Placeholder Image

字幕表 動画を再生する

  • A homomorphism between two groups does NOT have to be a one-to-one function. If it’s

  • not one-to-one, then there’s a group associated with the homomorphism which measures the degree

  • to which the function is not an injection. This group is called the kernel...

  • To really understand the kernel, there are a few properties you need to learn about homomorphisms:

  • homomorphisms send identities to identities, and inverses to inverses. Well begin by

  • proving these properties. Suppose we have two groups G and H, and a homomorphism F between

  • them. Recall that a function F is a homomorphism if F of (X-times-Y) equals (F-of-X) times

  • (F-of-Y). Don’t forget that the operation on the left hand side is for the group G,

  • while the operation on the right hand side is for the group H. These operations can be

  • different from one another.

  • Well first show why group homomorphisms send the identity in G to the identity in

  • H. So we don’t confuse the identities of these two groups with each other, let’s

  • denote the identity in G as 1-sub-G and the identity in H as 1-sub-H. We begin by picking

  • a random element in X that is different from the identity. Any element will do. From the

  • definition of the identity element, X times 1-sub-G equals X. Now look what happens if

  • we apply F to both sides of this equality. We get F-of-(X times 1-sub-G) equals F-of-X.

  • Since F is a homomorphism, we can write the left side as (F-of-X) times (F-of-1-sub-G).

  • Now F-of-X is an element in H; let’s call it Y. This gives us Y times (F-of-1-sub-G)

  • equals Y. And since H is a group, Y has an inverse. If we multiply both sides by the

  • inverse of Y we get F of 1-sub-G = 1-sub-H. This proves that homomorphisms send identity

  • elements to identity elements.

  • Next well show why homomorphisms send inverses to inverses. Suppose F-of-X equals Y. Note

  • that X is in G and Y is in H. We want to show that F-of-X-inverse = Y-inverse. To see why,

  • well use the fact that X times X-inverse equals 1-sub-G. If you apply F to both sides

  • you get F-of-(X times X-inverse) equals F-of-(1-sub-G). We just showed that F-of-(1-sub-G) = 1-sub-H.

  • And on the left, we can use the property of homomorphisms to get F-of-X times F-of-(X-inverse)

  • equals 1-sub-H... F-of-X equals Y, so we get Y times F-of-(X-inverse) equals 1-sub-H. Multiplying

  • both sides on the left by Y-inverse gives us F-of-(X-inverse) equals Y-inverse. This

  • proves that homomorphisms send inverses to inverses.

  • We now have all the tools we need to define and understand the kernel. Suppose that F

  • is not one-to-one. Then there are at least two elements in G which map to the same element

  • in H. To be concrete, suppose X1, X2, and so on are the elements which map to Y. This

  • means F-of-X1 equals Y, F-of-X2 equals Y, and so forth. Now watch what happens if we

  • multiply all of these equalities by F of X1-inverse... Since homomorphisms send inverses to inverses,

  • (F-of-X1-inverse) equals (Y-inverse). This allows us to simplify the right hand sides

  • Next, we can combine the left hand sides because F is a homomorphismBecause F is not one-to-one,

  • we find that there are multiple elements in G which all map to the identity in H. These

  • elements are called the kernel of F and we write the kernel like thisThis notation

  • emphasizes that the kernel is a property of the homomorphism, NOT the groups.

  • If I told you a homomorphism F was not an injection, it’s not obvious that there would

  • be more than one element that maps to the identity element. But there are! If F is not

  • one-to-one, then the kernel contains more than one element. Think of the kernel as a

  • way to measure the degree to which F fails to be one-to-one. We know for every homomorphism

  • the identity in G maps to the identity in H, so the kernel is never empty; it always

  • contains the identity 1-sub-G. And if the kernel only contains the identity, then F

  • is one-to-one.

  • The kernel is a subset of G, but it’s actually more than that. It’s also a SUBGROUP of

  • G. To test your understanding of homomorphisms and kernels, I’d like you to check that

  • the kernel is, in fact, a subgroup. If you get stuck, you can always ask a question in

  • the comment box below.

A homomorphism between two groups does NOT have to be a one-to-one function. If it’s

字幕と単語

ワンタップで英和辞典検索 単語をクリックすると、意味が表示されます

B1 中級

群同型のカーネル - 抽象代数 (The Kernel of a Group Homomorphism – Abstract Algebra)

  • 1 0
    林宜悉 に公開 2021 年 01 月 14 日
動画の中の単語