[GAP Forum] Manual Proof of GAP Results on Generators of Factor Groups

[Alexander Hulpke]

Dear Forum, dear Minhui Liu,

On May 2, 2014, at 5/2/14 7:11, Minghui Liu <matliumh at gmail.com> wrote:

> Dear GAP Forum,
> 
> I am trying to find generators of a factor group. I have input dozens
> of generators and relations and when I use the command
> 
> AbelianInvariants(F/relations);
> 
> the result was something like
> 
> 0, 0, 0, 0, 0, 0, 0, 2, 5.
> 
> After finding that the image of some generator, say F.1*F.2^5*F.3^3 has
> order 5 in the quotient group F/relations, is there any way we can manually
> write a proof that the order of the image of F.1*F.2^5*F.3^3 is equal to 5,
> perhaps with the help of GAP?

There also is a function `MaximalAbelianQuotient' that constructs a homomorphism G->G/G' which makes it easier to see what happens with particular elements of G.

If you want a manual proof, the calculation is essentially (both in what GAP does and how the result can be interpreted) a Smith Normal Form on a matrix of abelianized relators. It may be a pain to do so, but at least in principle you could do this by hand and thus see what happens.

> My point is, instead of writing "by calculation based on GAP, we have the
> order is equal to 5", it would be desirable to find a algebraic proof of
I don't think you would be able to do without some calculation or rewriting that in itself is not insightful, and when done by hand is prone to errors.

You could hide the (computer) calculations, e.g. just state:

The following map on the generators extends to a homomorphism from G to an abelian group (verify that the relators hold), and calculate the order of the image of your word in this abelian group.

And then show an explicit rewriting, expressing the 5th power of your element as a word in conjugates of commutators. 

This would be a proper proof that could be followed by hand, but frankly one without giving any insight how on earth you found the homomorphism and the expression.

Regards,

   Alexander Hulpke