All About Subgroups | Abstract Algebra

We introduce subgroups, the definition of subgroup, examples and non-examples of subgroups, and we prove that subgroups are groups. We also do an example proving a subset is a subgroup. If G is a group and H is a nonempty subset of G, we say H is a subgroup of G if H is closed with respect to inverses and closed under the group operation of G. #abstractalgebra #grouptheory
What are Groups?: • What is a Group? | Abs...
Abstract Algebra Course: • Abstract Algebra
Abstract Algebra Exercises:
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[@DrMathKinzie]

The asterisks usually are used to denote the non-zero rationals/reals/complexes, rather than the positive ones. In particular, there is no possible notion of "positive" for the complex numbers (no linear ordering that satisfies the usual properties of "

[@linonator]

Hey man! Just want to give a huge thanks to you. I just found your channel and from one video you helped unlock my understanding of something I was having trouble with. Thank you for create content like this. You really are making an impact on complete strangers for the better. Thanks again and keep up the great content

[@WrathofMath]

Thanks so much! Llet me know if you have any requests!

[@o_oyxy]

i love you bro youre carrying me

[@hamzaker829]

What a great channel ! THIS IS SOOOOO UNDERRATED

[@nazombie7935]

Thank you sire!! This video helped my understanding of subgroups so much

[@enderesting]

This video is the best explanation for subgroups on youtube :,)) Thank you so much! I found a lot of videos confusing because they never explain the definition of subgroup or give any example. This is perfect :,3

[@WrathofMath]

Thank you - so glad it helped! I am trying to make my Abstract Algebra playlist the best around! Will be working hard on it this summer. th-cam.com/play/PLztBpqftvzxVvdVmBMSM4PVeOsE5w1NnN.html

[@drapwill]

Absolutely love your channel thus far! Your explanations have unraveled some major knots in my mathematical thinking. Outstanding!

[@WrathofMath]

Thanks so much!

[@nilanjanabose3059]

You made this chapter very clear thank you❤ from india

[@xylynpasol9831]

Thank you so much! Very clear explanation.

[@WrathofMath]

Glad it was helpful! Check out the playlist for more if you haven't already! th-cam.com/play/PLztBpqftvzxVvdVmBMSM4PVeOsE5w1NnN.html

[@allstar965]

Thanks a lot! Very informative!

[@WrathofMath]

Thanks for watching! Isomorphic group lesson comes out tonight - lots more algebra to come!

[@punditgi]

More good stuff. Nice! 😃

[@WrathofMath]

Thank you! In a real abstract alg mood lately!

[@user-lz7kf8kd6n]

What note taking app are you using?

[@delzeltyna]

thank you!!!❤

[@WrathofMath]

Glad to help! Check out my abstract algebra playlist if you're looking for more, and let me know if you have any questions!: th-cam.com/play/PLztBpqftvzxVvdVmBMSM4PVeOsE5w1NnN.html

[@user-wt5xm9pz8p]

Really cool. Well explained. I enjoyed though

[@WrathofMath]

Awesome, thank you!

[@ronbitan2641]

Hello! According to the rule you gave where H has to be closed with respect to inverses and multiplication, ehy multiplication? What if the operation of the group is + for example? Does it still need to be closed under multiplication or does it depend on the operation?

[@WrathofMath]

As long as we're talking about groups, "multiplication" is just short for "the operation". We generally default to multiplicative notation and verbiage unless we know we're dealing with an additive group. Does that make sense? Definitely check out my abstract algebra playlist and the exercises playlist for more, here's a link to an unreleased video on subgroup tests you may find as useful follow up: th-cam.com/video/poQXf90tcFM/w-d-xo.html

[@liketsontobo8463]

great video bro! keep it up

[@WrathofMath]

Thank you - I will, let me know if you ever have any questions!

[@raulbeienheimer]

Thanks!

[@WrathofMath]

Glad to help - thanks so much for your support,
Raül! Let me know if you ever have any questions, I will be making a lot more Abstract Algebra content this summer.

[@Channel-zb1fi]

When you test for if a set is closed under some operation, do you have to test for the element multiplied by itself or do the elements have to be distinct? If that is the case then how do you go about dealing with sets containing only one element?

[@jonko82]

When you you want to prove closure under the operation you simply pick two arbitrary elements from the set. Those arbitrary elements may or may not be distinct.
In the example in the video loga and logb might be two distinct elements or they might be the same element. The proof holds either way.
When you want to disprove closure you just need to find any two elements of the set that, when combined, do not produce another element of the set. Those two elements may or may not be distinct also.

[@l.JAI.SHREE.RAM.l]

Sir, I have a doubt, at many places group is represented by G how can it be possible, G may be a set not group, group should always denoted by (G,*).
Please clear it🙏

[@laurenfung7265]

this video saved me from my math test !

[@WrathofMath]

So glad it helped!

[@SkinnyMMA]

Why do we consider multiplicative inverse of a and NOT additive inverse in 13:59

[@WrathofMath]

That's just log properties. We had -log a, and that is the same as log (a^-1). In general, xlog(y) = log(y^x). Does that answer your question?

[@SkinnyMMA]

@@WrathofMath oh actually I get it now... Btw, I knew about this log property, now I don't understand why on earth I asked this query 😅...
Thanks though for listening out

[@akuma1995]

at 14:20 wouldn't it mean H can't be a subgroup because the identity element 0 isn't in H?

[@SkinnyMMA]

"0" is in H. Like he said, because every element in H has it's inverse is in H .... And also H is closed, so you have to have 0.

[@neevetiasli]

I think log1 is the identity element here

[@dripici4371]

we fa pt facultate ca am nevoie poimaine la fai

[@guddu1958]

Don't you ever get tired 😴

[@WrathofMath]

Often!

[@jarrenblake2886]

Fun fact you lost 30% battery life during this vid

[@WrathofMath]

That's more a depressing fact haha. Time for a new iPad!