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Let a, b be elements in a group G and $c = a^{k_1}b^{l_1}a^{k_2}b^{k_2} \dots a^{k_n}b^{l_n}$ where $k_1 + k_2 + · · · + k_n = l_1 + l_2 + · · · + l_n = 0$. Show that $c$ belongs to the commutator subgroup $G´$, thus can be written as a product of commutators $x^{-1}y^{-1}xy$ with $x, y \in G$

So I have tried, but I need help, here is what I have done so far: Let S be the set of all elements of the form $a^{-1}b^{-1}ab$, $a,b \in G$. Let $G´$ be the set generated by $S$. Let $c$ be an element s.t

$c = a^{k_1}b^{l_1}a^{k_2}b^{k_2} \dots a^{k_n}b^{l_n}$

We know that $a^{-1}b^{-1}ab \in S \Rightarrow (a^{-1}b^{-1}ab)^m \in G´$.

This $(a^{-1}b^{-1}ab)^m = \underbrace {a^{-1}b^{-1}ab \dots a^{-1}b^{-1}ab} _{m\text{ factors}}$. We see that sum of the exponents add up to $0$. How can I proceed from here to show that $ c \in G´$?

Olba12
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1 Answers1

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Hint. Consider what happens to $c$ under the homomorphism from $G$ to the quotient group $G/G^{\prime}$ which, recall, is commutative.

EDIT (further hints). Write $\overline{G}=G/G^{\prime}$ and, for $x\in G$, let $\bar{x}=xG^{\prime} = f(x)$, where $f:G\to G/G^{\prime}$ is the canonical homomorphism. Then $$\bar{c} = f(c) = f(a^{k_1}b^{\ell_1}\cdots a^{k_n}b^{\ell_n}) = \bar{a}^{k_1}\bar{b}^{\ell_1}\cdots\bar{a}^{k_n}\bar{b}^{\ell_n}$$ in $\overline{G}$. Since $\overline{G}$ is commutative, we can move all the $\bar{a}$s to the front of the expression and all the $\bar{b}$s to the end, and so this becomes $$\bar{c} = \bar{a}^{k_1+\cdots +k_n}\bar{b}^{\ell_1+\cdots +\ell_n}$$ in $\overline{G}$. Can you finish it from here?

James
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  • Why do you wanna study this? I mean, how can u tell from the exercise that u wanna study the homomorphism between the two groups? – Olba12 Apr 06 '15 at 16:48
  • In algebra it is generally a good idea to let homomorphisms do the work for you. Here, you have a normal subgroup and you want to show that something lies in it. Since the subgroup $G^{\prime}$ is normal, you can form the quotient group $G/G^{\prime}$ and consider the canonical homomorphism from $G$ to this quotient. An element of $G$ belongs to $G^{\prime}$ if, and only if, it is in the kernel of this homomorphism; that is, if its image under the homomorphism is trivial. – James Apr 06 '15 at 17:01
  • Okey! I want to define a map $f:G \rightarrow G/G´$ by $f(a) = aG´$ And yes the quotient group is abelian since $aba^{-1}b^{-1} \in G´ $. And now i will try to show that f(c) = e. – Olba12 Apr 06 '15 at 17:10
  • Is my map defined correct? How can I evalute $cG´$? – Olba12 Apr 06 '15 at 17:30
  • Yes that is the correct map. Just evaluate the image of $c$ in $\overline{G} = G/G^{\prime}$, noting that this is an abelian group. – James Apr 06 '15 at 17:58
  • Im sorry, but could you explain the last part alittle more? – Olba12 Apr 06 '15 at 18:06
  • Too long for a comment; please see additional edits. – James Apr 06 '15 at 18:12
  • Oh! Now i understand what u meant with of c in $G¯¯¯¯=G/G′$. It was a bit confusing. Sorry for that. Thanks for the help! =) – Olba12 Apr 06 '15 at 18:17