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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?2012-04-21
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    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.2012-04-21
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    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?2012-04-21
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    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" $c=a^2$.2012-04-21
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    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).2012-04-21

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