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Homework Help: Opposite groups - show it is a group

  1. Aug 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose [tex](G, \circ)[/tex] is a group. Define an operation [tex]\star[/tex] on [tex]G[/tex] by [tex]a \star b = b \circ a[/tex] for all [tex]a,b \in G[/tex]. Show that [tex](G,\star)[/tex] is a group.

    3. The attempt at a solution

    So, I have to show that [tex](G,\star)[/tex] satisfies the associativity, identity, inverse and closure conditions:

    Associativity: Let a,b,c be elements in G.

    [tex](a \star b) \star c = (b \circ a) \star c[/tex]

    [tex]= c \circ (b \circ a) = (c \circ b) \circ a[/tex]

    [tex]= a \star (b \star c)[/tex].

    Identity: [tex](a \star e) = (e \circ a) = (a \circ e)=(e \star a) = a[/tex].

    Inverse: y is an inverse of a then [tex](a \star y)= (y \circ a) = (y \star a) = e[/tex].

    Are these correct so far??

    And for the closure, do I just need to show that any element of [tex](G,\star)[/tex] is in [tex]G[/tex]? I know that [tex](a \star b)=(b \circ a)[/tex] and [tex](b \circ a) \in (G, \circ)[/tex] so is that all I need to say?
     
  2. jcsd
  3. Aug 15, 2010 #2
    Re: Groups

    ya, thats it.
     
  4. Aug 16, 2010 #3
    Re: Groups

    Really? Even the explanation for the closure is correct?
     
  5. Aug 16, 2010 #4

    HallsofIvy

    User Avatar
    Science Advisor

    Re: Groups

    Yes, [tex](a \star b)=(b \circ a)[/tex] and, because G is a group, [itex]\circ[/itex] is closed so that [itex]b\circ a\in G[/itex] for all a,b so [itex]a\star b\in G[/itex] for all a, b.
     
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