The Kernel of a Group Homomorphism – Abstract Algebra

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

Milne, Algebra Course Notes (available free online)

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Teaching​ ​Assistant:​ ​​ ​Liliana​ ​de​ ​Castro
Written​ ​&​ ​Directed​ ​by​ ​Michael​ ​Harrison
Produced​ ​by​ ​Kimberly​ ​Hatch​ ​Harrison

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[senthilkumaranmahadevan]

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 great work ....

[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.

[petergartin]

I wish you were my abstract algebra prof.

[anamaria-oglo]

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

[chanfish]

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

[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!

[jeetendragour]

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

[mazenabdelbadea]

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

[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.

[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.

[hardikful]

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

[bluetaylor]

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!

[evanspaulmuwonge]

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.

[pishposh]

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

[Master]

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

[shivamagarwal]

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

[someshbarthwal]

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

[MrCardeso]

Beautifully presented! Thanks, Liliana and Socratica team!

[nikhilallenki]

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