ASKSAGE: Sage Q&A Forum - Individual question feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Wed, 10 May 2017 13:50:18 -0500How can you operate in a quotient of a group finitely presented?https://ask.sagemath.org/question/37511/how-can-you-operate-in-a-quotient-of-a-group-finitely-presented/Hello everyone, I have the following groups
G.<a,b> = FreeGroup()
H = G.quotient([a*b*a.inverse()*b.inverse()])
I would like SAGE to understand the product of to lateral clases in H? It is that possible?
Thanks in advance!Thu, 04 May 2017 12:56:13 -0500https://ask.sagemath.org/question/37511/how-can-you-operate-in-a-quotient-of-a-group-finitely-presented/Comment by dan_fulea for <p>Hello everyone, I have the following groups</p>
<pre><code>G.<a,b> = FreeGroup()
H = G.quotient([a*b*a.inverse()*b.inverse()])
</code></pre>
<p>I would like SAGE to understand the product of to lateral clases in H? It is that possible?</p>
<p>Thanks in advance!</p>
https://ask.sagemath.org/question/37511/how-can-you-operate-in-a-quotient-of-a-group-finitely-presented/?comment=37566#post-id-37566What kind of product should be constructed?
We have for instance:
sage: G.<a,b> = FreeGroup()
sage: H = G.quotient( [ a*b*a.inverse()*b.inverse() ] )
sage: H.is_abelian()
True
sage: h = H( a^3 * b^4 ) * H( a^-7 * b^2 )
sage: h
a^3*b^4*a^-7*b^2
sage: h == H(a^-4 * b^6)
True
sage: A.<s,t> = AbelianGroup( 2 )
sage: A
Multiplicative Abelian group isomorphic to Z x Z
sage: ( s^3 * t^4 ) * ( s^-7 * t^2 )
s^-4*t^6
The form `a^3*b^4*a^-7*b^2` for the above product computed in H is a possible one. There is no simplification, since there is no "canonical form" to simplify in a general quotient. But the equality `h == H(a^-4 * b^6)` could be verified.
Is there any reason to use `H` and not `A` as above?Wed, 10 May 2017 13:50:18 -0500https://ask.sagemath.org/question/37511/how-can-you-operate-in-a-quotient-of-a-group-finitely-presented/?comment=37566#post-id-37566