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Kernal of a subgroup

  1. Nov 17, 2011 #1
    I'm having a bit of a problem understanding the kernal of a subgroup. It appears to always be equal to the identity element. But that doesn't seem to make much sense. Anyone care to clarify this for me?
     
  2. jcsd
  3. Nov 17, 2011 #2

    Deveno

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    not always. and we don't get a kernel automatically as a subgroup, there has to be a homomorphism involved (the kernel is a subgroup that "goes with" the homomorphism). here is a simple example:

    let G = {e,a,b,ab} where a2 = b2 = e, and ab = ba,

    and let H = {-1,1}.

    define the map φ:G→H by:

    φ(e) = 1
    φ(a) = 1
    φ(b) = -1
    φ(ab) = -1.

    is φ a homomorphism? let's check:

    since φ maps e to 1, we don't have to check any products involving the identity of G (you can, if you really want to) and G is abelian, so we only have to actually check 3 products:

    a2, b2, and ab.

    φ(a2) = φ(e) = 1 = (1)(1) = φ(a)φ(a).
    φ(b2) = φ(e) = 1 = (-1)(-1) = φ(b)φ(b).
    φ(ab) = -1 = (1)(-1) = φ(a)φ(b).

    so yes, φ is indeed a homomorphism.

    what is ker(φ)? its {g in G : φ(g) = 1}, which is {e,a}, which is not just the identity of G.
     
  4. Nov 17, 2011 #3
    Thanks, yes, I didn't mention the homomorphism part.

    I understand your reply but I'm looking at {e,a}.

    a^2 = 1 so a = 1 I would think and so does e = 1.

    Are you saying because the kernal has two elements and they both aren't e that shows the kernal isn't always equal to the identity? I see that but it seems at the end of the day we still wind up with just ones in the kernal (represented by e and a).

    Maybe by seeing what I'm struggling with here my delima might be clearer.

    Like, is there ever a 2 or a 3 in the kernal?
     
  5. Nov 17, 2011 #4

    Deveno

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    here is a slightly more involved example:

    let G = Z, the integers under addition, and

    let H = Z/5Z, the integers under addition modulo 5.

    the homomorphism we have is φ(k) = k (mod 5).

    ker(φ) = { 5k : k in Z}, so, for example 5,10,15,-5,-10 are all in ker(φ), as well, of course 0.

    this is because 0 + 5Z = 5 + 5Z = 5Z, etc.

    the kernel of φ lives in G. φ takes everything in ker(φ), and shrinks it down to the identity of H.

    in other words, homomorphisms never make a group "bigger", but they sometimes make a group "smaller". the kernel measures "how much" shrinkage is going on (how much of G we "mod out").

    it is a well-known theorem, called the first isomorphism theorem (you may not have covered it yet), that if K = ker(φ), and gK is a coset of K in G, then φ takes the entire coset gK to just a single element, namely φ(g).
     
  6. Nov 17, 2011 #5
    Thaks. Cosets are in the next section.

    I'm going to have to spend some time with this to digest it, but I really appreciate it.
     
  7. Nov 19, 2011 #6
    OK, it took some scratch work but I think I understand it. Basically the kernal consists of the elements of the domain that by way of the mapping, show up as identity elements in the range.

    Thanks again for the help.
     
  8. Nov 20, 2011 #7

    Deveno

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    yes. for any mapping f:A→B, and any subset U of B, we can form the set:

    f-1(U) = {a in A: f(a) is in U}.

    such a set is called a pre-image (under f) of U.

    if f:G→H is a homomorphism, in the case of a kernel,

    ker(f) = f-1({eH}), the pre-image of the identity of H

    (so our "U" in this case is a 1-element set).
     
  9. Nov 22, 2011 #8
    The kernel of a homomorphism contains only the identity if and only if the homomorphism is injective (one to one). You have probably seen this as a theorem in your course and/or textbook. Of course, we can think of lots of homomorphisms which are not injective, for example what about [itex]\phi:\mathbb{Z}\rightarrow \mathbb{Z}_n[/itex] given by by [itex]\phi(k)=[k][/itex]. [itex]k\in \mathbb{Z}[/itex] is in the kernel of this homomorphism if and only if [itex][k]=0[/itex]. Since [itex][k]=0 \iff n|k [/itex], any multiple of n is in the kernel. This tells us that all multiples of n map to zero, or equivalently the homomorphism is not one to one.
    Edit: I just realized Deveno gave a specific example of this homomorphism already.
     
  10. Nov 22, 2011 #9
    Thanks. More is always better!
     
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