Difference between revisions of "Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup"
(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]] | ||
+ | # [[uses::Solvability is quotient-closed]] | ||
==Proof== | ==Proof== | ||
+ | |||
+ | '''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. | ||
+ | |||
+ | '''To prove''': <math>S</math> is a subgroup. | ||
+ | |||
+ | '''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. | ||
+ | |||
+ | ===Case that <math>p</math> divides <math>n</math>=== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | # 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. | ||
+ | # <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</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>. | ||
+ | # 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>=== | ||
{{fillin}} | {{fillin}} |
Revision as of 16:03, 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 Failed to parse (syntax error): rp^{\max(i,j)\} \le p^j . 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
PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]References
Textbook references
- The Theory of Groups by Marshall Hall, Jr., Page 145, Theorem 9.4.1, ^{More info}