Difference between revisions of "Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup"
Line 7: | Line 7: | ||
* [[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=== | ||
Line 19: | Line 19: | ||
# [[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]] | ||
+ | # [[uses::Solvability is subgroup-closed]] | ||
# [[uses::Hall subgroups exist in finite solvable]] | # [[uses::Hall subgroups exist in finite solvable]] | ||
− | + | # [[uses::Order of element divides order of group]] | |
==Proof== | ==Proof== | ||
Line 71: | Line 72: | ||
===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. | ||
+ | |- | ||
+ | | 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>. |
Revision as of 00:41, 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
- Minimal normal implies elementary abelian in finite solvable
- 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.
- Number of nth roots is a multiple of n
- Solvability is quotient-closed
- Solvability is subgroup-closed
- Hall subgroups exist in finite solvable
- 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 .
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) | <toggledisplay>By Fact (5), since is solvable, is solvable. By Fact (6), it must therefore have a -complement. |
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}