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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(New page: ==Statement== Suppose <math>G</math> is a fact about::finite solvable group, and <math>n</math> is a natural number dividing the order of <math>G</math>. If there are exactly <math>n<...)
 
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# [[uses::Minimal normal implies elementary abelian in finite solvable]]
 
# [[uses::Minimal normal implies elementary abelian in finite solvable]]
 
# [[uses::Number of nth roots is a multiple of n]]
 
# [[uses::Number of nth roots is a multiple of n]]
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# [[uses::Solvability is quotient-closed]]
  
 
==Proof==
 
==Proof==
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'''Given''': A group <math>G</math>. A natural number <math>n</math>. <math>S</math> is the set of <math>n^{th}</math> roots of unity in <math>G</math> (i.e., the set of elements of order dividing <math>n</math>, and <math>S</math> has <math>n</math> elements.
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'''To prove''': <math>S</math> is a subgroup.
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'''Proof''': By fact (1), <math>G</math> has a nontrivial elementary abelian normal subgroup <math>K</math> of order <math>p^i</math> for some prime power <math>p^i</math> dividing the order of <math>G</math>. We split the proof into two cases, based on whether <math>p</math> divides <math>n</math> or not.
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===Case that <math>p</math> divides <math>n</math>===
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In this case, <math>K \subseteq S</math>. Let <math>n = p^jm</math> and <math>|G| = g = p^kl</math> where <math>m,l</math> are relatively prime to <math>p</math>. Clearly, <math>j \le k</math> and <math>m|l</math>.
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# Let <math>u = p^{\max(j-i,0)}m</math>. Then, the order of <math>G/K</math> is divisible by <math>u</math>: This is clear from the fact that the order of <math>G/K</math> equals <math>p^{k-i}l</math>, and since <math>m|l</math> and <math>j \le k</math>, this follows.
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# <math>up</math> divides <math>n</math>: This follows essentially from the fact that exponent of <math>p</math> in <math>u</math> is strictly less than that in <math>n</math>.
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# There exists some natural number <math>r</math> such that <math>G/K</math> has <math>ru</math> elements of order dividing <math>u</math>: This follows from fact (2).
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# Let <math>\varphi:G \to G/K</math> be the quotient map. The set of elements <math>x \in G</math> such that <math>\varphi(x)^u = 1</math> has size <math>rup^i</math>: This follows from step (3).
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# If <math>\varphi(x)^u = 1</math>, we have <math>x^{up} = 1</math>: This follows from the fact that the order of any element in <math>K</math> is either <math>1</math> or <math>p</math>.
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# If <math>\varphi(x)^u = 1</math>, we have <math>x^n = 1</math>: This follows from the previous step and step (2), which says that <math>up|n</math>.
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# <math>r = 1</math>, <math>j \ge i</math> and <math>x^n = 1 \iff x^u \in K</math>: By steps (3) and (5), there are at least <math>rup^i</math> elements <math>x</math> for which <math>x^n = 1</math>. Thus, <math>rup^i \le n</math>. Simplifying, we obtain that <math>rp^{\max(i,j)\} \le p^j</math>. This forces <math>r = 1</math> and <math>j \ge i</math>. Further, since equality holds even under these assumptions, we conclude that the <math>rup^i</math> solutions to <math>x^u \in K</math> exhaust all solutions to <math>x^n = 1</math>.
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# The set of elements <math>y \in G/K</math> such that <math>y^u = 1</math> forms a subgroup. In particular, <math>S/K</math> is a subgroup: By the previous step, <math>r = 1</math>, so there are exactly <math>u</math> solutions in <math>G/K</math> to <math>y^u = 1</math>. By fact (3), <math>G/K</math> is solvable, so the induction hypothesis applies, and the set of solutions forms a subgroup.
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# <math>S</math> is a subgroup: It is the inverse image in <math>G</math> of a subgroup of <math>G/K</math>.
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===Case that <math>p</math> does not divide <math>n</math>===
  
 
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Revision as of 16:03, 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 Failed to parse (syntax error): 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

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References

Textbook references