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Group Theory

  1. Sep 17, 2005 #1


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    Group Theory, please help!

    Okay, so I'm stuck on a couple questions from my homework, and any guidance would be much appreciated.

    1. Prove that if G is a finite group with an even number of elements,
    then there is an element x in G such that x is not the identity and
    x^2 = e.

    I know there exists some element x in G because G is not empty. And because e (the identity element in G) is unique, x is not equal to e, so x is not the identity. But I can't see how to go from x doesn't equal e to x^2 = e.

    2. Prove that if (S,*) is a finite set with a binary operation that is
    associative, has an identity, and satisfies the cancellation laws,
    then (S,*) is a group.

    I know that for (S, *) to be a group, it must be associative, there must exist an identity element e in S wrt *, and every element in S must be invertible. The first two properties follow easily from the way (S,*) is defined, but I don't know how to show the last property holds. It makes sense to me that it's true when I look at the cancellation law (if a,b,c are in S and ab = ac, the b = c), and I've tried working backwards, but then I find myself wanting to create an inverse in S, and that seems wrong.
    Last edited: Sep 17, 2005
  2. jcsd
  3. Sep 17, 2005 #2


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    When x² = e, then x = x-1. However, if x² is not equal to e, then x and x-1 are distinct. If the order of G is even, then think about what happens if x and x-1 form distinct pairs for each non-identity x.
  4. Sep 18, 2005 #3
    Suppose the elements of S are [tex]x_1, \ldots, x_n[/tex]. Let [tex]x_k[/tex] be some arbitrary element. What happens when you consider the elements

    [tex]x_1 x_k, x_2 x_k, \cdots, x_n x_k[/tex]?

    Can two of them be equal to each other, for example?
  5. Sep 19, 2005 #4


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    I see if i let [tex]x_1x_k = x_2x_k = \cdots = x_nx_k[/tex] then, through the cancellation law, I get [tex]x_1 = x_2 = \cdots = x_n[/tex]. Then, because it is given that S has an identity element, if these are all the elements of S and they are all equal, then they must all equal that identity element. So, each element in S has an inverse. But how can I claim [tex]x_1x_k = x_2x_k = \cdots = x_nx_k[/tex] ?
  6. Sep 20, 2005 #5


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    You don't need all of them to be equal. But can any pair be equal? If any pair at all is equal, then there is an i and a j distinct from each other such that [itex]x_ix_k = x_jx_k[/itex] and by the right cancellation law, [itex]x_i = x_j[/itex]. But this contradicts the stipulation that [itex]x_i[/itex] and [itex]x_j[/itex] are distinct if i and j are distinct. So it cannot the the case that [itex]x_ix_k = x_jx_k[/itex] for any i and j. If S has n distinct elements, and [itex]x_1 x_k, x_2 x_k, \cdots, x_n x_k[/itex] are n distinct elements of S, then clearly [itex]S = \{x_1 x_k, x_2 x_k, \cdots, x_n x_k\}[/itex]. Since identity is an element of S, there is some, let's call it m, such that [itex]x_mx_k = e[/itex]. This should be more than enough, you can tie up the loose ends to complete the proof.
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