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 <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>. |
+ | # 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). | ||
+ | # 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>. | ||
+ | # 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). | ||
+ | # <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. | ||
+ | # 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. | ||
+ | # 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>. | ||
+ | # <math>H</math> has a <math>p</math>-complement, i.e., a Hall subgroup of order <math>n</math>: This follows from fact (4). | ||
+ | # 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. | ||
+ | |||
==References== | ==References== | ||
Revision as of 16:24, 5 March 2009
Contents
Statement
Suppose is a Finite solvable group (?), and is a natural number dividing the order of . If there are exactly elements of whose 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
- Minimal normal implies elementary abelian in finite solvable
- Number of nth roots is a multiple of n
- Solvability is quotient-closed
Proof
Given: A group . A natural number . is the set of roots of unity in (i.e., the set of elements of order dividing , and has elements.
To prove: is a subgroup.
Proof: By fact (1), has a nontrivial elementary abelian normal subgroup of order for some prime power dividing the order of . We split the proof into two cases, based on whether divides or not.
Case that divides
In this case, . Let and where are relatively prime to . Clearly, and .
- Let . Then, the order of is divisible by : This is clear from the fact that the order of equals , and since and , this follows.
- divides : This follows essentially from the fact that exponent of in is strictly less than that in .
- There exists some natural number such that has elements of order dividing : This follows from fact (2).
- Let be the quotient map. The set of elements such that has size : This follows from step (3).
- If , we have : This follows from the fact that the order of any element in is either or .
- If , we have : This follows from the previous step and step (2), which says that .
- , and : By steps (3) and (5), there are at least elements for which . Thus, . Simplifying, we obtain that . This forces and . Further, since equality holds even under these assumptions, we conclude that the solutions to exhaust all solutions to .
- The set of elements such that forms a subgroup. In particular, is a subgroup: By the previous step, , so there are exactly solutions in to . By fact (3), is solvable, so the induction hypothesis applies, and the set of solutions forms a subgroup.
- is a subgroup: It is the inverse image in of a subgroup of .
Case that does not divide
- divides the order of : This follows because divides the order of and is relatively prime to .
- The number of elements such that is for some natural number : This follows from fact (2).
- For every coset of in such that , there exists an element such that : Start with some element in the coset. Let . By definition, . Now, find an integer such that (this can be done since and are relatively prime). Let and let . Note that since is a power of , which in turn is a power of , commutes with . Thus, if , then since is a multiple of , and . Further, , so . Thus, we have found an element of whose order divides .
- There are at least solutions to in : This follows from steps (2) and (3).
- , and there are exactly solutions to in : This follows from the previous step and the assumption that there are exactly solutions.
- The set of solutions to in forms a subgroup, say : By fact (3), is solvable, so we can apply the induction hypothesis to the conclusion of the previous step.
- The inverse image, say of this subgroup of in is a group of order , where is relatively prime to .
- has a -complement, i.e., a Hall subgroup of order : This follows from fact (4).
- This Hall subgroup of order is precisely equal to the set of solutions to : All elements in the Hall subgroup of order are solutions to , and we know that there are exactly solutions. Hence, these must be the only solutions.
References
Textbook references
- The Theory of Groups by Marshall Hall, Jr., Page 145, Theorem 9.4.1, ^{More info}