Showing $\left/\left$ is finite

Question

I have had hard time to show that

$$G=\left<a,b,c : a^2=b^3=c^5=abc \right>/\left< abc \right>$$ is a finite group.

I guess that $G=\mathbb{Z}_2*\mathbb{Z}_3*\mathbb{Z}_5$. Even in this situation, it is not clear if it is finite or not. Could anybody help me for this problem? Thanks in advance!

Edit :

Apparently, this question has been answered in Proving Finiteness of Group from Presentation. However, the answer given in the link used the property of triangle group which is not given in Dummit and Foote algebra. Since the prerequisite of the given problem is the Dummit and Foote's Algebra book, I think there should be another way to solve it. So.. does anybody know how to solve this in different way with the answer of given link?

@ParclyTaxel No. What I stated above is correct. I mod the big group by a group generated by $abc$ — Byeong-Ho Bahn, Aug 08 '19 at 15:34
@LevBan Then that means all the group elements in the left pair of brackets are $=1$? — Parcly Taxel, Aug 08 '19 at 15:35
@ParclyTaxel I think then $a^2=b^3=c^5=abc=1$. But I could not extract good information from here.... Thanks for the link. That seems to say that the quotient group should be isomorphic to $A_5$. But unfortunately there is no proof.. — Byeong-Ho Bahn, Aug 08 '19 at 19:06
You should at least be able to produce a surjective homomorphism onto $A_5$. — YCor, Aug 09 '19 at 08:17
I think some context is required here. You say that the "the prerequisite of the given problem is the Dummit and Foote's Algebra book". Is this question in their book? (The phrasing of the question is slightly odd, giving it as as quotient of a presentation rather than just a presentation.) Also, The free product $\mathbb{Z}_2\ast\mathbb{Z}_3\ast\mathbb{Z}_5$ is infinite, as is any free product $A\ast B$ with $A, B$ non-trivial. — user1729, Aug 09 '19 at 13:51
Possibly ask your instructor. It may be that the previous cohort of students were given a presentation of $A_5$, and so the working to alter the presentation in your answer is sufficient. — user1729, Aug 09 '19 at 14:16

Byeong-Ho Bahn · Accepted Answer · 2019-08-12T13:48:37.460

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Thanks to ParclyTaxel and Ycor, I was able to figure out the answer hoping that there is no error.

Firstly, we know $G=\left<a,b,c : a^2=b^3=c^5=abc=1 \right>$. Now, note that we can reduce the number of generators and relations by following way. $$b=a^2bc^5=a(abc)c^4=ac^4=ac^{-1}$$ Now, renaming $a=p$ and $c^{-1}=q$, we have $$G=\left< p,q : p^2=q^5=(pq)^3=1 \right>$$

Then by the answer from this link, Group presentation of $A_5$ with two generators, we can prove that $G\cong A_5$ so $G$ is a finite group.

edited Aug 12 '19 at 13:48

answered Aug 09 '19 at 13:50

Byeong-Ho Bahn

2,968

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I think your line "$\varphi$ preserves the algebraic structure which means $\varphi$ is injective" should read "$\varphi$ preserves the algebraic structure which means $\varphi$ is a homomorphism". That is, you have defined a map and this is your verification that it is a homomorphism. – user1729 Aug 09 '19 at 13:55
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(It is not clear to me that this map is injective. Which, I believe, is where the subtlety in the problem lies.) – user1729 Aug 09 '19 at 13:55
@user1729 Thank you for the comment. I was thinking since the order of $\varphi(p)$ and $\varphi(q)$ are preserved, $Ker(\varphi)=\left<p^2,q^5,(pq)^3\right>=\left< 1 \right>.$ – Byeong-Ho Bahn Aug 09 '19 at 14:05
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Unfortunately that doesn't work. For example, The group $\langle a, b\mid a^{100}, b^{100}, (ab)^{100}\rangle$ is infinite, but surjects onto the finite group $\mathbb{Z}{100}\times\mathbb{Z}{100}$ in a way which preserves order. – user1729 Aug 09 '19 at 14:15
@user1729 Thanks. You are right. I will edit the answer! – Byeong-Ho Bahn Aug 09 '19 at 14:31
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Great, glad you worked it out! :-) – user1729 Aug 09 '19 at 14:57

Showing $\left/\left$ is finite

1 Answers1