socratica]

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

[@adiatarabi3786]

^^^^

[@WahranRai]

the grass is always greener elsewhere

[@ModeZt]

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]

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! :)

[@anamaria-og6lo]

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

[@Socratica]

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

[@LastvanLichtenGlorie]

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]

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

[@jeetendragour5140]

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

[@musicalBurr]

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]

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]

associative property need not be proved for subgroups.

[@evanspaulmuwonge1511]

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.

[@hardik1993ful]

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

[@mazenabdelbadea784]

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

[@bluetaylor7614]

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.

[@GelidGanef]

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]

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

[@69Solo]

+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]

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

[@kanikastudio2715]

True

[@ll-bc4gn]

you are the only mathematician that can make me understand abstract algebra so far.

[@pishposh54321]

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

[@Master1906]

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

[@BedrockBlocker]

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]

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]

underappreciated comment

[@whoknows00..]

Anyone watching this vedio now ?? Though vedio was posted 8y ago.. still helpful in 2024 too..

[@algerianinusa]

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

[@nikhilallenki943]

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

[@pittdancer85]

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!

[@Socratica]

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

[@sinisternightcore3489]

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.

[@jaimelima2420]

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.

[@papaonn]

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

[@2LightaManonFire]

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

[@joem8251]

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

[@someshbarthwal322]

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

[@shivamagarwal126]

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

[@nipun056]

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

[@MrCardeso]

Beautifully presented! Thanks, Liliana and Socratica team!

[@douggwyn9656]

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.

[@AnastasisKr]

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.

[@trinity-jaynehayward8509]

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! :)

[@Nakameguro97]

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]

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.

[@hunainghouri1768]

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

[@Dilip_Ghosh_BJP]

Super video. Short and *Concrete*

[@Ivana9910]

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

[@imagaynig6501]

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

[@RaviRanjan-ds4lz]

this lecture saved my time to understand this topic deeply

[@afanoromolyricsofficial]

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

[@Riesig88]

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

[@Socratica]

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

[@Riesig88]

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

[@Socratica]

+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!

[@thairameher3754]

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

[@cristopheririas1509]

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!

[@adamhendry945]

At 3:23, your kernel definition has an error. Every operation in a group has a corresponding unique identity for all elements, but not so for inverses. e.g. For integers under addition, -1 is the inverse of 1 and -2 the inverse of 2. Both add to the additive identity, 0, but -1 and -2 are not the same. Instead, every ELEMENT of a group has a unique inverse (again, for a given operation). You use x1^-1 for all elements on both sides of your equations. However, f(x1) * f(x1^-1) does not necessarily equal f(x2) * f(x1^-1). The correct statement is as follows: 1. Inverses map to inverses, as you previously showed, 2. Each element of a group has its own unique inverse (for a given operation), and 3. since the premise is x1, x2, etc., all map to y, then x1^-1, x2^-1, etc., all necessarily map to y^-1 since its inverse is unique.

[@Socratica]

The use of f(x_1^{-1}) on all expressions was a way to illustrate that different elements were mapped to the identity. This is a way to show that if f is not injective, then all the distinct elements which map to y can be used to generate a different set of distinct elements which map to 1. Since x1 and x2 are different, so are f(x1)*f(x1^-1) and f(x2)*f(x1^-1). This is what we wanted. Another way to think about this demonstration is that if {x1, x2, ...} all map to y, then {x1*x1^-1, x2*x1^-1, ...} all map to 1.

[@adamhendry945]

@@Socratica But you don't need to do that to illustrate that at all. It's incorrect and adds confusion. Multiple elements can map to the same y. Since every element has a unique inverse (for a given operation), and you previously proved that homomorphisms map inverses to inverses, it follows naturally that every inverse x1^-1, x2^-1, etc., maps to the same inverse, y^-1. Why did you use x1^-1 throughout? It makes no sense. You could have properly used the other inverses (x2^-1, x3^-1, etc.) and still made the same point without adding confusion or being incorrect.

[@adamhendry945]

Oh, wait wait wait, I see now. Using my approach, we would simply be reshowing what we already know: f(x * x^-1) = f(1G) = 1H. Your statement amounts to the fact that if there are multiple elements in G that map to the same y in H, then it can be shown that there exist multiple elements in G that map to 1H, and you do so by using the fact that f(x1^-1) = f(x2^-1) = f(x3^-1) = ... = y^-1. My apologies. Thank you for the clarification!

[@adityashekhar630]

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

[@Socratica]

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

[@aadeshnikam2643]

Socratica we're waiting for it..

[@charliedexter3202]

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

[@Socratica]

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

[@abidahaque853]

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...?

[@devesh09]

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

[@graykaufmann]

3:23 Why is f(xsup2) # f(xsup1 ^-1) = y # f(xsup1 ^-1) and not f(xsup2) # f(xsup2 ^-1) = y # f(xsup2 ^-1) ?

[@senahdongasso4580]

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

[@bhumisworld2803]

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

[@asitisj]

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

[@AshwaniMaurya-ph3vs]

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]

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

[@shuhaozhang7332]

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

[@RobElfrink]

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]

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

[@bobzheng5251]

Great explanation! Have more confidence for the incoming midterm

[@christopherellis2663]

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

[@ethanjensen7967]

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

[@MIRZAADNANBAIG]

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

[@dorveille1]

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.

[@spunpum]

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

[@chasr1843]

This women explains algegra very well

[@amitmishra-fe6yi]

Really your teaching style is so good ❤️❤️

[@steff420]

i love the creators of this channel

[@malenaalmasi1774]

I really love the clear notation.

[@chimwemwebanda6539]

Oh my gosh, you're a superhero! Thank you!

[@abdulrahmanabu-askar7011]

at 1:11 - 1:13 there is an error.

[@axeljebens2802]

Awesome. This channel is exceptional!

[@akrishna1729]

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

[@gfixler]

Is the phrase "sends inverses to inverses" equivalent to "preserves inverses?" The "sends/to" phrasing is used throughout the video, but I didn't hear anything about preserving. Should I not use the latter phrasing?

[@lbblackburn]

Yes. Mathematicians do say that homomorphisms "preserve inverses" and that's what the speaker means by "sends inverses to inverses."

[@gfixler]

Thanks, Leonard!

[@abrariscapoeira2613]

why these videos have scarry music in the BG? =because they are revealing scarry things of science🤣🤣

[@moularaoul643]

Thanks

[@cameronspalding9792]

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

[@rudyossanchez]

18 seconds in. I had a stroke

[@rudyossanchez]

supermanifold :) I did. however, this channel open my eye of just how ignorant I actually am. that's actually a good thing.

[@mappingtheshit]

Dawning-Kruger syndrom

[@harmenleusink7050]

@@mappingtheshit you mean effect ;-)

[@josephwilson47]

@@mappingtheshit you mean Dunning-Kruger :-)

[@charitylyngdoh8912]

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

[@lakhanpaul1458]

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

[@jeweljames6138]

Is it f(x2)*f(x2^-1) in the definition of kernel. In the video it is given f(x2)*f(x1^-1)

[@morgengabe1]

About the proof, is it not false if "f(1G)=/=1G", "1G" is not the identity in the group of "f(x)" and/or "1G=/=1H"? And is the Kernel not only a subgroup of "G->H"?

[@amywallquist3405]

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

[@Vivek-lu4eq]

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

[@paritoshmehta4048]

According to definition of homomorphism: F(1g*1g)=f(1g)$f(1g) Now, f(1g) is an idempotent element of {H, $} But the only idempotent element elements of a group is its identity, thus, f(1g) = 1h How does this sound?

[@PaoloLammens]

I think you're showing that ker(f) is a subset of G, but the challenge asks showing that it is a /subgroup/ of G; that is, a set equipped with an operation * (the same as G) which fulfills all four requirements of being classified as a group

[@havock0701]

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

[@yousify]

Beautiful explanation!!

[@coldassassin6615]

this is a really helpful video, thankyou!

[@phyziks878]

How to proof kernel is a normal subgroup ?And what is the use of modulus of kernel in mapping ?

[@saiganeshreddy7235]

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 ?

[@fazalhaq53]

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

[@kayri687]

Is it enough to prove kernel of the homomorphism is a subgroup of G by assuming ker(homom) is not a subgroup of G, so there exists some x not in G that maps to H but we know the homomorphism is surjective and only elements from G can map to elements in H, so the kernel has to be a subgroup of G

[@MuffinsAPlenty]

No, that's not good enough. There are _other_ reasons that a set could fail to be a subgroup besides simply not being a subset (such as not being closed under the operation or not every element having an inverse). Proof by contradiction here really isn't the way to go. It would be much more difficult than doing a direct proof using the definition of a subgroup.

[@greeshmamathew3663]

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]

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?

[@kusalweerasekara2305]

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..

[@yajaveri6633]

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

[@StanislavBashkirtsev]

So that's why in Linear Algebra they have two similar terms - Null Space and Kernel. One of them is native to LA, the other comes from Abstract Algebra?

[@cameronspalding9792]

Is a kernel a normal subgroup?

[@benjaminasmad9806]

Wow, what a great tutoring!

[@lianajaafar9475]

thank you very much this was very helpful

[@brunopinheiro5278]

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