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I'm trying to classify all groups of order $20$. I get two group presentation and want to identify isomorphism type. I think one of these isomorphic to $D_{10}$ and other isomorphic to $Dic_5$. Can anyone tell me to find exactly which one isomorphic to $D_{10}$ and other isomorphic to $F_5$? I might get wrong presentations.

These are my two presentation,

$$G_1 =\langle r,a,b\mid r^5=1,a^2=b^2=1,r^{3}br^{-3}=b,r^{3}ar^{-3}=a\rangle$$

and

$$G_2 =\langle r,s\mid r^5=1,s^4=1,r^3sr^{-3}=s\rangle $$

Thank you.

Mittens
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Alhabud
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    You should note that a lot of the relations given can be translated as certain elements commuting. Try to play around a bit with them and see if perhaps other elements turn out to also commute – Tobias Kildetoft Aug 09 '23 at 11:03
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    In $G_{2}$, do you agree that $s\in C_{G}(r^{3})=C_{G}(r)$, which means $o(rs)=20$? So $G_{2}$ is cyclic. – Ash Aug 09 '23 at 11:18
  • @user1729 Can you explain how $G_1$ is isomorphic to $D_{10}$? – Alhabud Aug 09 '23 at 12:02
  • @Ash I got it thank you. Can you explain anything about $G_1$? – Alhabud Aug 09 '23 at 12:05
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    @Alhabud $G_1$ is not isomorphic to $D_{10}$; it surjects onto the infinite dihedral group (add $r=1$ as a relation), and so isn't even finite! Where did you get the presentation from? – user1729 Aug 09 '23 at 12:11
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    GAP confirms that $G_2$ is $\Bbb Z_{20}$. – Shaun Aug 09 '23 at 12:47
  • TzGoGo does nothing to the presentation defining $G_1$, so, presumably, there is no straightforward way of showing whether it's $F_5$. – Shaun Aug 09 '23 at 12:51
  • Oops! I forgot about Frobenius groups; I thought you meant the free group of rank five! – Shaun Aug 09 '23 at 12:55
  • Still: StructureDescription is inconclusive on $G_1$. But that's because it is, indeed, infinite; see @user1729's comment above. – Shaun Aug 09 '23 at 12:57
  • @Alhabud I’ll tell you how we conclude that $G_{1}$ is not isomorphic to $D_{10}$. So by the same discussion $o(ra)=o(rb)=10$ and we know $\langle ra \rangle$ and $\langle rb \rangle$ are different subgroups isomorphic to $C_{10}$, but we know that $D_{10}$ has only one (unique) subgroup isomorphic to $C_{10}$. – Ash Aug 09 '23 at 13:01

2 Answers2

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This answer mainly refers to the first sentence "I'm trying to classify all groups of order 20".

The classification of groups of order $20$ is as follows. Every such group is isomorphic to exactly one of the following groups: $$ C_{20},C_2\times C_{10}, D_{10},F_{20},{\rm Dic}_{20}. $$ The proofs can be found here at this site, but there are many different aspects here in different posts.

The classifcation of abelian groups of order $20$ follows from the fundamental theorem, and the one of the non-abelian groups uses Sylow theorems and certain facts on semidirect products of groups.

I created a small collection of posts about the classification of groups of order $20$. The first link has a fairly complete proof in the answer.

  1. Find four groups of order 20 not isomorphic to each other.

  2. Four groups of order 20 that are not isomorphic

  3. Classifying groups of order $20$

  4. Classification of groups of order $20$

  5. Must a group of order $20$ have an element of order $10$?

  6. Prove G is a nonabelian group of order 20

  7. How can I show that $G$ is non abelian of order 20?

The last one, and the one before, have the presentation for the Frobenius group $F_{20}$, namely $$ G=\langle x,y|x^4=y^5=1,x^{-1}yx=y^2\rangle $$

To answer your second question, your group $G_2$ is isomorphic to $C_{20}$, and $G_1$ is an infinite group.

Dietrich Burde
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  • While these links may answer the question, it is better to include the essential parts of the answer here and provide the links for reference. – Anne Bauval Aug 28 '24 at 10:34
  • @AnneBauval I have included the result. But it is difficult to include the essential parts of the classification proof, without just copying, say, the fairly complete answer of the first link. – Dietrich Burde Aug 28 '24 at 11:46
  • The question is not that wide. It is about two particular group presentations. – Anne Bauval Aug 28 '24 at 12:07
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Your question doesn't seem to be your introductory sentence (mentioning your motivation) but "which [of your two groups $G_1,G_2$] is isomorphic to $D_{10}$ and other isomorphic to $F_5$" or "$Dic_5$".

The answer is: none. Indeed:

  • in $G_2$, the generator $s$ commutes with $r^3=r^{-2}$ hence also with $r^3r^{-2}=r$. Therefore, $$G_2=\langle r,s\mid r^5=1,s^4=1,rs=sr\rangle=\Bbb Z_5\times\Bbb Z_4\cong\Bbb Z_{20}.$$
  • the subgroup of $G_1$ generated by $a,b$ is the free product $\langle a,b\mid a^2=b^2=1\rangle=\Bbb Z_2*\Bbb Z_2$, hence $G_1$ is infinite. More precisely, since $a$ and $b$ commute with $r$ (by the same argument as above): $$G_1 =\langle r,a,b\mid r^5=1,a^2=b^2=1,ra=ar,rb=br\rangle=(\Bbb Z_2*\Bbb Z_2)\times\Bbb Z_5.$$
Anne Bauval
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