Mixing
Prove that $x$ is in all maximal subgroups of a group $G$ iff $x in X subseteq G$, where $X$ generates $G$,...
$begingroup$
Show that $x in G$ lies in the intersection of all maximal subgroups of G if and only if it has the following property: if $X subseteq G$ contains $x$ and generates $G$, then $Xsetminus{x}$ generates $G$.
Maximal subgroups are defined to be the largest possible subgroups that are generated by a certain set.
So the first part of the exercise is alright. If $x in bigcap(M_i)$, where $M_i$ is the set of all maximal subgroups, then the cyclic subgroup generated by x is obviously also in the group generated by $Xsetminus{x}$, so $G = langle {X} rangle = langle{X}setminus{x}rangle$.
For the converse however, I have got into a little bit of a problem. Take for example the Klein 4-group. $langle V_4 setminus {a}rangle = langle 1, b, c enspace | enspace b^2 = c^2 = 1, bc = cb = a rangle $, clearly generates $ {1,a,b,c}$, however $ a notin langle b rangle, langle c rangle$, so the converse is not always true?
Where have I gone wrong?
abstract-algebra group-theory
edited Jan 20 at 12:27
the_fox
2,90021537
asked Nov 15 '14 at 13:27
ejcqwejcqw
19518
$endgroup$
add a comment |
$begingroup$
Show that $x in G$ lies in the intersection of all maximal subgroups of G if and only if it has the following property: if $X subseteq G$ contains $x$ and generates $G$, then $Xsetminus{x}$ generates $G$.
Maximal subgroups are defined to be the largest possible subgroups that are generated by a certain set.
So the first part of the exercise is alright. If $x in bigcap(M_i)$, where $M_i$ is the set of all maximal subgroups, then the cyclic subgroup generated by x is obviously also in the group generated by $Xsetminus{x}$, so $G = langle {X} rangle = langle{X}setminus{x}rangle$.
For the converse however, I have got into a little bit of a problem. Take for example the Klein 4-group. $langle V_4 setminus {a}rangle = langle 1, b, c enspace | enspace b^2 = c^2 = 1, bc = cb = a rangle $, clearly generates $ {1,a,b,c}$, however $ a notin langle b rangle, langle c rangle$, so the converse is not always true?
Where have I gone wrong?
abstract-algebra group-theory
edited Jan 20 at 12:27
the_fox
2,90021537
asked Nov 15 '14 at 13:27
ejcqwejcqw
19518
$endgroup$
$begingroup$
In your example, $1$ is the only element that lies in the intersection of all the maximal subgroups, namely, $<a>, <b>, <c>$. I don't see any problem.
$endgroup$
– Krish
Nov 15 '14 at 13:39
$begingroup$
But the thing is, the exercise asks to prove the equivalence of the two statements. So if the latter part is satisfied (that if $a in V_4 Rightarrow V_4setminus{a}$ generates $V_4$) then $a$ must be in the intersection of all the maximal subgroups, which is false?
$endgroup$
– ejcqw
Nov 15 '14 at 13:44
1
$begingroup$
Take $X = {a, b }$. Then $X$ generates the group $V_4$. But $X - {a }$ does not generate $V_4$. The other part of the theorem must be true for all $X$ satisfying the property, not just for a particular set $X$.
$endgroup$
– Krish
Nov 15 '14 at 14:11
$begingroup$
I see why now, thanks! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
add a comment |
$begingroup$
Show that $x in G$ lies in the intersection of all maximal subgroups of G if and only if it has the following property: if $X subseteq G$ contains $x$ and generates $G$, then $Xsetminus{x}$ generates $G$.
Maximal subgroups are defined to be the largest possible subgroups that are generated by a certain set.
So the first part of the exercise is alright. If $x in bigcap(M_i)$, where $M_i$ is the set of all maximal subgroups, then the cyclic subgroup generated by x is obviously also in the group generated by $Xsetminus{x}$, so $G = langle {X} rangle = langle{X}setminus{x}rangle$.
For the converse however, I have got into a little bit of a problem. Take for example the Klein 4-group. $langle V_4 setminus {a}rangle = langle 1, b, c enspace | enspace b^2 = c^2 = 1, bc = cb = a rangle $, clearly generates $ {1,a,b,c}$, however $ a notin langle b rangle, langle c rangle$, so the converse is not always true?
Where have I gone wrong?
abstract-algebra group-theory
edited Jan 20 at 12:27
the_fox
2,90021537
asked Nov 15 '14 at 13:27
ejcqwejcqw
19518
$endgroup$
Show that $x in G$ lies in the intersection of all maximal subgroups of G if and only if it has the following property: if $X subseteq G$ contains $x$ and generates $G$, then $Xsetminus{x}$ generates $G$.
Maximal subgroups are defined to be the largest possible subgroups that are generated by a certain set.
So the first part of the exercise is alright. If $x in bigcap(M_i)$, where $M_i$ is the set of all maximal subgroups, then the cyclic subgroup generated by x is obviously also in the group generated by $Xsetminus{x}$, so $G = langle {X} rangle = langle{X}setminus{x}rangle$.
For the converse however, I have got into a little bit of a problem. Take for example the Klein 4-group. $langle V_4 setminus {a}rangle = langle 1, b, c enspace | enspace b^2 = c^2 = 1, bc = cb = a rangle $, clearly generates $ {1,a,b,c}$, however $ a notin langle b rangle, langle c rangle$, so the converse is not always true?
Where have I gone wrong?
abstract-algebra group-theory
abstract-algebra group-theory
edited Jan 20 at 12:27
the_fox
2,90021537
asked Nov 15 '14 at 13:27
ejcqwejcqw
19518
edited Jan 20 at 12:27
the_fox
2,90021537
asked Nov 15 '14 at 13:27
ejcqwejcqw
19518
edited Jan 20 at 12:27
the_fox
2,90021537
edited Jan 20 at 12:27
the_fox
2,90021537
edited Jan 20 at 12:27
the_fox
2,90021537
2,90021537
asked Nov 15 '14 at 13:27
ejcqwejcqw
19518
asked Nov 15 '14 at 13:27
ejcqwejcqw
19518
asked Nov 15 '14 at 13:27
ejcqwejcqw
19518
19518
$begingroup$
In your example, $1$ is the only element that lies in the intersection of all the maximal subgroups, namely, $<a>, <b>, <c>$. I don't see any problem.
$endgroup$
– Krish
Nov 15 '14 at 13:39
$begingroup$
But the thing is, the exercise asks to prove the equivalence of the two statements. So if the latter part is satisfied (that if $a in V_4 Rightarrow V_4setminus{a}$ generates $V_4$) then $a$ must be in the intersection of all the maximal subgroups, which is false?
$endgroup$
– ejcqw
Nov 15 '14 at 13:44
1
$begingroup$
Take $X = {a, b }$. Then $X$ generates the group $V_4$. But $X - {a }$ does not generate $V_4$. The other part of the theorem must be true for all $X$ satisfying the property, not just for a particular set $X$.
$endgroup$
– Krish
Nov 15 '14 at 14:11
$begingroup$
I see why now, thanks! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
add a comment |
$begingroup$
In your example, $1$ is the only element that lies in the intersection of all the maximal subgroups, namely, $<a>, <b>, <c>$. I don't see any problem.
$endgroup$
– Krish
Nov 15 '14 at 13:39
$begingroup$
But the thing is, the exercise asks to prove the equivalence of the two statements. So if the latter part is satisfied (that if $a in V_4 Rightarrow V_4setminus{a}$ generates $V_4$) then $a$ must be in the intersection of all the maximal subgroups, which is false?
$endgroup$
– ejcqw
Nov 15 '14 at 13:44
1
$begingroup$
Take $X = {a, b }$. Then $X$ generates the group $V_4$. But $X - {a }$ does not generate $V_4$. The other part of the theorem must be true for all $X$ satisfying the property, not just for a particular set $X$.
$endgroup$
– Krish
Nov 15 '14 at 14:11
$begingroup$
I see why now, thanks! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
$begingroup$
In your example, $1$ is the only element that lies in the intersection of all the maximal subgroups, namely, $<a>, <b>, <c>$. I don't see any problem.
$endgroup$
– Krish
Nov 15 '14 at 13:39
$begingroup$
In your example, $1$ is the only element that lies in the intersection of all the maximal subgroups, namely, $<a>, <b>, <c>$. I don't see any problem.
$endgroup$
– Krish
Nov 15 '14 at 13:39
$begingroup$
But the thing is, the exercise asks to prove the equivalence of the two statements. So if the latter part is satisfied (that if $a in V_4 Rightarrow V_4setminus{a}$ generates $V_4$) then $a$ must be in the intersection of all the maximal subgroups, which is false?
$endgroup$
– ejcqw
Nov 15 '14 at 13:44
$begingroup$
But the thing is, the exercise asks to prove the equivalence of the two statements. So if the latter part is satisfied (that if $a in V_4 Rightarrow V_4setminus{a}$ generates $V_4$) then $a$ must be in the intersection of all the maximal subgroups, which is false?
$endgroup$
– ejcqw
Nov 15 '14 at 13:44
1
1
$begingroup$
Take $X = {a, b }$. Then $X$ generates the group $V_4$. But $X - {a }$ does not generate $V_4$. The other part of the theorem must be true for all $X$ satisfying the property, not just for a particular set $X$.
$endgroup$
– Krish
Nov 15 '14 at 14:11
$begingroup$
Take $X = {a, b }$. Then $X$ generates the group $V_4$. But $X - {a }$ does not generate $V_4$. The other part of the theorem must be true for all $X$ satisfying the property, not just for a particular set $X$.
$endgroup$
– Krish
Nov 15 '14 at 14:11
$begingroup$
I see why now, thanks! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
$begingroup$
I see why now, thanks! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
add a comment |
1 Answer
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$begingroup$
Your definition of maximal subgroup is wrong.
A maximal subgroup $H$ of $G$ is a subgroup such that $Hne G$ ($H$ is a proper subgroup of $G$) and there is no subgroup $K$ of $G$ with $Hsubsetneq Ksubsetneq G$.
Any proper subgroup of a finite group is contained in a maximal subgroup, because the set of proper subgroups is ordered by inclusion and any finite ordered set has maximal elements.
Suppose $x$ belongs to all maximal subgroups and that $xin X$, where $X$ is a subset of $G$. Suppose $langle Xsetminus{x}ranglene G$; then this subgroup is contained in a maximal subgroup $H$ and $xin H$ by assumption, so $langle Xrangle=bigl<langle Xsetminus{x}ranglecup{x}bigr>subseteq H$. Therefore $langle Xranglene G$.
Conversely, suppose that $x$ is a nongenerator, that is, for any subset $Xsubseteq G$ containing $x$, if $langle Xrangle=G$, then $langle Xsetminus{x}rangle=G$. We want to show that this nongenerator $x$ belongs to all maximal subgroups of $G$. If $H$ is a maximal subgroup and $xnotin H$, then $langle Hcup{x}=G$, so, being $x$ a nongenerator, also $H=langle (Hcup{x})setminus{x}rangle=G$, which is a contradiction.
In the Klein group $V={1,a,b,c}$, the maximal subgroups are ${1,a}$, ${1,b}$ and ${1,c}$ and $1$ is the only element belonging to all maximal subgroups.
Note that just noting that ${a,b,c}$ generates $V$ but also ${b,c}$ generates $V$ doesn't mean $a$ is a nongenerator. This set ${a,b,c}$ is just one of the sets of generators containing $a$. Indeed, consider $X={a,b}$. Then $langle Xrangle=V$, but $langle Xsetminus{a}rangle=langle{b}rangle={1,b}ne V$. Thus $a$ is not a nongenerator.
answered Nov 15 '14 at 14:07
egregegreg
183k1486205
$endgroup$
$begingroup$
Really nice, thank you! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
add a comment |
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$begingroup$
Your definition of maximal subgroup is wrong.
A maximal subgroup $H$ of $G$ is a subgroup such that $Hne G$ ($H$ is a proper subgroup of $G$) and there is no subgroup $K$ of $G$ with $Hsubsetneq Ksubsetneq G$.
Any proper subgroup of a finite group is contained in a maximal subgroup, because the set of proper subgroups is ordered by inclusion and any finite ordered set has maximal elements.
Suppose $x$ belongs to all maximal subgroups and that $xin X$, where $X$ is a subset of $G$. Suppose $langle Xsetminus{x}ranglene G$; then this subgroup is contained in a maximal subgroup $H$ and $xin H$ by assumption, so $langle Xrangle=bigl<langle Xsetminus{x}ranglecup{x}bigr>subseteq H$. Therefore $langle Xranglene G$.
Conversely, suppose that $x$ is a nongenerator, that is, for any subset $Xsubseteq G$ containing $x$, if $langle Xrangle=G$, then $langle Xsetminus{x}rangle=G$. We want to show that this nongenerator $x$ belongs to all maximal subgroups of $G$. If $H$ is a maximal subgroup and $xnotin H$, then $langle Hcup{x}=G$, so, being $x$ a nongenerator, also $H=langle (Hcup{x})setminus{x}rangle=G$, which is a contradiction.
In the Klein group $V={1,a,b,c}$, the maximal subgroups are ${1,a}$, ${1,b}$ and ${1,c}$ and $1$ is the only element belonging to all maximal subgroups.
Note that just noting that ${a,b,c}$ generates $V$ but also ${b,c}$ generates $V$ doesn't mean $a$ is a nongenerator. This set ${a,b,c}$ is just one of the sets of generators containing $a$. Indeed, consider $X={a,b}$. Then $langle Xrangle=V$, but $langle Xsetminus{a}rangle=langle{b}rangle={1,b}ne V$. Thus $a$ is not a nongenerator.
answered Nov 15 '14 at 14:07
egregegreg
183k1486205
$endgroup$
$begingroup$
Really nice, thank you! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
add a comment |
$begingroup$
Your definition of maximal subgroup is wrong.
A maximal subgroup $H$ of $G$ is a subgroup such that $Hne G$ ($H$ is a proper subgroup of $G$) and there is no subgroup $K$ of $G$ with $Hsubsetneq Ksubsetneq G$.
Any proper subgroup of a finite group is contained in a maximal subgroup, because the set of proper subgroups is ordered by inclusion and any finite ordered set has maximal elements.
Suppose $x$ belongs to all maximal subgroups and that $xin X$, where $X$ is a subset of $G$. Suppose $langle Xsetminus{x}ranglene G$; then this subgroup is contained in a maximal subgroup $H$ and $xin H$ by assumption, so $langle Xrangle=bigl<langle Xsetminus{x}ranglecup{x}bigr>subseteq H$. Therefore $langle Xranglene G$.
Conversely, suppose that $x$ is a nongenerator, that is, for any subset $Xsubseteq G$ containing $x$, if $langle Xrangle=G$, then $langle Xsetminus{x}rangle=G$. We want to show that this nongenerator $x$ belongs to all maximal subgroups of $G$. If $H$ is a maximal subgroup and $xnotin H$, then $langle Hcup{x}=G$, so, being $x$ a nongenerator, also $H=langle (Hcup{x})setminus{x}rangle=G$, which is a contradiction.
In the Klein group $V={1,a,b,c}$, the maximal subgroups are ${1,a}$, ${1,b}$ and ${1,c}$ and $1$ is the only element belonging to all maximal subgroups.
Note that just noting that ${a,b,c}$ generates $V$ but also ${b,c}$ generates $V$ doesn't mean $a$ is a nongenerator. This set ${a,b,c}$ is just one of the sets of generators containing $a$. Indeed, consider $X={a,b}$. Then $langle Xrangle=V$, but $langle Xsetminus{a}rangle=langle{b}rangle={1,b}ne V$. Thus $a$ is not a nongenerator.
answered Nov 15 '14 at 14:07
egregegreg
183k1486205
$endgroup$
$begingroup$
Really nice, thank you! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
add a comment |
$begingroup$
Your definition of maximal subgroup is wrong.
A maximal subgroup $H$ of $G$ is a subgroup such that $Hne G$ ($H$ is a proper subgroup of $G$) and there is no subgroup $K$ of $G$ with $Hsubsetneq Ksubsetneq G$.
Any proper subgroup of a finite group is contained in a maximal subgroup, because the set of proper subgroups is ordered by inclusion and any finite ordered set has maximal elements.
Suppose $x$ belongs to all maximal subgroups and that $xin X$, where $X$ is a subset of $G$. Suppose $langle Xsetminus{x}ranglene G$; then this subgroup is contained in a maximal subgroup $H$ and $xin H$ by assumption, so $langle Xrangle=bigl<langle Xsetminus{x}ranglecup{x}bigr>subseteq H$. Therefore $langle Xranglene G$.
Conversely, suppose that $x$ is a nongenerator, that is, for any subset $Xsubseteq G$ containing $x$, if $langle Xrangle=G$, then $langle Xsetminus{x}rangle=G$. We want to show that this nongenerator $x$ belongs to all maximal subgroups of $G$. If $H$ is a maximal subgroup and $xnotin H$, then $langle Hcup{x}=G$, so, being $x$ a nongenerator, also $H=langle (Hcup{x})setminus{x}rangle=G$, which is a contradiction.
In the Klein group $V={1,a,b,c}$, the maximal subgroups are ${1,a}$, ${1,b}$ and ${1,c}$ and $1$ is the only element belonging to all maximal subgroups.
Note that just noting that ${a,b,c}$ generates $V$ but also ${b,c}$ generates $V$ doesn't mean $a$ is a nongenerator. This set ${a,b,c}$ is just one of the sets of generators containing $a$. Indeed, consider $X={a,b}$. Then $langle Xrangle=V$, but $langle Xsetminus{a}rangle=langle{b}rangle={1,b}ne V$. Thus $a$ is not a nongenerator.
answered Nov 15 '14 at 14:07
egregegreg
183k1486205
$endgroup$
Your definition of maximal subgroup is wrong.
A maximal subgroup $H$ of $G$ is a subgroup such that $Hne G$ ($H$ is a proper subgroup of $G$) and there is no subgroup $K$ of $G$ with $Hsubsetneq Ksubsetneq G$.
Any proper subgroup of a finite group is contained in a maximal subgroup, because the set of proper subgroups is ordered by inclusion and any finite ordered set has maximal elements.
Suppose $x$ belongs to all maximal subgroups and that $xin X$, where $X$ is a subset of $G$. Suppose $langle Xsetminus{x}ranglene G$; then this subgroup is contained in a maximal subgroup $H$ and $xin H$ by assumption, so $langle Xrangle=bigl<langle Xsetminus{x}ranglecup{x}bigr>subseteq H$. Therefore $langle Xranglene G$.
Conversely, suppose that $x$ is a nongenerator, that is, for any subset $Xsubseteq G$ containing $x$, if $langle Xrangle=G$, then $langle Xsetminus{x}rangle=G$. We want to show that this nongenerator $x$ belongs to all maximal subgroups of $G$. If $H$ is a maximal subgroup and $xnotin H$, then $langle Hcup{x}=G$, so, being $x$ a nongenerator, also $H=langle (Hcup{x})setminus{x}rangle=G$, which is a contradiction.
In the Klein group $V={1,a,b,c}$, the maximal subgroups are ${1,a}$, ${1,b}$ and ${1,c}$ and $1$ is the only element belonging to all maximal subgroups.
Note that just noting that ${a,b,c}$ generates $V$ but also ${b,c}$ generates $V$ doesn't mean $a$ is a nongenerator. This set ${a,b,c}$ is just one of the sets of generators containing $a$. Indeed, consider $X={a,b}$. Then $langle Xrangle=V$, but $langle Xsetminus{a}rangle=langle{b}rangle={1,b}ne V$. Thus $a$ is not a nongenerator.
answered Nov 15 '14 at 14:07
egregegreg
183k1486205
edited Jan 20 at 10:45
answered Nov 15 '14 at 14:07
egregegreg
183k1486205
answered Nov 15 '14 at 14:07
egregegreg
183k1486205
answered Nov 15 '14 at 14:07
egregegreg
183k1486205
183k1486205
$begingroup$
Really nice, thank you! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
add a comment |
$begingroup$
Really nice, thank you! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
$begingroup$
Really nice, thank you! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
$begingroup$
Really nice, thank you! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54
add a comment |
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$begingroup$
In your example, $1$ is the only element that lies in the intersection of all the maximal subgroups, namely, $<a>, <b>, <c>$. I don't see any problem.
$endgroup$
– Krish
Nov 15 '14 at 13:39
$begingroup$
But the thing is, the exercise asks to prove the equivalence of the two statements. So if the latter part is satisfied (that if $a in V_4 Rightarrow V_4setminus{a}$ generates $V_4$) then $a$ must be in the intersection of all the maximal subgroups, which is false?
$endgroup$
– ejcqw
Nov 15 '14 at 13:44
1
$begingroup$
Take $X = {a, b }$. Then $X$ generates the group $V_4$. But $X - {a }$ does not generate $V_4$. The other part of the theorem must be true for all $X$ satisfying the property, not just for a particular set $X$.
$endgroup$
– Krish
Nov 15 '14 at 14:11
$begingroup$
I see why now, thanks! :)
$endgroup$
– ejcqw
Nov 15 '14 at 16:54