Describing all homomorphisms of $S_n$ into $S_3$
Describe all homomorphisms of $S_n$ into $S_3$, $n geq 5$.
Somehow it is a bit unclear to me what describing means. However, I would say that describing is meant with respect to generators. We have $$S_n = leftlangle begin{pmatrix}1 & 2end{pmatrix},begin{pmatrix}1 & 2 & dots & nend{pmatrix}rightrangle$$ and $|S_3| = 6$. So theoretically there would be $36$ possible homomorphisms. But thats not really descriptive. In the former part of the exercise we've already proved that $text{Aut}(S_3) cong S_3$ but I do not really know if this helps.
abstract-algebra group-homomorphism
asked Oct 31 '16 at 21:37
TheGeekGreekTheGeekGreek
5,08831035
add a comment |
Describe all homomorphisms of $S_n$ into $S_3$, $n geq 5$.
Somehow it is a bit unclear to me what describing means. However, I would say that describing is meant with respect to generators. We have $$S_n = leftlangle begin{pmatrix}1 & 2end{pmatrix},begin{pmatrix}1 & 2 & dots & nend{pmatrix}rightrangle$$ and $|S_3| = 6$. So theoretically there would be $36$ possible homomorphisms. But thats not really descriptive. In the former part of the exercise we've already proved that $text{Aut}(S_3) cong S_3$ but I do not really know if this helps.
abstract-algebra group-homomorphism
asked Oct 31 '16 at 21:37
TheGeekGreekTheGeekGreek
5,08831035
Not all $36$ possibilities define a homomorphism. For instance, $(12)$ must be sent to $(1)$ or another transposition.
– lhf
Oct 31 '16 at 21:40
@Ihf Sorry, that is what I meant with theoretically.
– TheGeekGreek
Oct 31 '16 at 21:41
add a comment |
Describe all homomorphisms of $S_n$ into $S_3$, $n geq 5$.
Somehow it is a bit unclear to me what describing means. However, I would say that describing is meant with respect to generators. We have $$S_n = leftlangle begin{pmatrix}1 & 2end{pmatrix},begin{pmatrix}1 & 2 & dots & nend{pmatrix}rightrangle$$ and $|S_3| = 6$. So theoretically there would be $36$ possible homomorphisms. But thats not really descriptive. In the former part of the exercise we've already proved that $text{Aut}(S_3) cong S_3$ but I do not really know if this helps.
abstract-algebra group-homomorphism
asked Oct 31 '16 at 21:37
TheGeekGreekTheGeekGreek
5,08831035
Describe all homomorphisms of $S_n$ into $S_3$, $n geq 5$.
Somehow it is a bit unclear to me what describing means. However, I would say that describing is meant with respect to generators. We have $$S_n = leftlangle begin{pmatrix}1 & 2end{pmatrix},begin{pmatrix}1 & 2 & dots & nend{pmatrix}rightrangle$$ and $|S_3| = 6$. So theoretically there would be $36$ possible homomorphisms. But thats not really descriptive. In the former part of the exercise we've already proved that $text{Aut}(S_3) cong S_3$ but I do not really know if this helps.
abstract-algebra group-homomorphism
abstract-algebra group-homomorphism
asked Oct 31 '16 at 21:37
TheGeekGreekTheGeekGreek
5,08831035
asked Oct 31 '16 at 21:37
TheGeekGreekTheGeekGreek
5,08831035
asked Oct 31 '16 at 21:37
TheGeekGreekTheGeekGreek
5,08831035
asked Oct 31 '16 at 21:37
TheGeekGreekTheGeekGreek
5,08831035
asked Oct 31 '16 at 21:37
TheGeekGreekTheGeekGreek
5,08831035
5,08831035
Not all $36$ possibilities define a homomorphism. For instance, $(12)$ must be sent to $(1)$ or another transposition.
– lhf
Oct 31 '16 at 21:40
@Ihf Sorry, that is what I meant with theoretically.
– TheGeekGreek
Oct 31 '16 at 21:41
add a comment |
Not all $36$ possibilities define a homomorphism. For instance, $(12)$ must be sent to $(1)$ or another transposition.
– lhf
Oct 31 '16 at 21:40
@Ihf Sorry, that is what I meant with theoretically.
– TheGeekGreek
Oct 31 '16 at 21:41
Not all $36$ possibilities define a homomorphism. For instance, $(12)$ must be sent to $(1)$ or another transposition.
– lhf
Oct 31 '16 at 21:40
Not all $36$ possibilities define a homomorphism. For instance, $(12)$ must be sent to $(1)$ or another transposition.
– lhf
Oct 31 '16 at 21:40
@Ihf Sorry, that is what I meant with theoretically.
– TheGeekGreek
Oct 31 '16 at 21:41
@Ihf Sorry, that is what I meant with theoretically.
– TheGeekGreek
Oct 31 '16 at 21:41
add a comment |
1 Answer
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The trick here is to recall that the kernel of a group homomorphism is a normal subgroup. If $phi:S_n rightarrow S_3$ is a group homomorphism, we must have $S_n / ker(phi) cong operatorname{Im}(phi)$, were $ker(phi)$ is normal in $S_n$.
For $n geq 5$, the only nontrivial normal subgroup of $S_n$ is $A_n$, and $S_n / A_n cong mathbb{Z}_2$ (which is a subgroup of $S_3$, namely, a subgroup generated by a transposition, so a homomorphism $S_n rightarrow S_3$ is possible).
Lastly, don't forget about the trivial homomorphisms ;)
P.S. if you want to consider homomorphisms $S_4 rightarrow S_3$, it's useful to know that $S_4$ also has a normal subgroup isomorphic to $mathbb{Z}_2 times mathbb{Z}_2$. Given that $S_4 / mathbb{Z}_2 times mathbb{Z}_2$ has the same order as $S_3$, there is a possibility of an additional (surjective) homomorphism. Note that it suffices to check whether this quotient group is abelian.
answered Oct 31 '16 at 21:50
Kaj HansenKaj Hansen
27.2k43779
add a comment |
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1 Answer
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1 Answer
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The trick here is to recall that the kernel of a group homomorphism is a normal subgroup. If $phi:S_n rightarrow S_3$ is a group homomorphism, we must have $S_n / ker(phi) cong operatorname{Im}(phi)$, were $ker(phi)$ is normal in $S_n$.
For $n geq 5$, the only nontrivial normal subgroup of $S_n$ is $A_n$, and $S_n / A_n cong mathbb{Z}_2$ (which is a subgroup of $S_3$, namely, a subgroup generated by a transposition, so a homomorphism $S_n rightarrow S_3$ is possible).
Lastly, don't forget about the trivial homomorphisms ;)
P.S. if you want to consider homomorphisms $S_4 rightarrow S_3$, it's useful to know that $S_4$ also has a normal subgroup isomorphic to $mathbb{Z}_2 times mathbb{Z}_2$. Given that $S_4 / mathbb{Z}_2 times mathbb{Z}_2$ has the same order as $S_3$, there is a possibility of an additional (surjective) homomorphism. Note that it suffices to check whether this quotient group is abelian.
answered Oct 31 '16 at 21:50
Kaj HansenKaj Hansen
27.2k43779
add a comment |
The trick here is to recall that the kernel of a group homomorphism is a normal subgroup. If $phi:S_n rightarrow S_3$ is a group homomorphism, we must have $S_n / ker(phi) cong operatorname{Im}(phi)$, were $ker(phi)$ is normal in $S_n$.
For $n geq 5$, the only nontrivial normal subgroup of $S_n$ is $A_n$, and $S_n / A_n cong mathbb{Z}_2$ (which is a subgroup of $S_3$, namely, a subgroup generated by a transposition, so a homomorphism $S_n rightarrow S_3$ is possible).
Lastly, don't forget about the trivial homomorphisms ;)
P.S. if you want to consider homomorphisms $S_4 rightarrow S_3$, it's useful to know that $S_4$ also has a normal subgroup isomorphic to $mathbb{Z}_2 times mathbb{Z}_2$. Given that $S_4 / mathbb{Z}_2 times mathbb{Z}_2$ has the same order as $S_3$, there is a possibility of an additional (surjective) homomorphism. Note that it suffices to check whether this quotient group is abelian.
answered Oct 31 '16 at 21:50
Kaj HansenKaj Hansen
27.2k43779
add a comment |
The trick here is to recall that the kernel of a group homomorphism is a normal subgroup. If $phi:S_n rightarrow S_3$ is a group homomorphism, we must have $S_n / ker(phi) cong operatorname{Im}(phi)$, were $ker(phi)$ is normal in $S_n$.
For $n geq 5$, the only nontrivial normal subgroup of $S_n$ is $A_n$, and $S_n / A_n cong mathbb{Z}_2$ (which is a subgroup of $S_3$, namely, a subgroup generated by a transposition, so a homomorphism $S_n rightarrow S_3$ is possible).
Lastly, don't forget about the trivial homomorphisms ;)
P.S. if you want to consider homomorphisms $S_4 rightarrow S_3$, it's useful to know that $S_4$ also has a normal subgroup isomorphic to $mathbb{Z}_2 times mathbb{Z}_2$. Given that $S_4 / mathbb{Z}_2 times mathbb{Z}_2$ has the same order as $S_3$, there is a possibility of an additional (surjective) homomorphism. Note that it suffices to check whether this quotient group is abelian.
answered Oct 31 '16 at 21:50
Kaj HansenKaj Hansen
27.2k43779
The trick here is to recall that the kernel of a group homomorphism is a normal subgroup. If $phi:S_n rightarrow S_3$ is a group homomorphism, we must have $S_n / ker(phi) cong operatorname{Im}(phi)$, were $ker(phi)$ is normal in $S_n$.
For $n geq 5$, the only nontrivial normal subgroup of $S_n$ is $A_n$, and $S_n / A_n cong mathbb{Z}_2$ (which is a subgroup of $S_3$, namely, a subgroup generated by a transposition, so a homomorphism $S_n rightarrow S_3$ is possible).
Lastly, don't forget about the trivial homomorphisms ;)
P.S. if you want to consider homomorphisms $S_4 rightarrow S_3$, it's useful to know that $S_4$ also has a normal subgroup isomorphic to $mathbb{Z}_2 times mathbb{Z}_2$. Given that $S_4 / mathbb{Z}_2 times mathbb{Z}_2$ has the same order as $S_3$, there is a possibility of an additional (surjective) homomorphism. Note that it suffices to check whether this quotient group is abelian.
answered Oct 31 '16 at 21:50
Kaj HansenKaj Hansen
27.2k43779
edited Dec 4 '18 at 0:37
answered Oct 31 '16 at 21:50
Kaj HansenKaj Hansen
27.2k43779
answered Oct 31 '16 at 21:50
Kaj HansenKaj Hansen
27.2k43779
answered Oct 31 '16 at 21:50
Kaj HansenKaj Hansen
27.2k43779
27.2k43779
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Not all $36$ possibilities define a homomorphism. For instance, $(12)$ must be sent to $(1)$ or another transposition.
– lhf
Oct 31 '16 at 21:40
@Ihf Sorry, that is what I meant with theoretically.
– TheGeekGreek
Oct 31 '16 at 21:41