Difference between revisions of "Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup"

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(Case that p does not divide n)
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===Case that <math>p</math> does not divide <math>n</math>===
 
===Case that <math>p</math> does not divide <math>n</math>===
  
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# <math>n</math> divides the order of <math>G/K</math>: This follows because <math>n</math> divides the order of <math>G</math> and <math>n</math> is relatively prime to <math>p</math>.
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# The number of elements <math>z \in G/K</math> such that <math>z^n = 1</math> is <math>rn</math> for some natural number <math>r</math>: This follows from fact (2).
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# For every coset <math>z = xK</math> of <math>K</math> in <math>G</math> such that <math>z^n = 1</math>, there exists an element <math>y \in xK</math> such that <math>y^n = 1</math>: Start with some element <math>x</math> in the coset. Let <math>a = x^n</math>. By definition, <math>a \in K</math>. Now, find an integer <math>m</math> such that <math>p|mn - 1</math> (this can be done since <math>p</math> and <math>n</math> are relatively prime). Let <math>b = a^m</math> and let <math>y = xb^{-1}</math>. Note that since <math>b</math> is a power of <math>a</math>, which in turn is a power of <math>x</math>, <math>x</math> commutes with <math>b</math>. Thus, if <math>y = xb^{-1}</math>, then <math>y^n = (xb^{-1})^n = x^nb^{-n} = ab^{-n} = a^{1-mn} = 1</math> since <math>1 - mn</math> is a multiple of <math>p</math>, and <math>a^p = 1</math>. Further, <math>b^{-1} \in K</math>, so <math>y \in xK</math>. Thus, we have found an element of <math>xK</math> whose order divides <math>n</math>.
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# There are at least <math>rn</math> solutions to <math>x^n = 1</math> in <math>G</math>: This follows from steps (2) and (3).
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# <math>r = 1</math>, and there are exactly <math>n</math> solutions to <math>z^n = 1</math> in <math>G/K</math>: This follows from the previous step and the assumption that there are exactly <math>n</math> solutions.
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# The set of solutions to <math>z^n = 1</math> in <math>G/K</math> forms a subgroup, say <math>H/K</math>: By fact (3), <math>G/K</math> is solvable, so we can apply the induction hypothesis to the conclusion of the previous step.
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# The inverse image, say <math>H</math> of this subgroup <math>H/K</math> of <math>G/K</math> in <math>G</math> is a group of order <math>p^in</math>, where <math>n</math> is relatively prime to <math>p</math>.
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# <math>H</math> has a <math>p</math>-complement, i.e., a Hall subgroup of order <math>n</math>: This follows from fact (4).
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# This Hall subgroup of order <math>n</math> is precisely equal to the set of solutions to <math>x^n = 1</math>: All elements in the Hall subgroup of order <math>n</math> are solutions to <math>x^n = 1</math>, and we know that there are exactly <math>n</math> solutions. Hence, these must be the only solutions.
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==References==
 
==References==
  

Revision as of 16:24, 5 March 2009

Statement

Suppose G is a Finite solvable group (?), and n is a natural number dividing the order of G. If there are exactly n elements of G whose n^{th} power is the identity element, then these elements form a subgroup. This subgroup is clearly a normal subgroup. In fact, it is a characteristic subgroup, and even better, is a fully characteristic subgroup.

Facts used

  1. Minimal normal implies elementary abelian in finite solvable
  2. Number of nth roots is a multiple of n
  3. Solvability is quotient-closed

Proof

Given: A group G. A natural number n. S is the set of n^{th} roots of unity in G (i.e., the set of elements of order dividing n, and S has n elements.

To prove: S is a subgroup.

Proof: By fact (1), G has a nontrivial elementary abelian normal subgroup K of order p^i for some prime power p^i dividing the order of G. We split the proof into two cases, based on whether p divides n or not.

Case that p divides n

In this case, K \subseteq S. Let n = p^jm and |G| = g = p^kl where m,l are relatively prime to p. Clearly, j \le k and m|l.

  1. Let u = p^{\max(j-i,0)}m. Then, the order of G/K is divisible by u: This is clear from the fact that the order of G/K equals p^{k-i}l, and since m|l and j \le k, this follows.
  2. up divides n: This follows essentially from the fact that exponent of p in u is strictly less than that in n.
  3. There exists some natural number r such that G/K has ru elements of order dividing u: This follows from fact (2).
  4. Let \varphi:G \to G/K be the quotient map. The set of elements x \in G such that \varphi(x)^u = 1 has size rup^i: This follows from step (3).
  5. If \varphi(x)^u = 1, we have x^{up} = 1: This follows from the fact that the order of any element in K is either 1 or p.
  6. If \varphi(x)^u = 1, we have x^n = 1: This follows from the previous step and step (2), which says that up|n.
  7. r = 1, j \ge i and x^n = 1 \iff x^u \in K: By steps (3) and (5), there are at least rup^i elements x for which x^n = 1. Thus, rup^i \le n. Simplifying, we obtain that rp^{\max(i,j)} \le p^j. This forces r = 1 and j \ge i. Further, since equality holds even under these assumptions, we conclude that the rup^i solutions to x^u \in K exhaust all solutions to x^n = 1.
  8. The set of elements y \in G/K such that y^u = 1 forms a subgroup. In particular, S/K is a subgroup: By the previous step, r = 1, so there are exactly u solutions in G/K to y^u = 1. By fact (3), G/K is solvable, so the induction hypothesis applies, and the set of solutions forms a subgroup.
  9. S is a subgroup: It is the inverse image in G of a subgroup of G/K.

Case that p does not divide n

  1. n divides the order of G/K: This follows because n divides the order of G and n is relatively prime to p.
  2. The number of elements z \in G/K such that z^n = 1 is rn for some natural number r: This follows from fact (2).
  3. For every coset z = xK of K in G such that z^n = 1, there exists an element y \in xK such that y^n = 1: Start with some element x in the coset. Let a = x^n. By definition, a \in K. Now, find an integer m such that p|mn - 1 (this can be done since p and n are relatively prime). Let b = a^m and let y = xb^{-1}. Note that since b is a power of a, which in turn is a power of x, x commutes with b. Thus, if y = xb^{-1}, then y^n = (xb^{-1})^n = x^nb^{-n} = ab^{-n} = a^{1-mn} = 1 since 1 - mn is a multiple of p, and a^p = 1. Further, b^{-1} \in K, so y \in xK. Thus, we have found an element of xK whose order divides n.
  4. There are at least rn solutions to x^n = 1 in G: This follows from steps (2) and (3).
  5. r = 1, and there are exactly n solutions to z^n = 1 in G/K: This follows from the previous step and the assumption that there are exactly n solutions.
  6. The set of solutions to z^n = 1 in G/K forms a subgroup, say H/K: By fact (3), G/K is solvable, so we can apply the induction hypothesis to the conclusion of the previous step.
  7. The inverse image, say H of this subgroup H/K of G/K in G is a group of order p^in, where n is relatively prime to p.
  8. H has a p-complement, i.e., a Hall subgroup of order n: This follows from fact (4).
  9. This Hall subgroup of order n is precisely equal to the set of solutions to x^n = 1: All elements in the Hall subgroup of order n are solutions to x^n = 1, and we know that there are exactly n solutions. Hence, these must be the only solutions.

References

Textbook references