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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.

Shaun
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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

1 Answers1

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Here is a small collection of posts about the classification of groups of order $20$. It is too long for a comment.

  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 has the presentation for $F_5$, or sometimes denotes Frobenius $F_{20}$, namely $$ G=\langle x,y|x^4=y^5=1,x^{-1}yx=y^2\rangle $$

Dietrich Burde
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