# A finitely presented group

Given the presented group $$G=\Bigl\langle a,b\Bigm| a^2=c,\ b(c^2)b,\ ca(b^4)\Bigr\rangle,$$ determine the structure of the quotient $G/G'$,where G' is the derived subgroup of $G$ (i.e., the commutator subgroup of $G$).

Simple elimination shows $G$ is cyclic (as it's generated by $b$) of order as a divisor of $10$, how to then obtain $G/G'$? Note $G'$ is the derived group, i.e it's the commutator subgroup of $G$.

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P.S. That's a strange way to present the group; the introduction of $c$ just makes things seem more complicated. What's the point? – Arturo Magidin Apr 21 '12 at 5:05
Now we know two of the letters are redundant. As a textbook exercise probably the author aims the reader to first derive an explict presentation of the group without redundant words, just a matter of practice. – Harry Apr 21 '12 at 6:26
Just another general question: Given an abstractly presented group G with relations which are all satisfied by generators of another known group H, is there always a surjective homomorphism between the two groups which preserves the corresponding generators? I know by Von Dyck's theorem the known group H is a quotient of the abstract presented group G, but it does not imply there is a surjective homomorphism between G and H? – Harry Apr 21 '12 at 6:27
No, we don't know that "two of the letters are redundant", because we computed $G/G'$, not $G$ itself. $c$ is definitely redundant, because it is simply defined to be $a^2$, and is not even a generator. What I don't understand is why the group was not simply given as $$G=\langle a,b\mid ba^4b,\ a^3b^4\rangle$$instead of introducing the "abbreviation" $c=a^2$. – Arturo Magidin Apr 21 '12 at 20:20
If you have a presentation for $G$, and you know there is a group $H$ with elements that satisfy the relations, then there is an onto homomorphism from $G$ onto the subgroup of $H$ generated by those elements. If those elements generate $H$, then you have that the homomorphism is onto $H$. (If $S$ generates $G$, and you have a map from $G$ to $H$, then the image is generated by $f(S)$; if $f(S)$ contains a generating set for $H$, then the homomorphism is onto). – Arturo Magidin Apr 21 '12 at 20:22

Indeed, the group $G/G'$ is generated by $bG'$: let $\alpha$ denote the image of $a$ in $G/G'$ and $\beta$ the image of $b$. Then we have the relations $\alpha^4\beta^2 = \alpha^3\beta^4 = 1$; from there we obtain $$\beta^2 = \alpha^{-4} = \alpha^{-1}\alpha^{-3} = \alpha^{-1}\beta^{4},$$ so $\alpha = \beta^{2}$. And therefore $\alpha^4\beta^2 = \beta^8\beta^2 = \beta^{10}=1$. So the order of $\beta$ divides $10$. Therefore $G/G'$ is a quotient of $\langle x\mid x^{10}\rangle$, the cyclic group of order $10$.
Now consider the elements $x^2$ and $x$ in $K=\langle x\mid x^{10}\rangle$. We have $x\Bigl( (x^4)^2\Bigr)x=1$ and $x^4x^2(x^4) =1$. Therefore, there is a homomorphism $G\to K$ that maps $a$ to $x^2$ and $b$ to $x^{10}$, which trivially factors through $G/G'$. Therefore, $G/G'$ has the cyclic group of order $10$ as a quotient.
Since $G/G'$ is a quotient of the cyclic group of order $10$ and has the cyclic group of order $10$ as a quotient, it follows that $G/G'$ is cyclic of order $10$ (generated by $bG'$).