How do I find the abelianization of a group determined by generators and relations?

Asked Apr 10 '18 at 13:22

Active Apr 10 '18 at 14:40

Viewed 244 times

For example let $G=\langle a,b,c|abcbc=a^2cb^2=a^2cba^{-1}ca=1\rangle$

I know that $G^{ab} \cong \mathbb{Z}^n / \langle (_1) = · · · = (_) = 0\rangle.$ Where $$ maps $w$ to the column \begin{pmatrix} y_1\\\ y_2\\\ \vdots \\\ y_i \\\ \vdots \\\ y_k \end{pmatrix}with $y_i$ being the sum of the powers for $ x_i$ in $w$. So I get this matrix as an $Im$ of $L:ℤ^k→ ℤ^n$ mapping $ e_i→(w_i)$ \begin{pmatrix} 1 & 2 & 2 \\\ 2 & 2 & 1 \\\ 2 & 1 & 2 \end{pmatrix} I know that there are bases $_1, . . . , _ ∈ ℤ^k$ and $_1, . . . , _ ∈ ℤ^k$ so that $_ = \lambda__i$ if $i\leq n$ and $0$ otherwise

and i know that $\mathbb{Z}^n / Im = \prod^n_{i=1} \mathbb{Z}/\lambda \mathbb{Z}.$ But i'm not quite sure how to use it.

edited Apr 10 '18 at 14:40

asked Apr 10 '18 at 13:22

Alexander Kraynov

I don't understand your question. It would help if the math was a little bit cleaned up, see here for some help. (Also : I think the (or at least, a) correct spelling is "abelianization".) – Arnaud D. Apr 10 '18 at 13:52
This similar question could also be of interest. – Arnaud D. Apr 10 '18 at 13:55
Thanks, i didn't know the correct spelling. I have some problems writing negative powers with MathJax so i dont know how to fix some things – Alexander Kraynov Apr 10 '18 at 14:02
I cleaned it up as much as i can. I hope it's better now. – Alexander Kraynov Apr 10 '18 at 14:16
2

But that group is already abelian. Actually, it is the trivial group. – Tobias Kildetoft Apr 10 '18 at 14:26
@TobiasKildetoft : is it ? I found $G = \Bbb Z a \oplus \Bbb Z b/(5b - 2a = 0)$ – Nicolas Hemelsoet Apr 10 '18 at 14:28
@NicolasHemelsoet Last relation says that $b = c^{-1}$ and the one before that says that $a = c^3$. So the first one becomes $c^{-1}c^{-1}c^3 = 1$ which reduces to $c=1$ (and the group is generated by $c$ by the two previous relations). – Tobias Kildetoft Apr 10 '18 at 14:30
I might be confused but I think the last relation says $a^2 = c^5$. Or am I totally mistaken ? – Nicolas Hemelsoet Apr 10 '18 at 14:33
@NicolasHemelsoet It may be an error in the question, but it has $=$ all the way through, so all of those expressions are set to $1$. – Tobias Kildetoft Apr 10 '18 at 14:34
@TobiasKildetoft : oh I see ! You're right, sorry for this. – Nicolas Hemelsoet Apr 10 '18 at 14:35
I triple checked the original task and i didn't mistyped it. I can get another example if its trivial – Alexander Kraynov Apr 10 '18 at 14:37
Updated the task, I hope this one is fine. – Alexander Kraynov Apr 10 '18 at 14:41

How do I find the abelianization of a group determined by generators and relations?

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