Problem about group homomorphism of symmetric and alternating group?
For $n geq 2$,Let $S_n$ and $A_n$ be symmetric and Alternating group on $n$ letters. Let $C^*$ be multiplicative group of complex numbers. Then
There exist a non trivial homomorphism from $S_n$ to $C^*$
There exist a unique non trivial homomorphism from $S_n$ to $C^*$
For $n geq 3$ there exist a non trivial homomorphism from $A_n$ to $C^*$
For $n geq 5$ there in no non trivial homomorphism from $phi: A_n to C^*$
As-
(1) option is correct as map even cycle to $1$ and odd cycle to $-1$, we get homomorphism.
(4) Correct as if there exist such homomorphism, then if Ker$phi = 0$, then $A_n$ becomes embedded in $C^*$ which is not possible as it is non abelian. If $Ker phi$ is non trivial, then this situation is also not possible as $A_n$ is simple for $n geq 5$. So (3) is incorrect.
What about option (2)?
symmetric-groups group-homomorphism
edited Jun 19 '17 at 3:15
pjs36
15.7k32861
asked Jun 19 '17 at 2:43
Mittal G
1,188515
add a comment |
For $n geq 2$,Let $S_n$ and $A_n$ be symmetric and Alternating group on $n$ letters. Let $C^*$ be multiplicative group of complex numbers. Then
There exist a non trivial homomorphism from $S_n$ to $C^*$
There exist a unique non trivial homomorphism from $S_n$ to $C^*$
For $n geq 3$ there exist a non trivial homomorphism from $A_n$ to $C^*$
For $n geq 5$ there in no non trivial homomorphism from $phi: A_n to C^*$
As-
(1) option is correct as map even cycle to $1$ and odd cycle to $-1$, we get homomorphism.
(4) Correct as if there exist such homomorphism, then if Ker$phi = 0$, then $A_n$ becomes embedded in $C^*$ which is not possible as it is non abelian. If $Ker phi$ is non trivial, then this situation is also not possible as $A_n$ is simple for $n geq 5$. So (3) is incorrect.
What about option (2)?
symmetric-groups group-homomorphism
edited Jun 19 '17 at 3:15
pjs36
15.7k32861
asked Jun 19 '17 at 2:43
Mittal G
1,188515
0 is not in the multiplicative group.
– Mariano Suárez-Álvarez
Jun 19 '17 at 2:51
Sorry, i have made the change.
– Mittal G
Jun 19 '17 at 2:51
add a comment |
For $n geq 2$,Let $S_n$ and $A_n$ be symmetric and Alternating group on $n$ letters. Let $C^*$ be multiplicative group of complex numbers. Then
There exist a non trivial homomorphism from $S_n$ to $C^*$
There exist a unique non trivial homomorphism from $S_n$ to $C^*$
For $n geq 3$ there exist a non trivial homomorphism from $A_n$ to $C^*$
For $n geq 5$ there in no non trivial homomorphism from $phi: A_n to C^*$
As-
(1) option is correct as map even cycle to $1$ and odd cycle to $-1$, we get homomorphism.
(4) Correct as if there exist such homomorphism, then if Ker$phi = 0$, then $A_n$ becomes embedded in $C^*$ which is not possible as it is non abelian. If $Ker phi$ is non trivial, then this situation is also not possible as $A_n$ is simple for $n geq 5$. So (3) is incorrect.
What about option (2)?
symmetric-groups group-homomorphism
edited Jun 19 '17 at 3:15
pjs36
15.7k32861
asked Jun 19 '17 at 2:43
Mittal G
1,188515
For $n geq 2$,Let $S_n$ and $A_n$ be symmetric and Alternating group on $n$ letters. Let $C^*$ be multiplicative group of complex numbers. Then
There exist a non trivial homomorphism from $S_n$ to $C^*$
There exist a unique non trivial homomorphism from $S_n$ to $C^*$
For $n geq 3$ there exist a non trivial homomorphism from $A_n$ to $C^*$
For $n geq 5$ there in no non trivial homomorphism from $phi: A_n to C^*$
As-
(1) option is correct as map even cycle to $1$ and odd cycle to $-1$, we get homomorphism.
(4) Correct as if there exist such homomorphism, then if Ker$phi = 0$, then $A_n$ becomes embedded in $C^*$ which is not possible as it is non abelian. If $Ker phi$ is non trivial, then this situation is also not possible as $A_n$ is simple for $n geq 5$. So (3) is incorrect.
What about option (2)?
symmetric-groups group-homomorphism
symmetric-groups group-homomorphism
edited Jun 19 '17 at 3:15
pjs36
15.7k32861
asked Jun 19 '17 at 2:43
Mittal G
1,188515
edited Jun 19 '17 at 3:15
pjs36
15.7k32861
asked Jun 19 '17 at 2:43
Mittal G
1,188515
edited Jun 19 '17 at 3:15
pjs36
15.7k32861
edited Jun 19 '17 at 3:15
pjs36
15.7k32861
edited Jun 19 '17 at 3:15
pjs36
15.7k32861
15.7k32861
asked Jun 19 '17 at 2:43
Mittal G
1,188515
asked Jun 19 '17 at 2:43
Mittal G
1,188515
asked Jun 19 '17 at 2:43
Mittal G
1,188515
1,188515
0 is not in the multiplicative group.
– Mariano Suárez-Álvarez
Jun 19 '17 at 2:51
Sorry, i have made the change.
– Mittal G
Jun 19 '17 at 2:51
add a comment |
0 is not in the multiplicative group.
– Mariano Suárez-Álvarez
Jun 19 '17 at 2:51
Sorry, i have made the change.
– Mittal G
Jun 19 '17 at 2:51
0 is not in the multiplicative group.
– Mariano Suárez-Álvarez
Jun 19 '17 at 2:51
0 is not in the multiplicative group.
– Mariano Suárez-Álvarez
Jun 19 '17 at 2:51
Sorry, i have made the change.
– Mittal G
Jun 19 '17 at 2:51
Sorry, i have made the change.
– Mittal G
Jun 19 '17 at 2:51
add a comment |
2 Answers
2
active
oldest
votes
You're right about parts 1 and 4. Both parts 2 and 3 can be approached via the standard theorem that $G/N$ is abelian if and only if $G' le N$ (where $G'$ is the derived, or commutator, subgroup of $G$).
Part 3 can be answered by verifying that $S_n$ has no abelian quotients aside from $S_n / A_n cong C_2$. You only need to worry about $n = 3, 4$, assuming you know that $A_n$ is the unique proper, nontrivial normal subgroup of $S_n$ for $n ge 5$. For $n = 3$, the alternating group is the only normal subgroup, but for $n = 4$, there's the normal Klein four subgroup $V = {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}$ and quotient $S_4/V$ to think about.
But you should rethink part 2: Are you sure you can't find some $n$ for which $A_n$ has an abelian quotient (including, but not limited to, the case when $N = 1$)? Since part 4 applies and rules out such a thing happening for $n ge 5$, you should be looking at $n le 4$.
answered Jun 19 '17 at 3:14
pjs36
15.7k32861
Then what about the quotient group $S_4/V$? Is there exist such an homomorphism whose kernel is $V$?
– Mittal G
Jun 19 '17 at 4:20
I think in this case $S_4/V$ becomes abelian, then $A_4 subset V_4$ which is not possible.
– Mittal G
Jun 19 '17 at 7:01
$V$ is definitely normal in $S_4$, which means its the kernel of some homomorphism (just not, it'll turn out, to be a homomorphism to an abelian group). It turns out $S_4 / V cong S_3$ (see here for some details), the symmetric group on $3$ letters. The order is right, you just have to make sure that $S_4 / V$ is non-abelian, by finding to cosets $g_1V$ and $g_2V$ that don't commute.
– pjs36
Jun 19 '17 at 14:42
add a comment |
This is surely non-abelian since if it is abelian then it would be cyclic. That means there is element of order 6 in image set. As we know order of image element f(a) must divide the order of a. But in S4 maximum order of element is 4. So this is not possible. Hence it's non-abelian. But S3 is not subgroup of C* so in this case no homomorphism exist.
answered Nov 25 at 19:08
radhika jain
12
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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votes
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oldest
votes
You're right about parts 1 and 4. Both parts 2 and 3 can be approached via the standard theorem that $G/N$ is abelian if and only if $G' le N$ (where $G'$ is the derived, or commutator, subgroup of $G$).
Part 3 can be answered by verifying that $S_n$ has no abelian quotients aside from $S_n / A_n cong C_2$. You only need to worry about $n = 3, 4$, assuming you know that $A_n$ is the unique proper, nontrivial normal subgroup of $S_n$ for $n ge 5$. For $n = 3$, the alternating group is the only normal subgroup, but for $n = 4$, there's the normal Klein four subgroup $V = {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}$ and quotient $S_4/V$ to think about.
But you should rethink part 2: Are you sure you can't find some $n$ for which $A_n$ has an abelian quotient (including, but not limited to, the case when $N = 1$)? Since part 4 applies and rules out such a thing happening for $n ge 5$, you should be looking at $n le 4$.
answered Jun 19 '17 at 3:14
pjs36
15.7k32861
Then what about the quotient group $S_4/V$? Is there exist such an homomorphism whose kernel is $V$?
– Mittal G
Jun 19 '17 at 4:20
I think in this case $S_4/V$ becomes abelian, then $A_4 subset V_4$ which is not possible.
– Mittal G
Jun 19 '17 at 7:01
$V$ is definitely normal in $S_4$, which means its the kernel of some homomorphism (just not, it'll turn out, to be a homomorphism to an abelian group). It turns out $S_4 / V cong S_3$ (see here for some details), the symmetric group on $3$ letters. The order is right, you just have to make sure that $S_4 / V$ is non-abelian, by finding to cosets $g_1V$ and $g_2V$ that don't commute.
– pjs36
Jun 19 '17 at 14:42
add a comment |
You're right about parts 1 and 4. Both parts 2 and 3 can be approached via the standard theorem that $G/N$ is abelian if and only if $G' le N$ (where $G'$ is the derived, or commutator, subgroup of $G$).
Part 3 can be answered by verifying that $S_n$ has no abelian quotients aside from $S_n / A_n cong C_2$. You only need to worry about $n = 3, 4$, assuming you know that $A_n$ is the unique proper, nontrivial normal subgroup of $S_n$ for $n ge 5$. For $n = 3$, the alternating group is the only normal subgroup, but for $n = 4$, there's the normal Klein four subgroup $V = {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}$ and quotient $S_4/V$ to think about.
But you should rethink part 2: Are you sure you can't find some $n$ for which $A_n$ has an abelian quotient (including, but not limited to, the case when $N = 1$)? Since part 4 applies and rules out such a thing happening for $n ge 5$, you should be looking at $n le 4$.
answered Jun 19 '17 at 3:14
pjs36
15.7k32861
Then what about the quotient group $S_4/V$? Is there exist such an homomorphism whose kernel is $V$?
– Mittal G
Jun 19 '17 at 4:20
I think in this case $S_4/V$ becomes abelian, then $A_4 subset V_4$ which is not possible.
– Mittal G
Jun 19 '17 at 7:01
$V$ is definitely normal in $S_4$, which means its the kernel of some homomorphism (just not, it'll turn out, to be a homomorphism to an abelian group). It turns out $S_4 / V cong S_3$ (see here for some details), the symmetric group on $3$ letters. The order is right, you just have to make sure that $S_4 / V$ is non-abelian, by finding to cosets $g_1V$ and $g_2V$ that don't commute.
– pjs36
Jun 19 '17 at 14:42
add a comment |
You're right about parts 1 and 4. Both parts 2 and 3 can be approached via the standard theorem that $G/N$ is abelian if and only if $G' le N$ (where $G'$ is the derived, or commutator, subgroup of $G$).
Part 3 can be answered by verifying that $S_n$ has no abelian quotients aside from $S_n / A_n cong C_2$. You only need to worry about $n = 3, 4$, assuming you know that $A_n$ is the unique proper, nontrivial normal subgroup of $S_n$ for $n ge 5$. For $n = 3$, the alternating group is the only normal subgroup, but for $n = 4$, there's the normal Klein four subgroup $V = {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}$ and quotient $S_4/V$ to think about.
But you should rethink part 2: Are you sure you can't find some $n$ for which $A_n$ has an abelian quotient (including, but not limited to, the case when $N = 1$)? Since part 4 applies and rules out such a thing happening for $n ge 5$, you should be looking at $n le 4$.
answered Jun 19 '17 at 3:14
pjs36
15.7k32861
You're right about parts 1 and 4. Both parts 2 and 3 can be approached via the standard theorem that $G/N$ is abelian if and only if $G' le N$ (where $G'$ is the derived, or commutator, subgroup of $G$).
Part 3 can be answered by verifying that $S_n$ has no abelian quotients aside from $S_n / A_n cong C_2$. You only need to worry about $n = 3, 4$, assuming you know that $A_n$ is the unique proper, nontrivial normal subgroup of $S_n$ for $n ge 5$. For $n = 3$, the alternating group is the only normal subgroup, but for $n = 4$, there's the normal Klein four subgroup $V = {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}$ and quotient $S_4/V$ to think about.
But you should rethink part 2: Are you sure you can't find some $n$ for which $A_n$ has an abelian quotient (including, but not limited to, the case when $N = 1$)? Since part 4 applies and rules out such a thing happening for $n ge 5$, you should be looking at $n le 4$.
answered Jun 19 '17 at 3:14
pjs36
15.7k32861
answered Jun 19 '17 at 3:14
pjs36
15.7k32861
answered Jun 19 '17 at 3:14
pjs36
15.7k32861
answered Jun 19 '17 at 3:14
pjs36
15.7k32861
15.7k32861
Then what about the quotient group $S_4/V$? Is there exist such an homomorphism whose kernel is $V$?
– Mittal G
Jun 19 '17 at 4:20
I think in this case $S_4/V$ becomes abelian, then $A_4 subset V_4$ which is not possible.
– Mittal G
Jun 19 '17 at 7:01
$V$ is definitely normal in $S_4$, which means its the kernel of some homomorphism (just not, it'll turn out, to be a homomorphism to an abelian group). It turns out $S_4 / V cong S_3$ (see here for some details), the symmetric group on $3$ letters. The order is right, you just have to make sure that $S_4 / V$ is non-abelian, by finding to cosets $g_1V$ and $g_2V$ that don't commute.
– pjs36
Jun 19 '17 at 14:42
add a comment |
Then what about the quotient group $S_4/V$? Is there exist such an homomorphism whose kernel is $V$?
– Mittal G
Jun 19 '17 at 4:20
I think in this case $S_4/V$ becomes abelian, then $A_4 subset V_4$ which is not possible.
– Mittal G
Jun 19 '17 at 7:01
$V$ is definitely normal in $S_4$, which means its the kernel of some homomorphism (just not, it'll turn out, to be a homomorphism to an abelian group). It turns out $S_4 / V cong S_3$ (see here for some details), the symmetric group on $3$ letters. The order is right, you just have to make sure that $S_4 / V$ is non-abelian, by finding to cosets $g_1V$ and $g_2V$ that don't commute.
– pjs36
Jun 19 '17 at 14:42
Then what about the quotient group $S_4/V$? Is there exist such an homomorphism whose kernel is $V$?
– Mittal G
Jun 19 '17 at 4:20
Then what about the quotient group $S_4/V$? Is there exist such an homomorphism whose kernel is $V$?
– Mittal G
Jun 19 '17 at 4:20
I think in this case $S_4/V$ becomes abelian, then $A_4 subset V_4$ which is not possible.
– Mittal G
Jun 19 '17 at 7:01
I think in this case $S_4/V$ becomes abelian, then $A_4 subset V_4$ which is not possible.
– Mittal G
Jun 19 '17 at 7:01
$V$ is definitely normal in $S_4$, which means its the kernel of some homomorphism (just not, it'll turn out, to be a homomorphism to an abelian group). It turns out $S_4 / V cong S_3$ (see here for some details), the symmetric group on $3$ letters. The order is right, you just have to make sure that $S_4 / V$ is non-abelian, by finding to cosets $g_1V$ and $g_2V$ that don't commute.
– pjs36
Jun 19 '17 at 14:42
$V$ is definitely normal in $S_4$, which means its the kernel of some homomorphism (just not, it'll turn out, to be a homomorphism to an abelian group). It turns out $S_4 / V cong S_3$ (see here for some details), the symmetric group on $3$ letters. The order is right, you just have to make sure that $S_4 / V$ is non-abelian, by finding to cosets $g_1V$ and $g_2V$ that don't commute.
– pjs36
Jun 19 '17 at 14:42
add a comment |
This is surely non-abelian since if it is abelian then it would be cyclic. That means there is element of order 6 in image set. As we know order of image element f(a) must divide the order of a. But in S4 maximum order of element is 4. So this is not possible. Hence it's non-abelian. But S3 is not subgroup of C* so in this case no homomorphism exist.
answered Nov 25 at 19:08
radhika jain
12
add a comment |
This is surely non-abelian since if it is abelian then it would be cyclic. That means there is element of order 6 in image set. As we know order of image element f(a) must divide the order of a. But in S4 maximum order of element is 4. So this is not possible. Hence it's non-abelian. But S3 is not subgroup of C* so in this case no homomorphism exist.
answered Nov 25 at 19:08
radhika jain
12
add a comment |
This is surely non-abelian since if it is abelian then it would be cyclic. That means there is element of order 6 in image set. As we know order of image element f(a) must divide the order of a. But in S4 maximum order of element is 4. So this is not possible. Hence it's non-abelian. But S3 is not subgroup of C* so in this case no homomorphism exist.
answered Nov 25 at 19:08
radhika jain
12
This is surely non-abelian since if it is abelian then it would be cyclic. That means there is element of order 6 in image set. As we know order of image element f(a) must divide the order of a. But in S4 maximum order of element is 4. So this is not possible. Hence it's non-abelian. But S3 is not subgroup of C* so in this case no homomorphism exist.
answered Nov 25 at 19:08
radhika jain
12
answered Nov 25 at 19:08
radhika jain
12
answered Nov 25 at 19:08
radhika jain
12
answered Nov 25 at 19:08
radhika jain
12
12
add a comment |
add a comment |
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0 is not in the multiplicative group.
– Mariano Suárez-Álvarez
Jun 19 '17 at 2:51
Sorry, i have made the change.
– Mittal G
Jun 19 '17 at 2:51