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]] | ||
− | * [[ | + | * [[At most n elements of order dividing n implies every finite subgroup is cyclic]] |
===Related conjectures=== | ===Related conjectures=== | ||
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==Facts used== | ==Facts used== | ||
− | + | {{facts used table disclaimer}} | |
− | + | ||
− | + | {| class="sortable" border="1" | |
− | + | ! Fact no. !! Statement !! Steps in the proof where it is used !! Qualitative description of how it is used !! What does it rely on? !! Difficulty level !! Other applications | |
− | # [[uses::Hall subgroups exist in finite solvable]] | + | |- |
+ | | 1 || [[uses::Minimal normal implies elementary abelian in finite solvable]] || Setting up the elementary abelian normal subgroup <math>K</math> of <math>G</math>, happens before we split into cases. || The proof is inductive, and the goal is to induct from <math>G/K</math> to <math>G</math>. || Basic/intermediate group theory || {{#show:minimal normal implies elementary abelian in finite solvable| ?Difficulty level}} || {{uses short|minimal normal implies elementary abelian in finite solvable}} | ||
+ | |- | ||
+ | | 2 || [[uses::Lagrange's theorem]]: We in particular are interested in the version which states that if <math>A</math> is a subgroup of <math>B</math>, <math>|B/A| = |B|/|A|</math>, where <math>B/A</math> is the quotient set. When <math>A</math> is a normal subgroup, <math>B/A</math> is the [[quotient group]]. || Step (4) of the <math>p</math> divides <math>n</math>, Step (1) of the other case || We use it to compute the order of the quotient group <math>G/K</math>, from which we ultimately induct. The idea is to try to show that the inductive hypothesis conditions apply to <math>G/K</math>. || Basic group theory || {{#show:Lagrange's theorem| ?Difficulty level}} || {{uses short|Lagrange's theorem}} | ||
+ | |- | ||
+ | | 3 || [[uses::Number of nth roots is a multiple of n]] (when <math>n</math> divides the group order) || Step (6) of <math>p|n</math> case, Step (2) of the other case || Applied to the quotient group to show that the number of roots of a certain kind of <math>G/K</math> is at ''least'' a certain amount, which is then played off against it being at most a certain amount. || Basic/intermediate group theory || {{#show: number of nth roots is a multiple of n| ?Difficulty level}} || {{uses short|number of nth roots is a multiple of n}} | ||
+ | |- | ||
+ | | 4 || [[uses::Solvability is quotient-closed]] || Step (13) of <math>p|n</math> case, Step (6) of other case || Show conditions prevail to apply inductive hypothesis to quotient group <math>G/K</math> || Basic group theory || {{#show: solvability is quotient-closed| ?Difficulty level}} || {{uses short|solvability is quotient-closed}} | ||
+ | |- | ||
+ | | 5 || [[uses::Solvability is subgroup-closed]] || Step (8) of <math>p</math> does not divide <math>n</math> case || Show solvability of a subgroup of <math>G</math> || Basic group theory ||{{#show: solvability is subgroup-closed| ?Difficulty level}} || {{uses short|solvablility is subgroup-closed}} | ||
+ | |- | ||
+ | | 6 || [[uses::Hall subgroups exist in finite solvable]] || Step (8) of <math>p</math> does not divide <math>n</math> case || Find a subgroup of order <math>n</math> in a subgroup where the <math>p'</math>-part is <math>n</math> || Advanced || {{#show: Hall subgroups exist in finite solvable| ?Difficulty level}} || {{uses short|Hall subgroups exist in finite solvable}} | ||
+ | |- | ||
+ | | 7 || [[uses::Order of element divides order of group]] || Step (9) of <math>p</math> does not divide <math>n</math> case || After finding subgroup of order <math>n</math>, show it is precisely the subgroup we want || Basic group theory || {{#show:order of element divides order of group| ?Difficulty level}} || {{uses short|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>=== | ||
− | + | {| class="sortable" border="1" | |
− | + | ! No. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation | |
− | + | |- | |
− | + | | 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>. | |
− | + | |- | |
− | + | | 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 | |
− | + | |- | |
− | + | | 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> | |
− | + | |- | |
− | + | | 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 | + | |- |
+ | | 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> | ||
+ | |- | ||
+ | | 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> | ||
+ | |- | ||
+ | | 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> | ||
+ | |- | ||
+ | | 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.</toggledisplay> | ||
+ | |- | ||
+ | | 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>. | ||
+ | |} | ||
+ | ===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>. |
Latest revision as of 03:00, 27 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
- At most 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
The table below lists key facts used directly and explicitly in the proof. Fact numbers as used in the table may be referenced in the proof. This table need not list facts used indirectly, i.e., facts that are used to prove these facts, and it need not list facts used implicitly through assumptions embedded in the choice of terminology and language.
Fact no. | Statement | Steps in the proof where it is used | Qualitative description of how it is used | What does it rely on? | Difficulty level | Other applications |
---|---|---|---|---|---|---|
1 | Minimal normal implies elementary abelian in finite solvable | Setting up the elementary abelian normal subgroup of , happens before we split into cases. | The proof is inductive, and the goal is to induct from to . | Basic/intermediate group theory | click here | |
2 | Lagrange's theorem: We in particular are interested in the version which states that if is a subgroup of , , where is the quotient set. When is a normal subgroup, is the quotient group. | Step (4) of the divides , Step (1) of the other case | We use it to compute the order of the quotient group , from which we ultimately induct. The idea is to try to show that the inductive hypothesis conditions apply to . | Basic group theory | 2 | click here |
3 | Number of nth roots is a multiple of n (when divides the group order) | Step (6) of case, Step (2) of the other case | Applied to the quotient group to show that the number of roots of a certain kind of is at least a certain amount, which is then played off against it being at most a certain amount. | Basic/intermediate group theory | click here | |
4 | Solvability is quotient-closed | Step (13) of case, Step (6) of other case | Show conditions prevail to apply inductive hypothesis to quotient group | Basic group theory | click here | |
5 | Solvability is subgroup-closed | Step (8) of does not divide case | Show solvability of a subgroup of | Basic group theory | click here | |
6 | Hall subgroups exist in finite solvable | Step (8) of does not divide case | Find a subgroup of order in a subgroup where the -part is | Advanced | click here | |
7 | Order of element divides order of group | Step (9) of does not divide case | After finding subgroup of order , show it is precisely the subgroup we want | Basic group theory | click here |
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 .
PROOF OF INDUCTIVE STEP:
Given: A finite solvable group of order . 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 | has order | Fact (2) | has order | Step (3) | [SHOW MORE] |
5 | Let . Then, divides the order of . | Steps (3), (4) | [SHOW MORE] | ||
6 | There exists some natural number such that has elements of order dividing | Fact (3) | Step (5) | [SHOW MORE] | |
7 | Let be the quotient map. The set of elements such that is the identity element of has size | has order | Step (6) | [SHOW MORE] | |
8 | If is the identity element in , then is the identity element in . | Step (1) | [SHOW MORE] | ||
9 | divides . | divides | Steps (3), (5) | [SHOW MORE] | |
12 | and is the identity element iff . In other words, . | Steps (5), (11) | [SHOW MORE] | ||
13 | The set of elements such that is the identity element forms a subgroup. In particular, is a subgroup | Fact (4) | inductive hypothesis, is solvable | Step (12) | [SHOW MORE] |
14 | is a subgroup | Step (13) | [SHOW MORE] |
Case that does not divide
No. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | divides the order of | Fact (2) | divides the order of , has order , does not divide | By Fact (2), the order of is which is . We're given that divides , so since is relatively prime to , we must also have divides . | |
2 | The number of elements such that is the identity is for some natural number | Fact (3) | Step (1) | Previous step+Fact direct | |
3 | For every coset of in such that is the identity in , there exists an element such that is the identity in | does not divide | [SHOW MORE] | ||
4 | There are at least solutions to being the identity in | Steps (2), (3) | Step-combination direct | ||
5 | , and there are exactly solutions to being the identity in | There are exactly elements of order dividing | Steps (2), (4) | [SHOW MORE] | |
6 | The set of elements in of order dividing forms a subgroup of order in | Fact (4) | is solvable, induction hypothesis | Step (5) | [SHOW MORE] |
7 | The inverse image in of the subgroup of obtained in Step (6) is a subgroup of , say . It has order . | Fact (2) | has order . | [SHOW MORE] | |
8 | has a -complement, i.e., a Hall subgroup of order . Call this -complement . | Facts (5), (6) | does not divide , is solvable | Step (7) | [SHOW MORE] |
9 | , i.e., is precisely the set of elements of order dividing . Since it's a subgroup, this completes the proof. | Fact (7) | There are exactly elements of order dividing . | <toggledisplay>Since has order dividing , all elements in there also have order dividing by Fact (7). So . By size considerations, we have . |
Proof that the subgroup is normal, characteristic and fully invariant
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}