Find a nontrivial presentation of this group of permutations: the identity with $(1234), (13)(24), (4321),...
0
The Question:
Find a nontrivial presentation of the permutation group $G$ consisting of the identity with $(1234), (13)(24), (4321), (13), (24), (12)(34),$ and $(14)(23)$.
My Attempt:
Under the mapping $amapsto (13), bmapsto (24), cmapsto (1234)$, I have $$G=langle a, b, cmid a^2, b^2, (ab)^2, c^4, (ac)^2rangle,$$ but I don't think that quite captures it all; is it right?
Please help :)
group-theory permutations group-presentation
asked Jul 27 '17 at 12:20
ShaunShaun
8,820113681
add a comment |
0
The Question:
Find a nontrivial presentation of the permutation group $G$ consisting of the identity with $(1234), (13)(24), (4321), (13), (24), (12)(34),$ and $(14)(23)$.
My Attempt:
Under the mapping $amapsto (13), bmapsto (24), cmapsto (1234)$, I have $$G=langle a, b, cmid a^2, b^2, (ab)^2, c^4, (ac)^2rangle,$$ but I don't think that quite captures it all; is it right?
Please help :)
group-theory permutations group-presentation
asked Jul 27 '17 at 12:20
ShaunShaun
8,820113681
1
That's simply the dihedral group of order 8, so you have $G=langle c,amid a^2=c^4=e, aca^{-1}=c^{-1}rangle$.
– user424862
Jul 27 '17 at 12:27
@user424862 Of course! I don't know why I didn't recognise it.
– Shaun
Jul 27 '17 at 12:31
@user424862 Is my attempt okay?
– Shaun
Jul 27 '17 at 12:46
1
This is not really an attempt, it is just a guess. You need to be able to prove that your answer is correct. As it happens, it is not correct, It would be correct with the additional realtor $abc^{-2}$, and then you could leave out the relator $c^4$. But it is much easier to prove that the $2$-generator presentation suggested by user424862 is correct.
– Derek Holt
Jul 27 '17 at 13:10
As user424862 has already pointed out it's $D_8$. I would think of the problem in the following manner. Firstly, since the given group has eight elements, we look at the isomorphism classes of Groups of order eight. Then try finding a group isomorphic to the one given. Which is not so difficult to do as you can eliminate most of the groups in the isomorphism classes just by having a look. Once you have an easier(familiar) representation of the group given to you, you can work with the problem of determining the generator set.
– Naive
Jul 27 '17 at 14:46
add a comment |
0
0
0
1
The Question:
Find a nontrivial presentation of the permutation group $G$ consisting of the identity with $(1234), (13)(24), (4321), (13), (24), (12)(34),$ and $(14)(23)$.
My Attempt:
Under the mapping $amapsto (13), bmapsto (24), cmapsto (1234)$, I have $$G=langle a, b, cmid a^2, b^2, (ab)^2, c^4, (ac)^2rangle,$$ but I don't think that quite captures it all; is it right?
Please help :)
group-theory permutations group-presentation
asked Jul 27 '17 at 12:20
ShaunShaun
8,820113681
The Question:
Find a nontrivial presentation of the permutation group $G$ consisting of the identity with $(1234), (13)(24), (4321), (13), (24), (12)(34),$ and $(14)(23)$.
My Attempt:
Under the mapping $amapsto (13), bmapsto (24), cmapsto (1234)$, I have $$G=langle a, b, cmid a^2, b^2, (ab)^2, c^4, (ac)^2rangle,$$ but I don't think that quite captures it all; is it right?
Please help :)
group-theory permutations group-presentation
group-theory permutations group-presentation
asked Jul 27 '17 at 12:20
ShaunShaun
8,820113681
asked Jul 27 '17 at 12:20
ShaunShaun
8,820113681
edited Nov 30 '18 at 2:19
Shaun
asked Jul 27 '17 at 12:20
ShaunShaun
8,820113681
asked Jul 27 '17 at 12:20
ShaunShaun
8,820113681
asked Jul 27 '17 at 12:20
ShaunShaun
8,820113681
8,820113681
1
That's simply the dihedral group of order 8, so you have $G=langle c,amid a^2=c^4=e, aca^{-1}=c^{-1}rangle$.
– user424862
Jul 27 '17 at 12:27
@user424862 Of course! I don't know why I didn't recognise it.
– Shaun
Jul 27 '17 at 12:31
@user424862 Is my attempt okay?
– Shaun
Jul 27 '17 at 12:46
1
This is not really an attempt, it is just a guess. You need to be able to prove that your answer is correct. As it happens, it is not correct, It would be correct with the additional realtor $abc^{-2}$, and then you could leave out the relator $c^4$. But it is much easier to prove that the $2$-generator presentation suggested by user424862 is correct.
– Derek Holt
Jul 27 '17 at 13:10
As user424862 has already pointed out it's $D_8$. I would think of the problem in the following manner. Firstly, since the given group has eight elements, we look at the isomorphism classes of Groups of order eight. Then try finding a group isomorphic to the one given. Which is not so difficult to do as you can eliminate most of the groups in the isomorphism classes just by having a look. Once you have an easier(familiar) representation of the group given to you, you can work with the problem of determining the generator set.
– Naive
Jul 27 '17 at 14:46
add a comment |
1
That's simply the dihedral group of order 8, so you have $G=langle c,amid a^2=c^4=e, aca^{-1}=c^{-1}rangle$.
– user424862
Jul 27 '17 at 12:27
@user424862 Of course! I don't know why I didn't recognise it.
– Shaun
Jul 27 '17 at 12:31
@user424862 Is my attempt okay?
– Shaun
Jul 27 '17 at 12:46
1
This is not really an attempt, it is just a guess. You need to be able to prove that your answer is correct. As it happens, it is not correct, It would be correct with the additional realtor $abc^{-2}$, and then you could leave out the relator $c^4$. But it is much easier to prove that the $2$-generator presentation suggested by user424862 is correct.
– Derek Holt
Jul 27 '17 at 13:10
As user424862 has already pointed out it's $D_8$. I would think of the problem in the following manner. Firstly, since the given group has eight elements, we look at the isomorphism classes of Groups of order eight. Then try finding a group isomorphic to the one given. Which is not so difficult to do as you can eliminate most of the groups in the isomorphism classes just by having a look. Once you have an easier(familiar) representation of the group given to you, you can work with the problem of determining the generator set.
– Naive
Jul 27 '17 at 14:46
1
1
That's simply the dihedral group of order 8, so you have $G=langle c,amid a^2=c^4=e, aca^{-1}=c^{-1}rangle$.
– user424862
Jul 27 '17 at 12:27
That's simply the dihedral group of order 8, so you have $G=langle c,amid a^2=c^4=e, aca^{-1}=c^{-1}rangle$.
– user424862
Jul 27 '17 at 12:27
@user424862 Of course! I don't know why I didn't recognise it.
– Shaun
Jul 27 '17 at 12:31
@user424862 Of course! I don't know why I didn't recognise it.
– Shaun
Jul 27 '17 at 12:31
@user424862 Is my attempt okay?
– Shaun
Jul 27 '17 at 12:46
@user424862 Is my attempt okay?
– Shaun
Jul 27 '17 at 12:46
1
1
This is not really an attempt, it is just a guess. You need to be able to prove that your answer is correct. As it happens, it is not correct, It would be correct with the additional realtor $abc^{-2}$, and then you could leave out the relator $c^4$. But it is much easier to prove that the $2$-generator presentation suggested by user424862 is correct.
– Derek Holt
Jul 27 '17 at 13:10
This is not really an attempt, it is just a guess. You need to be able to prove that your answer is correct. As it happens, it is not correct, It would be correct with the additional realtor $abc^{-2}$, and then you could leave out the relator $c^4$. But it is much easier to prove that the $2$-generator presentation suggested by user424862 is correct.
– Derek Holt
Jul 27 '17 at 13:10
As user424862 has already pointed out it's $D_8$. I would think of the problem in the following manner. Firstly, since the given group has eight elements, we look at the isomorphism classes of Groups of order eight. Then try finding a group isomorphic to the one given. Which is not so difficult to do as you can eliminate most of the groups in the isomorphism classes just by having a look. Once you have an easier(familiar) representation of the group given to you, you can work with the problem of determining the generator set.
– Naive
Jul 27 '17 at 14:46
As user424862 has already pointed out it's $D_8$. I would think of the problem in the following manner. Firstly, since the given group has eight elements, we look at the isomorphism classes of Groups of order eight. Then try finding a group isomorphic to the one given. Which is not so difficult to do as you can eliminate most of the groups in the isomorphism classes just by having a look. Once you have an easier(familiar) representation of the group given to you, you can work with the problem of determining the generator set.
– Naive
Jul 27 '17 at 14:46
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1
That's simply the dihedral group of order 8, so you have $G=langle c,amid a^2=c^4=e, aca^{-1}=c^{-1}rangle$.
– user424862
Jul 27 '17 at 12:27
@user424862 Of course! I don't know why I didn't recognise it.
– Shaun
Jul 27 '17 at 12:31
@user424862 Is my attempt okay?
– Shaun
Jul 27 '17 at 12:46
1
This is not really an attempt, it is just a guess. You need to be able to prove that your answer is correct. As it happens, it is not correct, It would be correct with the additional realtor $abc^{-2}$, and then you could leave out the relator $c^4$. But it is much easier to prove that the $2$-generator presentation suggested by user424862 is correct.
– Derek Holt
Jul 27 '17 at 13:10
As user424862 has already pointed out it's $D_8$. I would think of the problem in the following manner. Firstly, since the given group has eight elements, we look at the isomorphism classes of Groups of order eight. Then try finding a group isomorphic to the one given. Which is not so difficult to do as you can eliminate most of the groups in the isomorphism classes just by having a look. Once you have an easier(familiar) representation of the group given to you, you can work with the problem of determining the generator set.
– Naive
Jul 27 '17 at 14:46