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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− | | 1 || <math>K | + | | 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 || | + | | 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. | + | | 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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===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
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, a homomorph-containing subgroup and a variety-containing subgroup.
Related facts
- Number of nth roots 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
Related conjectures
- Frobenius conjecture on nth roots: The conjecture states that the assumption of solvability can be dropped.
Facts used
- Minimal normal implies elementary abelian in finite solvable
- Number of nth roots is a multiple of n
- Solvability is quotient-closed
- Hall subgroups exist in finite solvable
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
No. | Assertion/construction | Facts used | Given data/assumptions used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | The order of every element in is either or | is elementary abelian of order | |||
2 | divides , comprises all elements whose power is the identity | Step (1) | [SHOW MORE] | ||
3 | Let and where are relatively prime to . Then, and . | divides the order of | [SHOW MORE] | ||
4 | Let . Then, the order of is divisible by . | [SHOW MORE] | |||
5 | divides | [SHOW MORE] | |||
6 | There exists some natural number such that has elements of order dividing | Fact (2) | Step (1) | ||
7 | Let be the quotient map. The set of elements such that is the identity element has size | step (3) | |||
8 | If is the identity element, then | ||||
9 | If , we have : This follows from the previous step and step (2), which says that . | ||||
10 | and is the identity element iff | [SHOW MORE] | |||
11 | The set of elements such that is the identity element forms a subgroup. In particular, is a subgroup | Step (10) | [SHOW MORE] | ||
12 | is a subgroup | Step (11) | [SHOW MORE] |
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.
Proof that the subgroup is normal, characteristic and fully characteristic
This follows from the fact that if is a homomorphism from to any subgroup of , then . Hence, if , so is .
References
Textbook references
- The Theory of Groups by Marshall Hall, Jr., Page 145, Theorem 9.4.1, ^{More info}