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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==Statement==
 
==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</math> elements of <math>G</math> whose <math>n^{th}</math> 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]].
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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</math> elements of <math>G</math> whose <math>n^{th}</math> 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]].
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==Related facts==
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* [[Number of nth roots is a multiple of n]]
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* [[Number of nth roots of any conjugacy class is a multiple of n]]
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* [[At most n elements of order dividing n implies every finite subgroup is cyclic]]
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===Related conjectures===
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* [[Frobenius conjecture on nth roots]]: The conjecture states that the assumption of solvability can be dropped.
  
 
==Facts used==
 
==Facts used==
  
# [[uses::Minimal normal implies elementary abelian in finite solvable]]
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{{facts used table disclaimer}}
# [[uses::Number of nth roots is a multiple of n]]
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# [[uses::Solvability is quotient-closed]]
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{| class="sortable" border="1"
# [[uses::Hall subgroups exist in finite solvable]]
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! 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
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|-
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| 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}}
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|-
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| 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}}
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|-
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| 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}}
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|-
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| 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}}
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|-
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| 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}}
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|-
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| 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}}
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|-
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| 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}}
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|}
  
 
==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.
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{{proof by induction}}
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We do the proof by induction. Specifically, we assume that the statement holds true for all finite solvable groups of strictly smaller orders than <math>N</math>.
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'''PROOF OF INDUCTIVE STEP''':
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'''Given''': A finite solvable group <math>G</math> of order <math>N</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.
 
'''To prove''': <math>S</math> is a subgroup.
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===Case that <math>p</math> divides <math>n</math>===
 
===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>.
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{| class="sortable" border="1"
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! No. !! Assertion/construction !! Facts used !! Given data/assumptions used !! Previous steps used !! Explanation
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|-
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| 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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|-
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| 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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|-
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| 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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|-
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| 4 || <math>G/K</math> has order <math>p^{k-i}l</math> || Fact (2) || <math>K</math> has order <math>p^i</math> || Step (3) || <toggledisplay>By fact (2), <math>|G/K| = |G|/|K|</math>, which, by the notation of Step (3), is <math>p^kl/p^i = p^{k-i}l</math>. </toggledisplay>
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|-
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| 5 || Let <math>u = p^{\max(j-i,0)}m</math>. Then, <math>u</math> divides the order of <math>G/K</math>. || || || Steps (3), (4) || <toggledisplay>From Step (3), <math>j - i \le k - i</math>. Also, <math>0 \le k - i</math>. So <math>\max(j-i,0) \le k - i</math>. Also, <math>m|l</math> by Step (3). Combining, we get that <math>u = p^{\max(j-i,0)}m</math> divides <math>p^{k-i}l</math>, which is the order of <math>G/K</math> by Step (4).</toggledisplay>
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|-
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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 (3) || || Step (5) || <toggledisplay>By Fact (3), the number of elements of order dividing <math>u</math> is a multiple of <math>u</math>, so it is <math>ru</math> for some natural number <math>r</math>.</toggledisplay>
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|-
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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 of <math>G/K</math> has size <math>rup^i</math> ||  || <math>K</math> has order <math>p^i</math> || Step (6) || <toggledisplay>By Step (6), the number of cosets of <math>K</math> in <math>G</math> containing elements <math>x</math> for which <math>\varphi(x)^u</math> is the identity element is <math>ru</math>. Since <math>K</math> has order <math>p^i</math>, each such coset has size <math>p^i</math>, so the total number of elements is <math>rup^i</math>.</toggledisplay>
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|-
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| 8 || If <math>\varphi(x)^u </math> is the identity element in <math>G/K</math>, then <math>x^{up}</math> is the identity element in <math>G/K</math>. || || || Step (1) || <toggledisplay>If <math>\varphi(x)^u</math> is the identity element in <math>G/K</math>, then, since <math>\varphi(x)^u = \varphi(x^u)</math>, we obtain that <math>x^u \in K</math>. By Step (1), <math>(x^u)^p</math> is the identity element of <math>G</math>, so <math>x^{up}</math> is the identity element of <math>G</math>.</toggledisplay>
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|-
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| 9 ||  <math>up</math> divides <math>n</math>. || || <math>p</math> divides <math>n</math> || Steps (3), (5) || <toggledisplay>We note that since <math>K</math> is nontrivial, <math>p^i > 1</math>, so <math>i > 0</math>, <math>j - i < j</math>. Also, since <math>p</math> divides <math>n</math>, <math>j > 0</math>, so <math>0 < j</math>. The upshot is that <math>\max(j-i,0) < j</math>. Since <math>n = p^jm</math> and <math>u = p^{\max(j-i,0)}m</math>, we obtain that <math>up<
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| 10 || If <math>\varphi(x)^u</math> is the identity element in <math>G/K</math>, then <math>x \in S</math>, i.e., <math>x</math> has order dividing <math>n</math>. || || || Steps (8), (9) || Step-combination direct.
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|-
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| 11 || <math>rup^i \le n</math> || || There are exactly <math>n</math> elements in <math>G</math> of order dividing <math>n</math>, i.e., <math>S</math> has size <math>n</math>. || || Steps (7), (10) || <toggledisplay>The set of elements <math>x \in G</math> for which <math>\varphi(x)^u</math> is the identity is contained in <math>S</math> by Step (10) and has size <math>rup^i</math> by Step (7). Since <math>S</math> has size <math>n</math>, we get <math>rup^i \le n</math>.</toggledisplay>
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|-
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| 12 || <math>r = 1, j \ge i</math> and <math>x^n</math> is the identity element iff <math>x^u \in K</math>. In other words, <math>S = \{ x \mid x^u \in K \}</math>. || || || Steps (5), (11) ||<toggledisplay>We have <math>rup^i \le n</math>. Plugging in the expression for <math>u</math> and canceling the factor of <math>m</math>, we get <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> give all of <math>S</math>.</toggledisplay>
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|-
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| 13 || 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 || Fact (4) || inductive hypothesis, <math>G</math> is solvable || Step (12) || <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</math> being the identity element. By fact (4), <math>G/K</math> is solvable, so the induction hypothesis applies, and the set of solutions forms a subgroup.</toggledisplay>
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|-
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| 14 || <math>S</math> is a subgroup  || || || Step (13) || <toggledisplay>It is the inverse image in <math>G</math> of a subgroup of <math>G/K</math></toggledisplay>
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|}
  
# 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.
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===Case that <math>p</math> does not divide <math>n</math>===
# <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>===
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{| class="sortable" border="1"
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! No. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
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|-
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| 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>.
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|-
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| 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
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|-
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| 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>
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| 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
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| 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>
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|-
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| 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>
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|-
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| 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>
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|-
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| 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>
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|-
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| 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>.
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|}
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===Proof that the subgroup is normal, characteristic and fully invariant===
  
# <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>.
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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>.
# 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==

Latest revision as of 03:00, 27 May 2011

Statement

Suppose G is a Finite solvable group (?), and n is a natural number dividing the order of G. If there are exactly n elements of G whose n^{th} 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

Related conjectures

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 K of G, happens before we split into cases. The proof is inductive, and the goal is to induct from G/K to G. Basic/intermediate group theory click here
2 Lagrange's theorem: We in particular are interested in the version which states that if A is a subgroup of B, |B/A| = |B|/|A|, where B/A is the quotient set. When A is a normal subgroup, B/A is the quotient group. Step (4) of the p divides n, Step (1) of the other case We use it to compute the order of the quotient group G/K, from which we ultimately induct. The idea is to try to show that the inductive hypothesis conditions apply to G/K. Basic group theory 2 click here
3 Number of nth roots is a multiple of n (when n divides the group order) Step (6) of p|n case, Step (2) of the other case Applied to the quotient group to show that the number of roots of a certain kind of G/K 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 p|n case, Step (6) of other case Show conditions prevail to apply inductive hypothesis to quotient group G/K Basic group theory click here
5 Solvability is subgroup-closed Step (8) of p does not divide n case Show solvability of a subgroup of G Basic group theory click here
6 Hall subgroups exist in finite solvable Step (8) of p does not divide n case Find a subgroup of order n in a subgroup where the p'-part is n Advanced click here
7 Order of element divides order of group Step (9) of p does not divide n case After finding subgroup of order n, 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 N.

PROOF OF INDUCTIVE STEP:

Given: A finite solvable group G of order N. A natural number n. S is the set of n^{th} roots of unity in G (i.e., the set of elements of order dividing n, and S has n elements.

To prove: S is a subgroup.

Proof: By fact (1), G has a nontrivial elementary abelian normal subgroup K of order p^i for some prime power p^i dividing the order of G. We split the proof into two cases, based on whether p divides n or not.

Case that p divides n

No. Assertion/construction Facts used Given data/assumptions used Previous steps used Explanation
1 The order of every element in K is either 1 or p K is elementary abelian of order p^i
2 K \subseteq S p divides n, S comprises all elements whose n^{th} power is the identity Step (1) [SHOW MORE]
3 Let n = p^jm and |G| = N = p^kl where m,l are relatively prime to p. Then, j \le k and m|l. n divides the order of G [SHOW MORE]
4 G/K has order p^{k-i}l Fact (2) K has order p^i Step (3) [SHOW MORE]
5 Let u = p^{\max(j-i,0)}m. Then, u divides the order of G/K. Steps (3), (4) [SHOW MORE]
6 There exists some natural number r such that G/K has ru elements of order dividing u Fact (3) Step (5) [SHOW MORE]
7 Let \varphi:G \to G/K be the quotient map. The set of elements x \in G such that \varphi(x)^u is the identity element of G/K has size rup^i K has order p^i Step (6) [SHOW MORE]
8 If \varphi(x)^u is the identity element in G/K, then x^{up} is the identity element in G/K. Step (1) [SHOW MORE]
9 up divides n. p divides n Steps (3), (5) [SHOW MORE]
12 r = 1, j \ge i and x^n is the identity element iff x^u \in K. In other words, S = \{ x \mid x^u \in K \}. Steps (5), (11) [SHOW MORE]
13 The set of elements y \in G/K such that y^u is the identity element forms a subgroup. In particular, S/K is a subgroup Fact (4) inductive hypothesis, G is solvable Step (12) [SHOW MORE]
14 S is a subgroup Step (13) [SHOW MORE]

Case that p does not divide n

No. Assertion/construction Facts used Given data used Previous steps used Explanation
1 n divides the order of G/K Fact (2) n divides the order of G, K has order p^i, p does not divide n By Fact (2), the order of G/K is |G|/|K| which is N/p^i. We're given that n divides N, so since n is relatively prime to p, we must also have n divides N/p^i.
2 The number of elements z \in G/K such that z^n is the identity is rn for some natural number r Fact (3) Step (1) Previous step+Fact direct
3 For every coset z = xK of K in G such that z^n is the identity in G/K, there exists an element y \in xK such that y^n is the identity in G p does not divide n [SHOW MORE]
4 There are at least rn solutions to x^n being the identity in G Steps (2), (3) Step-combination direct
5 r = 1, and there are exactly n solutions to z^n being the identity in G/K There are exactly n elements of order dividing n Steps (2), (4) [SHOW MORE]
6 The set of elements in G/K of order dividing n forms a subgroup of order n in G/K Fact (4) G is solvable, induction hypothesis Step (5) [SHOW MORE]
7 The inverse image in G of the subgroup of G/K obtained in Step (6) is a subgroup of G, say H. It has order p^in. Fact (2) K has order p^i. [SHOW MORE]
8 H has a p-complement, i.e., a Hall subgroup of order n. Call this p-complement L. Facts (5), (6) p does not divide n, G is solvable Step (7) [SHOW MORE]
9 L = S, i.e., L is precisely the set of elements of order dividing n. Since it's a subgroup, this completes the proof. Fact (7) There are exactly n elements of order dividing n. <toggledisplay>Since L has order dividing n, all elements in there also have order dividing n by Fact (7). So L \subseteq S. By size considerations, we have L = S.

Proof that the subgroup is normal, characteristic and fully invariant

This follows from the fact that if \alpha is a homomorphism from G to any subgroup of G, then (\alpha(g))^n = \alpha(g^n). Hence, if g^n = 1, so is (\alpha(g))^n.

References

Textbook references