Proving The Fundamental Homomorphism Theorem | Abstract Algebra

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We introduce and prove the fundamental homomorphism theorem (also called the first isomorphism theorem), which states that every homomorphic image of a group G is isomorphic to the quotient group of G by the kernel of f, given that f is a homomorphism from G onto H. To prove this we will use some prior results. We previously proved that a quotient group of G is a homomorphic image of G, so we now have that a group is a quotient group of G if and only if it is a homomorphic image of G (or isomorphic to one). So homomorphic images and quotient groups are somewhat interchangeable. #abstractalgebra #grouptheory

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Комментарии
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So helpful! Had to wrap my head around this theorem and quotient groups... your teaching style is very soothing and nice to follow 🙂

kenzo
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I needed this a few years ago 😢. Good work! I love supporting math videos!

hellaradandy
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Thank you so much for the whole series! You are a blessing from heaven haha

kale
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Mate thanks for the help and aspiration, i passed the algebraic structures subject and i am waiting the results of number theory one, you where a great reason for it and of understanding better the fascinating world of algebra. Thanks again keep going !!!

cantorbernoulli
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6:14 im trying to follow along but this abstract stuff is hard for me... i dont understand the statement here: "Thus, when we construct quotient groups of a group G, not only are we constructing homomorphic images of G, but we are constructing all the homomorphic images of G (up to isomorphism)"

I am confused about this because quotient groups of G are sets of cosets of H, whereas the fundamental theorem isnt a statement about the cosets of H, but rather the cosets of the kernel of the thing that gets you from G to H.. so i dont see how those theorems are related or how we know we are getting all the homomorphic images from that. i think i understood the theorem up to that point but i didnt understand that implication.

nathanisbored
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Waiting to see this theorem "put into action". I guess you need some more Patreon subscribers, eh?

MrCoreyTexas
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When you prove that G/H is a homomorphic image of G, doesn't H have to be a normal subgroup?

zilanh