ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 29 Jun 2020 10:57:38 +0200Find elements of left coset ZZ_6/{0,3}https://ask.sagemath.org/question/52082/find-elements-of-left-coset-zz_603/ If I have a group G = (ZZ_6,+), i.e. the integers modulo 6 under addition, and a subgroup of G, H = ({0,3}, +), how can I find G/H with sage?Fri, 19 Jun 2020 07:28:22 +0200https://ask.sagemath.org/question/52082/find-elements-of-left-coset-zz_603/Answer by dan_fulea for <p>If I have a group G = (ZZ_6,+), i.e. the integers modulo 6 under addition, and a subgroup of G, H = ({0,3}, +), how can I find G/H with sage?</p>
https://ask.sagemath.org/question/52082/find-elements-of-left-coset-zz_603/?answer=52252#post-id-52252Note that the sage instances corresponding to the *ring* $(\Bbb Z/6,+,\cdot)$, and to the cyclic *abelian group* $(C_6,+)$ differ, as it also happens in the mathematical world. The quotient in the category of rings is built w.r.t. an ideal, in our case the ideal is generated by $3\in \Bbb Z/6$. In the category of abelian groups we mod out a subgroup, in our case the subgroup is generated by $3$. Let us construct the quotient in both cases.
In the category of rings, we introduce the ring $R$ by either of...
sage: Integers(6)
Ring of integers modulo 6
sage: Zmod(6)
Ring of integers modulo 6
sage: IntegerModRing(6)
Ring of integers modulo 6
then consider its ideal $J=(3)$, and the quotient:
sage: R = Zmod(6)
sage: R
Ring of integers modulo 6
sage: J = R.ideal(3)
sage: J
Principal ideal (3) of Ring of integers modulo 6
sage: list(R)
[0, 1, 2, 3, 4, 5]
sage: Q = R.quotient(J)
sage: Q
Ring of integers modulo 3
sage: for k in R:
....: print(f"{k} modulo J is {Q(k)}")
....:
0 modulo J is 0
1 modulo J is 1
2 modulo J is 2
3 modulo J is 0
4 modulo J is 1
5 modulo J is 2
sage:
In the category of abelian groups (and/or in the category of groups) one can construct:
sage: C6.<a> = AbelianGroup(1, [6])
sage: C6
Multiplicative Abelian group isomorphic to C6
sage: # or
sage: C6 = CyclicPermutationGroup(6)
sage: a = C6.gens()[0]
sage: a.order()
6
sage: a
(1,2,3,4,5,6)
And the quotient is...
sage: H = C6.subgroup([a^3])
sage: Q = C6.quotient(H)
sage: Q
Permutation Group with generators [(1,2,3)]
Unfortunately, the version using `C6.<a> = AbelianGroup(1, [6])` was leading to a sage crash on my machine while trying to build the quotient with respect to the subgroup `H` constructed mot-a-mot as above. So just use the construction involving a permutation group.
Mon, 29 Jun 2020 10:57:38 +0200https://ask.sagemath.org/question/52082/find-elements-of-left-coset-zz_603/?answer=52252#post-id-52252