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Proof of Cauchy's Theorem for Finite Groups

  1. Jul 11, 2012 #1
    I know that this is a lot, but I would love some help. My trouble is at the end of Part II.

    Theorem (Cauchy’s Theorem): Let G be a finite group, and let p be a prime divisor of the order of G, then G has element of order p.
    Proof: Suppose G is a finite group. Let the order of G be k. Let p be any prime divisor of k. Our goal is to prove that G has an element of order p. We shall do so by dividing our claim into two parts: one where G is abelian and one where G is not abelian.

    Part I: G is abelian.

    First, we shall suppose G is abelian. We will prove our claim for this case by induction. If k=1, then our claim is true trivially. Now, suppose that our claim is true for abelian finite groups of order n, where n is less than k.

    Take any a in G, where a ≠ e. Then, if the order of a is p, then we are done. Suppose the order of a is a multiple of p. That is, aqp=e for some integer q. But then, the order of aq is p and, again, we are done. Thus, suppose the order of a is not equal to a multiple of p. Now, let us consider the group G/<a>. First, let |<a>|=n (and consequently, n will also be the order of a) and note that, by Lagrange’s Theorem, n|k. We recall that this quotient group is the group of all cosets of <a> in G. Thus, |G/<a>|=(G:<a>)= k/n. Next, note that if n=k, the n is a multiple of p, which is a contradiction! Thus, n<k. Now, let k=ln, where l is an integer. Since p|k, then p|ln, but p does not divide n, hence, p|l, and l is the order of G/<a>. Now, we have that G/<a> has order l<k and prime divisor p. Thus, by our assumption G/<a> has an element of order p.


    From a previous exercise, we have shown that if H is a normal subgroup of a group G, then the order of the element Ha in G/H is a divisor of the order of a in G. Thus, since p is the order of an element in G/<a>, say x<a>, p must also be a divisor of an element in G, namely x. Then, let m be the order of x. Then, m=qp for some integer q. That is, xm=xqp=e. This implies that xq in G has order p, which is our desire result.

    Part II: G is not abelian.


    Now, we suppose G is not abelian with order k. Again, we will proceed with induction. The case for k=1 remains true by default. Thus, suppose that our claim is true for any finite nonabelian group G with order n<k. Let C be the center of G for each a in G, let Ca be the centralizer. Let us recall our class equation for G. If |C|=c, then, |G|=k=c+ks+….+ kt.

    Suppose is p is not a factor of | Ca| for any a in G, where a is not in the center. Consider the quotient group G/ Ca. There a one-to-one correspondence between the set of all conjugates of a and the set of all cosets of Ca. That is, the set of all cosets of the centralizer give rises to a partition of G. Now, let n be the order of Ca. Then, | G/ Ca| = (G: Ca)= k/n. Now, if n=k, then p divides n, which is a contradiction. Hence, n<k. Let k=ln for some integer l. We note that l is the order of G/ Ca. We also note that p|l since p|k and p|ln, but p does not divide n. Since p divides | G/Ca|=(G/Ca) for each a in G, a ,not the center, every conjugacy class is a multiple of p [Remember, our bijection between the conjugacy classes of a and the cosets of Ca]. Then, |G|=k=c+ks+….+ kt =c+q1p+…qnp for integers q1,…,qn.

    If we solve for c we observe that we get k-q1p…-qnp=c. We can factor the prime p out of everything on the right. Leaving p times an integer equal to c. Thus, p is a factor of c. .......
    _______________________End Attempt__________________________

    I am having a hard time expressing why G has an element of order p from this. Any hints? By the way, after I finish this, would anyone care to share a "slick" verison of this proof.

    Also, this proof is developed from a series of questions in Pinter's book. I wanted to make one cohesive proof to make it my own in a sense. :) At least I tried!


    **Edit: I wanted to add that this is a type up from my notebook. There may be typos!
     
  2. jcsd
  3. Jul 11, 2012 #2
    I didn't check Part I because you said the trouble was in Part II. After the part
    , I would proceed as follows:
    Firstly, suppose [itex]c[/itex] is divisible by [itex]p[/itex]. Because [itex]C \neq G[/itex], we may use the induction hypothesis to find an element of order [itex]p[/itex] in [itex]C[/itex]. This element also has order [itex]p[/itex] in [itex]G[/itex].
    Suppose then that [itex]c[/itex] is not divisible by [itex]p[/itex]. From the class equation you can conclude that there must be a conjugacy class [itex]Cl(a)[/itex] with [itex]a \notin C[/itex] such that [itex]|Cl(a)|[/itex] is not divisible by [itex]p[/itex]. We have the equality (from your one-to-one correspondence) [itex]k = |Cl(a)| \cdot |C_a|[/itex]. Thus [itex]|C_a|[/itex] is divisible by [itex]p[/itex]. Because [itex]a \notin C[/itex], [itex]C_a \neq G[/itex], and we may apply the induction hypothesis to [itex]C_a[/itex].

    The proof you give is the 'slickest' one I know of that uses only very elementary techniques. A more general result, from which Cauchy's theorem follows directly, is Sylow's theorem (there are actually 3). Sylow's first theorem states the following:
    Let [itex]G[/itex] be a finite group. Write [itex]|G| = p^r m[/itex], where [itex]p[/itex] is a certain prime number and [itex]r,m \in \mathbb{N}[/itex]. For each [itex]0 \leq s \leq r[/itex] there exists a subgroup [itex]H[/itex] of [itex]G[/itex] with [itex]|H| = p^s[/itex].
     
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