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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* [[Number of nth roots is a multiple of n]]
 
* [[Number of nth roots is a multiple of n]]
 
* [[Number of nth roots of any conjugacy class is a multiple of n]]
 
* [[Number of nth roots of any conjugacy class is a multiple of n]]
* [[Exactly n elements of order dividing n implies every finite subgroup is cyclic]]
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* [[At most n elements of order dividing n implies every finite subgroup is cyclic]]
  
 
===Related conjectures===
 
===Related conjectures===
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# [[uses::Number of nth roots is a multiple of n]]
 
# [[uses::Number of nth roots is a multiple of n]]
 
# [[uses::Solvability is quotient-closed]]
 
# [[uses::Solvability is quotient-closed]]
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# [[uses::Solvability is subgroup-closed]]
 
# [[uses::Hall subgroups exist in finite solvable]]
 
# [[uses::Hall subgroups exist in finite solvable]]
 
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# [[uses::Order of element divides order of group]]
 
==Proof==
 
==Proof==
  
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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>===
  
# <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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{| class="sortable" border="1"
# 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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! No. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
# 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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| 1 || <math>n</math> divides the order of <math>G/K</math> || Fact (2) || <math>n</math> divides the order of <math>G</math>, <math>K</math> has order <math>p^i</math>, <math>p</math> does not divide <math>n</math> || || By Fact (2), the order of <math>G/K</math> is <math>|G|/|K|</math> which is <math>N/p^i</math>. We're given that <math>n</math> divides <math>N</math>, so since <math>n</math> is relatively prime to <math>p</math>, we must also have <math>n</math> divides <math>N/p^i</math>.
# <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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| 2 || The number of elements <math>z \in G/K</math> such that <math>z^n</math> is the identity is <math>rn</math> for some natural number <math>r</math> || Fact (3) || || Step (1) || Previous step+Fact direct
# 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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| 3 || For every coset <math>z = xK</math> of <math>K</math> in <math>G</math> such that <math>z^n</math> is the identity in <math>G/K</math>, there exists an element <math>y \in xK</math> such that <math>y^n</math> is the identity in <math>G</math> || || <math>p</math> does not divide <math>n</math> || || <toggledisplay>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>.</toggledisplay>
# 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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|-
 
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| 4 || There are at least <math>rn</math> solutions to <math>x^n</math> being the identity in <math>G</math> || || || Steps (2), (3) || Step-combination direct
===Proof that the subgroup is normal, characteristic and fully characteristic===
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|-
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| 5 || <math>r = 1</math>, and there are exactly <math>n</math> solutions to <math>z^n</math> being the identity in <math>G/K</math> || || There are exactly <math>n</math> elements of order dividing <math>n</math> || Steps (2), (4) || <toggledisplay>By Step (4) and the given data, we get <math>rn \le n</math>. Since <math>r \ge 1</math>, we are forced to have <math>r = 1</math>. Plugging in Step (2) gives the other half of the assertion.</toggledisplay>
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|-
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| 6 || The set of elements in <math>G/K</math> of order dividing <math>n</math> forms a subgroup of order <math>n</math> in <math>G/K</math> || Fact (4) || <math>G</math> is solvable, induction hypothesis || Step (5) || <toggledisplay>By Fact (4), <math>G/K</math> is solvable. Combining Step (5) and the induction hypothesis, we are done.</toggledisplay>
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|-
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| 7|| The inverse image in <math>G</math> of the subgroup of <math>G/K</math> obtained in Step (6) is a subgroup of <math>G</math>, say <math>H</math>. It has order <math>p^in</math>. || Fact (2) || <math>K</math> has order <math>p^i</math>.  || || <toggledisplay>The inverse image of a subgroup is a subgroup. Call the inverse image <math>H</math>. Then, by Step (6), <math>H/K</math> has size <math>n</math>, so by Fact (2), <math>|H| = |K||H/K| = p^in</math>.</toggledisplay>
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|-
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| 8 || <math>H</math> has a <math>p</math>-complement, i.e., a Hall subgroup of order <math>n</math>. Call this <math>p</math>-complement <math>L</math>. || Facts (5), (6) || <math>p</math> does not divide <math>n</math>, <math>G</math> is solvable || Step (7) || <toggledisplay>By Fact (5), since <math>G</math> is solvable, <math>H</math> is solvable. By Fact (6), it must therefore have a <math>p</math>-complement.
 +
|-
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| 9 || <math>L = S</math>, i.e., <math>L</math> is precisely the set of elements of order dividing <math>n</math>. Since it's a subgroup, this completes the proof. || Fact (7) || There are exactly <math>n</math> elements of order dividing <math>n</math>. || || <toggledisplay>Since <math>L</math> has order dividing <math>n</math>, all elements in there also have order dividing <math>n</math> by Fact (7). So <math>L \subseteq S</math>. By size considerations, we have <math>L = S</math>.
 +
|}
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===Proof that the subgroup is normal, characteristic and fully invariant===
  
 
This follows from the fact that if <math>\alpha</math> is a homomorphism from <math>G</math> to any subgroup of <math>G</math>, then <math>(\alpha(g))^n = \alpha(g^n)</math>. Hence, if <math>g^n = 1</math>, so is <math>(\alpha(g))^n</math>.
 
This follows from the fact that if <math>\alpha</math> is a homomorphism from <math>G</math> to any subgroup of <math>G</math>, then <math>(\alpha(g))^n = \alpha(g^n)</math>. Hence, if <math>g^n = 1</math>, so is <math>(\alpha(g))^n</math>.

Revision as of 00:41, 27 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. Lagrange's theorem: We in particular are interested in the version which states that if A is a subgroup of B, |B/A| = |B|/|A|, where B/A is the quotient set. When A is a normal subgroup, B/A is the quotient group.
  3. Number of nth roots is a multiple of n
  4. Solvability is quotient-closed
  5. Solvability is subgroup-closed
  6. Hall subgroups exist in finite solvable
  7. Order of element divides order of group

Proof

This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).

We do the proof by induction. Specifically, we assume that the statement holds true for all finite solvable groups of strictly smaller orders than N.

PROOF OF INDUCTIVE STEP:

Given: A finite solvable group G of order N. 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 G/K has order p^{k-i}l Fact (2) K has order p^i Step (3) [SHOW MORE]
5 Let u = p^{\max(j-i,0)}m. Then, u divides the order of G/K. Steps (3), (4) [SHOW MORE]
6 There exists some natural number r such that G/K has ru elements of order dividing u Fact (3) Step (5) [SHOW MORE]
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 of G/K has size rup^i K has order p^i Step (6) [SHOW MORE]
8 If \varphi(x)^u is the identity element in G/K, then x^{up} is the identity element in G/K. Step (1) [SHOW MORE]
9 up divides n. p divides n Steps (3), (5) [SHOW MORE]
12 r = 1, j \ge i and x^n is the identity element iff x^u \in K. In other words, S = \{ x \mid x^u \in K \}. Steps (5), (11) [SHOW MORE]
13 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 Fact (4) inductive hypothesis, G is solvable Step (12) [SHOW MORE]
14 S is a subgroup Step (13) [SHOW MORE]

Case that p does not divide n

No. Assertion/construction Facts used Given data used Previous steps used Explanation
1 n divides the order of G/K Fact (2) n divides the order of G, K has order p^i, p does not divide n By Fact (2), the order of G/K is |G|/|K| which is N/p^i. We're given that n divides N, so since n is relatively prime to p, we must also have n divides N/p^i.
2 The number of elements z \in G/K such that z^n is the identity is rn for some natural number r Fact (3) Step (1) Previous step+Fact direct
3 For every coset z = xK of K in G such that z^n is the identity in G/K, there exists an element y \in xK such that y^n is the identity in G p does not divide n [SHOW MORE]
4 There are at least rn solutions to x^n being the identity in G Steps (2), (3) Step-combination direct
5 r = 1, and there are exactly n solutions to z^n being the identity in G/K There are exactly n elements of order dividing n Steps (2), (4) [SHOW MORE]
6 The set of elements in G/K of order dividing n forms a subgroup of order n in G/K Fact (4) G is solvable, induction hypothesis Step (5) [SHOW MORE]
7 The inverse image in G of the subgroup of G/K obtained in Step (6) is a subgroup of G, say H. It has order p^in. Fact (2) K has order p^i. [SHOW MORE]
8 H has a p-complement, i.e., a Hall subgroup of order n. Call this p-complement L. Facts (5), (6) p does not divide n, G is solvable Step (7) <toggledisplay>By Fact (5), since G is solvable, H is solvable. By Fact (6), it must therefore have a p-complement.
9 L = S, i.e., L is precisely the set of elements of order dividing n. Since it's a subgroup, this completes the proof. Fact (7) There are exactly n elements of order dividing n. <toggledisplay>Since L has order dividing n, all elements in there also have order dividing n by Fact (7). So L \subseteq S. By size considerations, we have L = S.

Proof that the subgroup is normal, characteristic and fully invariant

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