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Closure of groups

  1. Sep 14, 2013 #1
    Let G be a group and my book defines closure as: For all a,bε G the element a*b is a well defined element of G. Then G is called a group. When they say well defined element does that mean I have to show a*b is well defined and it is a element of the group? Or do I just show a*b is closed under *(the operation)?
  2. jcsd
  3. Sep 14, 2013 #2
    I've always found the closure axiom a bit silly. It's implied if you just write [itex]*: G\times G\to G[/itex]. All it means is that, given [itex]a,b\in G[/itex], there's a thing named [itex]a*b[/itex], and that whatever this thing is, it belongs to [itex]G[/itex].
  4. Sep 14, 2013 #3
    Thanks. I saw that other abstract algebra books have it defined as how you said it. My book apparently has it defined a little differently.
  5. Sep 14, 2013 #4
    This will often depend on the context. * is by definition a binary function *:GxG→G, and a function is well-defined by definition.. I think the problem is best illustrated by examples:

    Let G = {[x]: x is an integer not a multiple of 3}, where [x] = {integers y s.t. x~y}, where we write x~y if x and y leave the same remainder upon division by 3 (alternatively, x-y is divisible by 3). The elements of G are sets called equivalence classes of Z modulo 3, and we can easily verify that G={[1],[2]}. Now define a binary operation on G by [a]*=[ab]. At first glance it might not be obvious that * is well-defined since [a] and have many different representations, and [ab] might depend on which of these representations we choose. For example [2]=[5] and [1]=[31], so we better make sure that we get [2]*[1]=[5]*[31] with how we defined *! We can verify that [2]*[1]=[2]=[155]=[5]*[31], since 155 leaves remainder 2 upon division by 3. Indeed, we can prove that [a]*=[ab] gives the same element of G no matter how we choose to write [a] and , i.e. * is well defined!

    Edit: G={[0],[1],[2]} is not a group ([0] is not invertible), edited so that G={[1],[2]}
    Last edited: Sep 14, 2013
  6. Sep 14, 2013 #5
    Yes its just like showing a function is well defined. I wasn't sure if it just suffices to show that (G,*) is closed under the operation *.
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