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 divides n)
(Proof)
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! No. !! Assertion/construction !! Facts used !! Given data/assumptions used !! Previous steps used !! Explanation
 
! No. !! Assertion/construction !! Facts used !! Given data/assumptions used !! Previous steps used !! Explanation
 
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| 1 || <math>K \subseteq S</math> || || <math>p</math> divides <math>n</math>, <math>K</math> is elementary abelian of order <math>p^i</math>, <math>S</math> comprises all elements whose <math>n^{th}</math> power is the identity || || <toggledisplay>Since <math>K</math> is elementary abelian of order <math>p^i</math>, this in particular means it has exponent <math>p</math>, so every element in it has order <math>1</math> or <math>p</math>. <math>S</math> contains all elements of order dividing <math>n</math>. Thus, it must contain all elements of <math>K</math>.</toggledisplay>
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| 1 || The order of every element in <math>K</math> is either <math>1</math> or <math>p</math> || || <math>K</math> is elementary abelian of order <math>p^i</math> || ||
 
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| 2 || Let <math>n = p^jm</math> and <math>|G| = N = p^kl</math> where <math>m,l</math> are relatively prime to <math>p</math>. Then, <math>j \le k</math> and <math>m|l</math>. || || <math>n</math> divides the order of <math>G</math> || || <toggledisplay>Since <math>n</math> divides the order <math>N</math> of <math>G</math>, the <math>p</math>-part of <math>n</math> must divide the <math>p</math>-part of <math>N</math>, and the <math>p'</math>-part of <math>n</math> must divide the <math>p'</math>-part of <math>N</math>.</toggledisplay>
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| 2 || <math>K \subseteq S</math> || || <math>p</math> divides <math>n</math>, <math>S</math> comprises all elements whose <math>n^{th}</math> power is the identity || Step (1) || <toggledisplay>Since <math>K</math> is elementary abelian of order <math>p^i</math>, this in particular means it has exponent <math>p</math>, so every element in it has order <math>1</math> or <math>p</math>. <math>S</math> contains all elements of order dividing <math>n</math>. Thus, it must contain all elements of <math>K</math>.</toggledisplay>
 
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| 3 || Let <math>u = p^{\max(j-i,0)}m</math>. Then, the order of <math>G/K</math> is divisible by <math>u</math>. || || || || <toggledisplay>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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| 3 || Let <math>n = p^jm</math> and <math>|G| = N = p^kl</math> where <math>m,l</math> are relatively prime to <math>p</math>. Then, <math>j \le k</math> and <math>m|l</math>. || || <math>n</math> divides the order of <math>G</math> || || <toggledisplay>Since <math>n</math> divides the order <math>N</math> of <math>G</math>, the <math>p</math>-part of <math>n</math> must divide the <math>p</math>-part of <math>N</math>, and the <math>p'</math>-part of <math>n</math> must divide the <math>p'</math>-part of <math>N</math>.</toggledisplay>
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| 4 || Let <math>u = p^{\max(j-i,0)}m</math>. Then, the order of <math>G/K</math> is divisible by <math>u</math>. || || || || <toggledisplay>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.</toggledisplay>
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| 5 || <math>up</math> divides <math>n</math> || || || || <toggledisplay>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>.</toggledisplay>
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| 6 || 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> || Fact (2) || || Step (1) ||
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| 7 || 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</math> is the identity element has size <math>rup^i</math> || || || step (3) ||
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| 8 || If <math>\varphi(x)^u = 1</math> is the identity element, then <math>x^{up} = 1</math> || || || ||
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| 9||  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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| 10 || <math>r = 1, j \ge i</math> and <math>x^n</math> is the identity element iff <math>x^u \in K</math> || || || ||<toggledisplay>Clean up: 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>.</toggledisplay>
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| 11 || The set of elements <math>y \in G/K</math> such that <math>y^u </math> is the identity element forms a subgroup. In particular, <math>S/K</math> is a subgroup || || || Step (10) || <toggledisplay>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.</toggledisplay>
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| 12 || <math>S</math> is a subgroup  || || || Step (11) || <toggledisplay>It is the inverse image in <math>G</math> of a subgroup of <math>G/K</math></toggledisplay>
 
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(remaining stuff needs to be cleaned up. Numbers may be off).
 
 
# <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>.
 
# 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).
 
# 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).
 
# 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>.
 
# 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>.
 
# <math>r = 1, 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>.
 
# 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.
 
# <math>S</math> is a subgroup: It is the inverse image in <math>G</math> of a subgroup of <math>G/K</math>.
 
  
 
===Case that <math>p</math> does not divide <math>n</math>===
 
===Case that <math>p</math> does not divide <math>n</math>===

Revision as of 22:53, 26 May 2011

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, a homomorph-containing subgroup and a variety-containing subgroup.

Related facts

Related conjectures

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
  4. Hall subgroups exist in finite solvable

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

No. Assertion/construction Facts used Given data/assumptions used Previous steps used Explanation
1 The order of every element in K is either 1 or p K is elementary abelian of order p^i
2 K \subseteq S p divides n, S comprises all elements whose n^{th} power is the identity Step (1) [SHOW MORE]
3 Let n = p^jm and |G| = N = p^kl where m,l are relatively prime to p. Then, j \le k and m|l. n divides the order of G [SHOW MORE]
4 Let u = p^{\max(j-i,0)}m. Then, the order of G/K is divisible by u. [SHOW MORE]
5 up divides n [SHOW MORE]
6 There exists some natural number r such that G/K has ru elements of order dividing u Fact (2) Step (1)
7 Let \varphi:G \to G/K be the quotient map. The set of elements x \in G such that \varphi(x)^u is the identity element has size rup^i step (3)
8 If \varphi(x)^u = 1 is the identity element, then x^{up} = 1
9 If \varphi(x)^u = 1, we have x^n = 1: This follows from the previous step and step (2), which says that up|n.
10 r = 1, j \ge i and x^n is the identity element iff x^u \in K [SHOW MORE]
11 The set of elements y \in G/K such that y^u is the identity element forms a subgroup. In particular, S/K is a subgroup Step (10) [SHOW MORE]
12 S is a subgroup Step (11) [SHOW MORE]

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.

Proof that the subgroup is normal, characteristic and fully characteristic

This follows from the fact that if \alpha is a homomorphism from G to any subgroup of G, then (\alpha(g))^n = \alpha(g^n). Hence, if g^n = 1, so is (\alpha(g))^n.

References

Textbook references