Mixing
Group generated by $(1234567)$ and $(26)(34)$ is of order 168.
$begingroup$
Let $G$ be a subgroup of $S_7$ generated by $(1234567)$ and $(26)(34)$. Show that $|G| = 168$.
This is a question from Algebra by Hungerford (page.112, exercise 15). And I just have no idea of dealing with this kind of problem about the structure of large number finite groups. Could you please also recommend some reference books?
abstract-algebra finite-groups symmetric-groups
asked Jan 16 at 11:41
X.T ChenX.T Chen
918
$endgroup$
add a comment |
$begingroup$
Let $G$ be a subgroup of $S_7$ generated by $(1234567)$ and $(26)(34)$. Show that $|G| = 168$.
This is a question from Algebra by Hungerford (page.112, exercise 15). And I just have no idea of dealing with this kind of problem about the structure of large number finite groups. Could you please also recommend some reference books?
abstract-algebra finite-groups symmetric-groups
asked Jan 16 at 11:41
X.T ChenX.T Chen
918
$endgroup$
$begingroup$
See the solutions for this homework, question $(3)$. The homepage is here.
$endgroup$
– Dietrich Burde
Jan 16 at 11:53
add a comment |
$begingroup$
Let $G$ be a subgroup of $S_7$ generated by $(1234567)$ and $(26)(34)$. Show that $|G| = 168$.
This is a question from Algebra by Hungerford (page.112, exercise 15). And I just have no idea of dealing with this kind of problem about the structure of large number finite groups. Could you please also recommend some reference books?
abstract-algebra finite-groups symmetric-groups
asked Jan 16 at 11:41
X.T ChenX.T Chen
918
$endgroup$
Let $G$ be a subgroup of $S_7$ generated by $(1234567)$ and $(26)(34)$. Show that $|G| = 168$.
This is a question from Algebra by Hungerford (page.112, exercise 15). And I just have no idea of dealing with this kind of problem about the structure of large number finite groups. Could you please also recommend some reference books?
abstract-algebra finite-groups symmetric-groups
abstract-algebra finite-groups symmetric-groups
asked Jan 16 at 11:41
X.T ChenX.T Chen
918
asked Jan 16 at 11:41
X.T ChenX.T Chen
918
asked Jan 16 at 11:41
X.T ChenX.T Chen
918
asked Jan 16 at 11:41
X.T ChenX.T Chen
918
asked Jan 16 at 11:41
X.T ChenX.T Chen
918
918
$begingroup$
See the solutions for this homework, question $(3)$. The homepage is here.
$endgroup$
– Dietrich Burde
Jan 16 at 11:53
add a comment |
$begingroup$
See the solutions for this homework, question $(3)$. The homepage is here.
$endgroup$
– Dietrich Burde
Jan 16 at 11:53
$begingroup$
See the solutions for this homework, question $(3)$. The homepage is here.
$endgroup$
– Dietrich Burde
Jan 16 at 11:53
$begingroup$
See the solutions for this homework, question $(3)$. The homepage is here.
$endgroup$
– Dietrich Burde
Jan 16 at 11:53
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
And now for something really out there...
Consider the seven nonzero points in $F_2^3$, where $F_2$ is the field with three elements. Each element of the 168-element simple group $GL_3(F_2)$ acts on these seven points as a permutation. If we can realize these two elements as invertible matrices, $G$ will be a subgroup of the full linear group, with order dividing $168$.
That simple group is part of the $PSL_n(K)$ family; working over $F_2$ allows us to ignore both the "special" (determinant $1$) and "projective" (quotient by the constant multiples of $I$) parts of that.
So then, first, how do we get something of order $7$? Well, $F_2^3$ can be given a multiplication operator to make it a field. The nonzero elements in that field form a group of order $7$, and any of them other than the identity will have order $7$, acting on the points as a $7$-cycle. Let $x$ be one of these nontrivial elements, and choose a basis $(1,x,x^2)$ for $F_2^3$. There is some polynomial relation writing $x^3$ as a linear combination of the smaller ones - this will be an irreducible polynomial, the minimal polynomial of $x$. We have two choices - either $x^3+x^2+1=0$ or $x^3+x+1=0$. Let's go with the former. Then, with respect to that basis, multiplication by $x$ has the matrix $X=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}$. The seven elements we cycle through are, in order, $1=begin{bmatrix}1\0\0end{bmatrix}$, $x=begin{bmatrix}0\1\0end{bmatrix}$, $x^2=begin{bmatrix}0\0\1end{bmatrix}$, $x^3=begin{bmatrix}1\0\1end{bmatrix}$, $x^4=begin{bmatrix}1\1\1end{bmatrix}$, $x^5=begin{bmatrix}1\1\0end{bmatrix}$, and $x^6=begin{bmatrix}0\1\1end{bmatrix}$.
OK, now how can we get that $(26)(34)$ permutation? Which element is labeled "1" is an arbitrary choice, so we can start that anywhere in the cycle. For simplicity, make those labels the exponent of $x$ ($1=x^0=x^7$). So then, we need an invertible matrix that fixes $begin{bmatrix}1\0\0end{bmatrix}$ and $begin{bmatrix}0\1\0end{bmatrix}$, while sending $begin{bmatrix}0\0\1end{bmatrix}$ to $begin{bmatrix}0\1\1end{bmatrix}$. That matrix is $A=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}$. Checking its action on the other elements - $Ax^3=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\0\1end{bmatrix}=begin{bmatrix}1\1\1end{bmatrix}=x^4$, $Ax^4=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\1end{bmatrix}=begin{bmatrix}1\0\1end{bmatrix}=x^3$, $Ax^5=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\0end{bmatrix}=begin{bmatrix}1\1\0end{bmatrix}=x^5$, and $Ax^6=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}0\1\1end{bmatrix}=begin{bmatrix}0\0\1end{bmatrix}=x^2$. Confirmed; these matrices have the proper action, and we have realized this group of permutations as a subgroup of $SL_3(F_2)$.
So then, all that remains is to show that the group generated by these two matrices $A$ and $X$ is the whole linear group $G=SL_3(F_2)$. To make our lives easier, we note that $G$ cannot have any nontrivial subgroups with index $le 5$; if $H$ is a subgroup with index $k$, $G$ acts on the conjugates of $H$ by conjugation, giving a homomorphism from $G$ to the symmetric group on $k$ elements. The kernel of that is a normal subgroup with index at most $k!$ in $G$, the intersection of all conjugates of $H$. Since we know $G$ is simple, it can't have any nontrivial normal subgroups of order $le 5!=120$, and there are no nontrivial subgroups of index $le 5$.
In particular, there are no subgroups of index $2$ or $4$. If we can find an element of order $3$ in the subgroup generated by $A$ and $X$, that subgroup will have order divisible by $2cdot 3cdot 7 = frac{168}{4}$, and will have to be all of $G$.
This is easy. $XA=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix} =begin{bmatrix}0&0&1\1&0&0\0&1&0end{bmatrix}$ has order $3$, and the subgroup generated by $A$ and $X$ must be all of $SL_3(F_2)$. We have found not only the group's order but it's full structure, the unique simple group of order $168$.
answered Jan 16 at 20:48
jmerryjmerry
9,8171225
$endgroup$
$begingroup$
When I saw "168" I figured it had to be the same group. Thanks for making it explicit!
$endgroup$
– Cheerful Parsnip
Jan 16 at 20:52
add a comment |
$begingroup$
Here is a short outline.
Set $x=(1 2 3 4 5 6 7)$, $y=(2 6)(3 4)$ and $S = <x>$. Clearly $G = <x, y> leq Alt(7)$. Hence $S$ is a Sylow $7$-subgroup of $G$. $Alt(7)$ has 6! elements of order $7$ and therefore 5! Sylow $7$-subgroups. Hence $|N_{Alt(7)}(S)| = 21$. Set $T_1 = S^{y}$ and $T_{i+1} = T_{i}^{x}$. The set ${S, T_{1}, ldots, T_{7}}$ is $G$-invariant. Hence it constitutes the set of all Sylow $7$-subgroups of $G$. In particular $|G : N_{G}(S)| = 8$. Since $S leq N_{G}(S) leq N_{Alt(7)}(S)$ either $|N_{G}(S)| = 7$ or $|N_{G}(S)| = 21$. As a result there is only two options for $|G|$. Now determine the order of $x cdot y$.
answered Jan 16 at 19:29
user515430user515430
17613
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
And now for something really out there...
Consider the seven nonzero points in $F_2^3$, where $F_2$ is the field with three elements. Each element of the 168-element simple group $GL_3(F_2)$ acts on these seven points as a permutation. If we can realize these two elements as invertible matrices, $G$ will be a subgroup of the full linear group, with order dividing $168$.
That simple group is part of the $PSL_n(K)$ family; working over $F_2$ allows us to ignore both the "special" (determinant $1$) and "projective" (quotient by the constant multiples of $I$) parts of that.
So then, first, how do we get something of order $7$? Well, $F_2^3$ can be given a multiplication operator to make it a field. The nonzero elements in that field form a group of order $7$, and any of them other than the identity will have order $7$, acting on the points as a $7$-cycle. Let $x$ be one of these nontrivial elements, and choose a basis $(1,x,x^2)$ for $F_2^3$. There is some polynomial relation writing $x^3$ as a linear combination of the smaller ones - this will be an irreducible polynomial, the minimal polynomial of $x$. We have two choices - either $x^3+x^2+1=0$ or $x^3+x+1=0$. Let's go with the former. Then, with respect to that basis, multiplication by $x$ has the matrix $X=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}$. The seven elements we cycle through are, in order, $1=begin{bmatrix}1\0\0end{bmatrix}$, $x=begin{bmatrix}0\1\0end{bmatrix}$, $x^2=begin{bmatrix}0\0\1end{bmatrix}$, $x^3=begin{bmatrix}1\0\1end{bmatrix}$, $x^4=begin{bmatrix}1\1\1end{bmatrix}$, $x^5=begin{bmatrix}1\1\0end{bmatrix}$, and $x^6=begin{bmatrix}0\1\1end{bmatrix}$.
OK, now how can we get that $(26)(34)$ permutation? Which element is labeled "1" is an arbitrary choice, so we can start that anywhere in the cycle. For simplicity, make those labels the exponent of $x$ ($1=x^0=x^7$). So then, we need an invertible matrix that fixes $begin{bmatrix}1\0\0end{bmatrix}$ and $begin{bmatrix}0\1\0end{bmatrix}$, while sending $begin{bmatrix}0\0\1end{bmatrix}$ to $begin{bmatrix}0\1\1end{bmatrix}$. That matrix is $A=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}$. Checking its action on the other elements - $Ax^3=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\0\1end{bmatrix}=begin{bmatrix}1\1\1end{bmatrix}=x^4$, $Ax^4=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\1end{bmatrix}=begin{bmatrix}1\0\1end{bmatrix}=x^3$, $Ax^5=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\0end{bmatrix}=begin{bmatrix}1\1\0end{bmatrix}=x^5$, and $Ax^6=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}0\1\1end{bmatrix}=begin{bmatrix}0\0\1end{bmatrix}=x^2$. Confirmed; these matrices have the proper action, and we have realized this group of permutations as a subgroup of $SL_3(F_2)$.
So then, all that remains is to show that the group generated by these two matrices $A$ and $X$ is the whole linear group $G=SL_3(F_2)$. To make our lives easier, we note that $G$ cannot have any nontrivial subgroups with index $le 5$; if $H$ is a subgroup with index $k$, $G$ acts on the conjugates of $H$ by conjugation, giving a homomorphism from $G$ to the symmetric group on $k$ elements. The kernel of that is a normal subgroup with index at most $k!$ in $G$, the intersection of all conjugates of $H$. Since we know $G$ is simple, it can't have any nontrivial normal subgroups of order $le 5!=120$, and there are no nontrivial subgroups of index $le 5$.
In particular, there are no subgroups of index $2$ or $4$. If we can find an element of order $3$ in the subgroup generated by $A$ and $X$, that subgroup will have order divisible by $2cdot 3cdot 7 = frac{168}{4}$, and will have to be all of $G$.
This is easy. $XA=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix} =begin{bmatrix}0&0&1\1&0&0\0&1&0end{bmatrix}$ has order $3$, and the subgroup generated by $A$ and $X$ must be all of $SL_3(F_2)$. We have found not only the group's order but it's full structure, the unique simple group of order $168$.
answered Jan 16 at 20:48
jmerryjmerry
9,8171225
$endgroup$
$begingroup$
When I saw "168" I figured it had to be the same group. Thanks for making it explicit!
$endgroup$
– Cheerful Parsnip
Jan 16 at 20:52
add a comment |
$begingroup$
And now for something really out there...
Consider the seven nonzero points in $F_2^3$, where $F_2$ is the field with three elements. Each element of the 168-element simple group $GL_3(F_2)$ acts on these seven points as a permutation. If we can realize these two elements as invertible matrices, $G$ will be a subgroup of the full linear group, with order dividing $168$.
That simple group is part of the $PSL_n(K)$ family; working over $F_2$ allows us to ignore both the "special" (determinant $1$) and "projective" (quotient by the constant multiples of $I$) parts of that.
So then, first, how do we get something of order $7$? Well, $F_2^3$ can be given a multiplication operator to make it a field. The nonzero elements in that field form a group of order $7$, and any of them other than the identity will have order $7$, acting on the points as a $7$-cycle. Let $x$ be one of these nontrivial elements, and choose a basis $(1,x,x^2)$ for $F_2^3$. There is some polynomial relation writing $x^3$ as a linear combination of the smaller ones - this will be an irreducible polynomial, the minimal polynomial of $x$. We have two choices - either $x^3+x^2+1=0$ or $x^3+x+1=0$. Let's go with the former. Then, with respect to that basis, multiplication by $x$ has the matrix $X=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}$. The seven elements we cycle through are, in order, $1=begin{bmatrix}1\0\0end{bmatrix}$, $x=begin{bmatrix}0\1\0end{bmatrix}$, $x^2=begin{bmatrix}0\0\1end{bmatrix}$, $x^3=begin{bmatrix}1\0\1end{bmatrix}$, $x^4=begin{bmatrix}1\1\1end{bmatrix}$, $x^5=begin{bmatrix}1\1\0end{bmatrix}$, and $x^6=begin{bmatrix}0\1\1end{bmatrix}$.
OK, now how can we get that $(26)(34)$ permutation? Which element is labeled "1" is an arbitrary choice, so we can start that anywhere in the cycle. For simplicity, make those labels the exponent of $x$ ($1=x^0=x^7$). So then, we need an invertible matrix that fixes $begin{bmatrix}1\0\0end{bmatrix}$ and $begin{bmatrix}0\1\0end{bmatrix}$, while sending $begin{bmatrix}0\0\1end{bmatrix}$ to $begin{bmatrix}0\1\1end{bmatrix}$. That matrix is $A=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}$. Checking its action on the other elements - $Ax^3=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\0\1end{bmatrix}=begin{bmatrix}1\1\1end{bmatrix}=x^4$, $Ax^4=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\1end{bmatrix}=begin{bmatrix}1\0\1end{bmatrix}=x^3$, $Ax^5=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\0end{bmatrix}=begin{bmatrix}1\1\0end{bmatrix}=x^5$, and $Ax^6=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}0\1\1end{bmatrix}=begin{bmatrix}0\0\1end{bmatrix}=x^2$. Confirmed; these matrices have the proper action, and we have realized this group of permutations as a subgroup of $SL_3(F_2)$.
So then, all that remains is to show that the group generated by these two matrices $A$ and $X$ is the whole linear group $G=SL_3(F_2)$. To make our lives easier, we note that $G$ cannot have any nontrivial subgroups with index $le 5$; if $H$ is a subgroup with index $k$, $G$ acts on the conjugates of $H$ by conjugation, giving a homomorphism from $G$ to the symmetric group on $k$ elements. The kernel of that is a normal subgroup with index at most $k!$ in $G$, the intersection of all conjugates of $H$. Since we know $G$ is simple, it can't have any nontrivial normal subgroups of order $le 5!=120$, and there are no nontrivial subgroups of index $le 5$.
In particular, there are no subgroups of index $2$ or $4$. If we can find an element of order $3$ in the subgroup generated by $A$ and $X$, that subgroup will have order divisible by $2cdot 3cdot 7 = frac{168}{4}$, and will have to be all of $G$.
This is easy. $XA=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix} =begin{bmatrix}0&0&1\1&0&0\0&1&0end{bmatrix}$ has order $3$, and the subgroup generated by $A$ and $X$ must be all of $SL_3(F_2)$. We have found not only the group's order but it's full structure, the unique simple group of order $168$.
answered Jan 16 at 20:48
jmerryjmerry
9,8171225
$endgroup$
$begingroup$
When I saw "168" I figured it had to be the same group. Thanks for making it explicit!
$endgroup$
– Cheerful Parsnip
Jan 16 at 20:52
add a comment |
$begingroup$
And now for something really out there...
Consider the seven nonzero points in $F_2^3$, where $F_2$ is the field with three elements. Each element of the 168-element simple group $GL_3(F_2)$ acts on these seven points as a permutation. If we can realize these two elements as invertible matrices, $G$ will be a subgroup of the full linear group, with order dividing $168$.
That simple group is part of the $PSL_n(K)$ family; working over $F_2$ allows us to ignore both the "special" (determinant $1$) and "projective" (quotient by the constant multiples of $I$) parts of that.
So then, first, how do we get something of order $7$? Well, $F_2^3$ can be given a multiplication operator to make it a field. The nonzero elements in that field form a group of order $7$, and any of them other than the identity will have order $7$, acting on the points as a $7$-cycle. Let $x$ be one of these nontrivial elements, and choose a basis $(1,x,x^2)$ for $F_2^3$. There is some polynomial relation writing $x^3$ as a linear combination of the smaller ones - this will be an irreducible polynomial, the minimal polynomial of $x$. We have two choices - either $x^3+x^2+1=0$ or $x^3+x+1=0$. Let's go with the former. Then, with respect to that basis, multiplication by $x$ has the matrix $X=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}$. The seven elements we cycle through are, in order, $1=begin{bmatrix}1\0\0end{bmatrix}$, $x=begin{bmatrix}0\1\0end{bmatrix}$, $x^2=begin{bmatrix}0\0\1end{bmatrix}$, $x^3=begin{bmatrix}1\0\1end{bmatrix}$, $x^4=begin{bmatrix}1\1\1end{bmatrix}$, $x^5=begin{bmatrix}1\1\0end{bmatrix}$, and $x^6=begin{bmatrix}0\1\1end{bmatrix}$.
OK, now how can we get that $(26)(34)$ permutation? Which element is labeled "1" is an arbitrary choice, so we can start that anywhere in the cycle. For simplicity, make those labels the exponent of $x$ ($1=x^0=x^7$). So then, we need an invertible matrix that fixes $begin{bmatrix}1\0\0end{bmatrix}$ and $begin{bmatrix}0\1\0end{bmatrix}$, while sending $begin{bmatrix}0\0\1end{bmatrix}$ to $begin{bmatrix}0\1\1end{bmatrix}$. That matrix is $A=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}$. Checking its action on the other elements - $Ax^3=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\0\1end{bmatrix}=begin{bmatrix}1\1\1end{bmatrix}=x^4$, $Ax^4=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\1end{bmatrix}=begin{bmatrix}1\0\1end{bmatrix}=x^3$, $Ax^5=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\0end{bmatrix}=begin{bmatrix}1\1\0end{bmatrix}=x^5$, and $Ax^6=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}0\1\1end{bmatrix}=begin{bmatrix}0\0\1end{bmatrix}=x^2$. Confirmed; these matrices have the proper action, and we have realized this group of permutations as a subgroup of $SL_3(F_2)$.
So then, all that remains is to show that the group generated by these two matrices $A$ and $X$ is the whole linear group $G=SL_3(F_2)$. To make our lives easier, we note that $G$ cannot have any nontrivial subgroups with index $le 5$; if $H$ is a subgroup with index $k$, $G$ acts on the conjugates of $H$ by conjugation, giving a homomorphism from $G$ to the symmetric group on $k$ elements. The kernel of that is a normal subgroup with index at most $k!$ in $G$, the intersection of all conjugates of $H$. Since we know $G$ is simple, it can't have any nontrivial normal subgroups of order $le 5!=120$, and there are no nontrivial subgroups of index $le 5$.
In particular, there are no subgroups of index $2$ or $4$. If we can find an element of order $3$ in the subgroup generated by $A$ and $X$, that subgroup will have order divisible by $2cdot 3cdot 7 = frac{168}{4}$, and will have to be all of $G$.
This is easy. $XA=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix} =begin{bmatrix}0&0&1\1&0&0\0&1&0end{bmatrix}$ has order $3$, and the subgroup generated by $A$ and $X$ must be all of $SL_3(F_2)$. We have found not only the group's order but it's full structure, the unique simple group of order $168$.
answered Jan 16 at 20:48
jmerryjmerry
9,8171225
$endgroup$
And now for something really out there...
Consider the seven nonzero points in $F_2^3$, where $F_2$ is the field with three elements. Each element of the 168-element simple group $GL_3(F_2)$ acts on these seven points as a permutation. If we can realize these two elements as invertible matrices, $G$ will be a subgroup of the full linear group, with order dividing $168$.
That simple group is part of the $PSL_n(K)$ family; working over $F_2$ allows us to ignore both the "special" (determinant $1$) and "projective" (quotient by the constant multiples of $I$) parts of that.
So then, first, how do we get something of order $7$? Well, $F_2^3$ can be given a multiplication operator to make it a field. The nonzero elements in that field form a group of order $7$, and any of them other than the identity will have order $7$, acting on the points as a $7$-cycle. Let $x$ be one of these nontrivial elements, and choose a basis $(1,x,x^2)$ for $F_2^3$. There is some polynomial relation writing $x^3$ as a linear combination of the smaller ones - this will be an irreducible polynomial, the minimal polynomial of $x$. We have two choices - either $x^3+x^2+1=0$ or $x^3+x+1=0$. Let's go with the former. Then, with respect to that basis, multiplication by $x$ has the matrix $X=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}$. The seven elements we cycle through are, in order, $1=begin{bmatrix}1\0\0end{bmatrix}$, $x=begin{bmatrix}0\1\0end{bmatrix}$, $x^2=begin{bmatrix}0\0\1end{bmatrix}$, $x^3=begin{bmatrix}1\0\1end{bmatrix}$, $x^4=begin{bmatrix}1\1\1end{bmatrix}$, $x^5=begin{bmatrix}1\1\0end{bmatrix}$, and $x^6=begin{bmatrix}0\1\1end{bmatrix}$.
OK, now how can we get that $(26)(34)$ permutation? Which element is labeled "1" is an arbitrary choice, so we can start that anywhere in the cycle. For simplicity, make those labels the exponent of $x$ ($1=x^0=x^7$). So then, we need an invertible matrix that fixes $begin{bmatrix}1\0\0end{bmatrix}$ and $begin{bmatrix}0\1\0end{bmatrix}$, while sending $begin{bmatrix}0\0\1end{bmatrix}$ to $begin{bmatrix}0\1\1end{bmatrix}$. That matrix is $A=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}$. Checking its action on the other elements - $Ax^3=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\0\1end{bmatrix}=begin{bmatrix}1\1\1end{bmatrix}=x^4$, $Ax^4=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\1end{bmatrix}=begin{bmatrix}1\0\1end{bmatrix}=x^3$, $Ax^5=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}1\1\0end{bmatrix}=begin{bmatrix}1\1\0end{bmatrix}=x^5$, and $Ax^6=begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}begin{bmatrix}0\1\1end{bmatrix}=begin{bmatrix}0\0\1end{bmatrix}=x^2$. Confirmed; these matrices have the proper action, and we have realized this group of permutations as a subgroup of $SL_3(F_2)$.
So then, all that remains is to show that the group generated by these two matrices $A$ and $X$ is the whole linear group $G=SL_3(F_2)$. To make our lives easier, we note that $G$ cannot have any nontrivial subgroups with index $le 5$; if $H$ is a subgroup with index $k$, $G$ acts on the conjugates of $H$ by conjugation, giving a homomorphism from $G$ to the symmetric group on $k$ elements. The kernel of that is a normal subgroup with index at most $k!$ in $G$, the intersection of all conjugates of $H$. Since we know $G$ is simple, it can't have any nontrivial normal subgroups of order $le 5!=120$, and there are no nontrivial subgroups of index $le 5$.
In particular, there are no subgroups of index $2$ or $4$. If we can find an element of order $3$ in the subgroup generated by $A$ and $X$, that subgroup will have order divisible by $2cdot 3cdot 7 = frac{168}{4}$, and will have to be all of $G$.
This is easy. $XA=begin{bmatrix}0&0&1\1&0&0\0&1&1end{bmatrix}begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix} =begin{bmatrix}0&0&1\1&0&0\0&1&0end{bmatrix}$ has order $3$, and the subgroup generated by $A$ and $X$ must be all of $SL_3(F_2)$. We have found not only the group's order but it's full structure, the unique simple group of order $168$.
answered Jan 16 at 20:48
jmerryjmerry
9,8171225
answered Jan 16 at 20:48
jmerryjmerry
9,8171225
answered Jan 16 at 20:48
jmerryjmerry
9,8171225
answered Jan 16 at 20:48
jmerryjmerry
9,8171225
9,8171225
$begingroup$
When I saw "168" I figured it had to be the same group. Thanks for making it explicit!
$endgroup$
– Cheerful Parsnip
Jan 16 at 20:52
add a comment |
$begingroup$
When I saw "168" I figured it had to be the same group. Thanks for making it explicit!
$endgroup$
– Cheerful Parsnip
Jan 16 at 20:52
$begingroup$
When I saw "168" I figured it had to be the same group. Thanks for making it explicit!
$endgroup$
– Cheerful Parsnip
Jan 16 at 20:52
$begingroup$
When I saw "168" I figured it had to be the same group. Thanks for making it explicit!
$endgroup$
– Cheerful Parsnip
Jan 16 at 20:52
add a comment |
$begingroup$
Here is a short outline.
Set $x=(1 2 3 4 5 6 7)$, $y=(2 6)(3 4)$ and $S = <x>$. Clearly $G = <x, y> leq Alt(7)$. Hence $S$ is a Sylow $7$-subgroup of $G$. $Alt(7)$ has 6! elements of order $7$ and therefore 5! Sylow $7$-subgroups. Hence $|N_{Alt(7)}(S)| = 21$. Set $T_1 = S^{y}$ and $T_{i+1} = T_{i}^{x}$. The set ${S, T_{1}, ldots, T_{7}}$ is $G$-invariant. Hence it constitutes the set of all Sylow $7$-subgroups of $G$. In particular $|G : N_{G}(S)| = 8$. Since $S leq N_{G}(S) leq N_{Alt(7)}(S)$ either $|N_{G}(S)| = 7$ or $|N_{G}(S)| = 21$. As a result there is only two options for $|G|$. Now determine the order of $x cdot y$.
answered Jan 16 at 19:29
user515430user515430
17613
$endgroup$
add a comment |
$begingroup$
Here is a short outline.
Set $x=(1 2 3 4 5 6 7)$, $y=(2 6)(3 4)$ and $S = <x>$. Clearly $G = <x, y> leq Alt(7)$. Hence $S$ is a Sylow $7$-subgroup of $G$. $Alt(7)$ has 6! elements of order $7$ and therefore 5! Sylow $7$-subgroups. Hence $|N_{Alt(7)}(S)| = 21$. Set $T_1 = S^{y}$ and $T_{i+1} = T_{i}^{x}$. The set ${S, T_{1}, ldots, T_{7}}$ is $G$-invariant. Hence it constitutes the set of all Sylow $7$-subgroups of $G$. In particular $|G : N_{G}(S)| = 8$. Since $S leq N_{G}(S) leq N_{Alt(7)}(S)$ either $|N_{G}(S)| = 7$ or $|N_{G}(S)| = 21$. As a result there is only two options for $|G|$. Now determine the order of $x cdot y$.
answered Jan 16 at 19:29
user515430user515430
17613
$endgroup$
add a comment |
$begingroup$
Here is a short outline.
Set $x=(1 2 3 4 5 6 7)$, $y=(2 6)(3 4)$ and $S = <x>$. Clearly $G = <x, y> leq Alt(7)$. Hence $S$ is a Sylow $7$-subgroup of $G$. $Alt(7)$ has 6! elements of order $7$ and therefore 5! Sylow $7$-subgroups. Hence $|N_{Alt(7)}(S)| = 21$. Set $T_1 = S^{y}$ and $T_{i+1} = T_{i}^{x}$. The set ${S, T_{1}, ldots, T_{7}}$ is $G$-invariant. Hence it constitutes the set of all Sylow $7$-subgroups of $G$. In particular $|G : N_{G}(S)| = 8$. Since $S leq N_{G}(S) leq N_{Alt(7)}(S)$ either $|N_{G}(S)| = 7$ or $|N_{G}(S)| = 21$. As a result there is only two options for $|G|$. Now determine the order of $x cdot y$.
answered Jan 16 at 19:29
user515430user515430
17613
$endgroup$
Here is a short outline.
Set $x=(1 2 3 4 5 6 7)$, $y=(2 6)(3 4)$ and $S = <x>$. Clearly $G = <x, y> leq Alt(7)$. Hence $S$ is a Sylow $7$-subgroup of $G$. $Alt(7)$ has 6! elements of order $7$ and therefore 5! Sylow $7$-subgroups. Hence $|N_{Alt(7)}(S)| = 21$. Set $T_1 = S^{y}$ and $T_{i+1} = T_{i}^{x}$. The set ${S, T_{1}, ldots, T_{7}}$ is $G$-invariant. Hence it constitutes the set of all Sylow $7$-subgroups of $G$. In particular $|G : N_{G}(S)| = 8$. Since $S leq N_{G}(S) leq N_{Alt(7)}(S)$ either $|N_{G}(S)| = 7$ or $|N_{G}(S)| = 21$. As a result there is only two options for $|G|$. Now determine the order of $x cdot y$.
answered Jan 16 at 19:29
user515430user515430
17613
edited Jan 16 at 21:54
answered Jan 16 at 19:29
user515430user515430
17613
answered Jan 16 at 19:29
user515430user515430
17613
answered Jan 16 at 19:29
user515430user515430
17613
17613
add a comment |
add a comment |
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See the solutions for this homework, question $(3)$. The homepage is here.
$endgroup$
– Dietrich Burde
Jan 16 at 11:53