Isomorphic Groups and Isomorphisms in Group Theory | Abstract Algebra

[@mcmoodoo]

The author clearly has a skill of providing clear explanations! Well done, sir!

[@WrathofMath]

Many thanks!

[@KermitTheHermit.]

Understood perfectly! Thank you for a different perspective. Was stuck with the textbook definition for long. Thanks again🙏🏻😊

[@WrathofMath]

Glad to help, thanks for watching!

[@VijitChandna]

First! Tommorow is my exam and I had commented on his channel about this topic and he sent me an unlisted link! Thank you so much :)

[@wenzhang365]

Please keep up the good work, thank you!

[@JTan-fq6vy]

Thanks for the great video! Is there any theory that deals with the generalization of this isomorphism? For example, if I want to verify an equivalent relation between two mathematical objects with arbitrary properties (not specifically the ones of binary operator for groups), is there a modification of the definition in time 1:50 that can give a generalized notion of isomorphism?

[@EatYourVegs]

Really well done video. Thanks this will help me in my upcoming test

[@lamyamalcolm-uu8zp]

Very skillful and talented, thank you so much. You videos help me a lot with my studies here.

[@WrathofMath]

Glad to hear it, thanks for watching!

[@gabriels.i.780]

Thank you so much! Very clear and rich explanation. I would like to ask...Isomorphism seems pretty restrictive as a way to study identity/similarity between groups. Is there any concept in abstract algebra that can account for "weaker" forms of similarity? Thanks!

[@WrathofMath]

Great question and the answer is a big yes! kzbin.info/www/bejne/qHvTpmVon7KYf7s&pp=ygUSaG9tb21vcnBoaWMgZ3JvdXBz

[@claytonbenignus4688]

I Get It!!! You could say that if you take a Zane Grey Novel and transform a few words (Rancher's Daughter = Martian Princess; Rifle = Disintegrator; Stage Coach = Rocket Shoip; The Cavalry = Star Fleet; etc.), you get a Star Trek Episode . . .

[@Ancientladd]

Hi, I believe the Group table for the multiplicative operation might be wrong. a *a = e multiplying a by itself gives the identity element. I could be wrong, if someone can explain why?

[@alexdrougkas208]

Thank you so much for the lecture. Keep up the great quality of work!

[@WrathofMath]

Thank you Alex!

[@marioeraso3674]

Awesome! I understood!

[@2kreskimatmy]

do you get into Cayley's theorem in some video?

[@madomalene5898]

For the first example,how did you obtain the second table. What rules were you using to perform the multiplication

[@lemonandgaming6013]

he just renamed all elements and the operation of g1. that is how he got g2. then he proved that these groups are isomorphic (the same), which is trivial since one is a renaming of the other

[@kabirbhattacharyya9014]

One thing I was wondering.... How does isomorphism "transforms" a group to another? Like how I thought.... It's like a bridge to Each storeys of two almost identical buildings. One is red, one is blue.. etc. what exactly I'm calling the "Isomorphism"? Also could you help me with the "transforms" a group to another?

[@JPL454]

It means it sends each element of G to a single element (injective) of H, with all elements of H being the image of some element of G (surjective). This for an isomorphism from G onto H. It's basically just the same idea of a bijective function that has the property f(a)f(b)=f(ab) for every a,b from the group G

[@kabirbhattacharyya9014]

@JPL454 thanks. My exams ended over 1.5 months ago, but thanks anyways🙌

[@JPL454]

@@kabirbhattacharyya9014 Oh your welcome😅, hope that they went well

[@ahasdasetodu6304]

The second theorem mentions a set of all groups but from my understanding of set theory such a thing would lead to contradictions the same way a set of all sets does. Wouldn't it be better to say a class of all groups?

[@algierithm1994]

Hello what notepad are you using? Thanks

[@WrathofMath]

Notability!

[@algierithm1994]

@@WrathofMath thanks much. btw, I love your videos.

[@gp2111]

For the portion where you discuss ways to find groups that are NOT isomorphic, you give 4 criteria but I'm curious what the difference between #2 and #3 are? If a G1 has an element of order n, does that not make it cyclic, which would be the same as #2?

[@WrathofMath]

Thanks for watching and for the question! Perhaps you're confused because you think I mean 'n' to be the order of the group? I simply mean n to be a finite number, and a group having an element of finite order does not force it to be cyclic. Does that answer your question?

[@gp2111]

@@WrathofMath Gotcha! It does answer my question. Thanks.

[@jaaaaaaaaaaaac6986]

very nice video！you should put this into your list, can't find this one in the list.

[@WrathofMath]

In the playlist? Weird, I see it. It is right after Permutation groups and before Order of Elements in a Group! I have spreadsheets on spreadsheets to keep all my playlists organized haha!

[@MrCoreyTexas]

For anyone interested, you should look up the Wikipedia on the Klein Group with 4 elements

[@WIRELESStransport]

Thanks so much for this

[@WrathofMath]

Glad to help - thanks for watching!

[@SphereofTime]

1:42

[@VijitChandna]

The chapters seem to say homomorphism for some reason

[@WrathofMath]

Looks like they're correct in the description, will probably just take some time to update hopefully!

[@scito1541]

so isomorphism is a homomorphism that is a bijection right?

[@scito1541]

if anyone wants to know i asked Bing AI and it basically said yes

[@WrathofMath]

Exactly!