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I have to prove or give counter example "Is it true that if $|G_1|$ and $|G_2|$ are dihedral groups of order $|G_1|=|G_2|$ then G1≅G2 "

$D_{2n}=<a,b\quad | \quad a^n=b^2=1 \quad ba=a^{-1}b>$ Dihedral group order of 2n

I am confused between two cases

1) There exist two dihedral groups with same order?

For example sym(n) (symmetric groups) for n $\in N$ There exist only one symmetric group of order n i.e Sym(3) the uniqe symmetric group of order 6. why it is not true for dihedral groups?

2) If there exist two different dihedral groups of same order

Quadratic groups of order 8 $Q_8$ can be written by $Q_8=C_4.C_2$ where $ C_n$ is cyclic group of order n . Why $Q_8$ is not dihedral group.

I think $Q_8$ beacuse $Q_8$ not isomorphic to $D_8$ ($D_8$ is dihedral of order 8 )

Thanks for any help

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    $\begingroup$ How is 1) different from 2) ? You are asking: is it true that if $G_1$ and $G_2$ are dihedral groups of order $|G_1|=|G_2|$ then $G_1 \cong G_2$? You can phrase your question like that. Several comments you make seem irrelevant. Yes, it is true. By the way, what makes you so sure there is only one symmetric group of a given degree? You might want to think a little about that first. (I am not saying it's not true, but thinking about it might help you clarify what your task really is.) $\endgroup$
    – the_fox
    Commented May 17, 2020 at 11:50
  • $\begingroup$ First, you have to tell us what definition of dihedral group you are using, senapideci. $\endgroup$ Commented May 17, 2020 at 12:02
  • $\begingroup$ Thank you for advice. I edited. Can you give me a hint about how can it be true? @the_fox $\endgroup$
    – senapideci
    Commented May 17, 2020 at 12:18
  • $\begingroup$ I edited. @GerryMyerson $\endgroup$
    – senapideci
    Commented May 17, 2020 at 12:20

2 Answers 2

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For different $n,m$ the dihedral groups $D_n$ and $D_m$ have different order, because they have order $2n\neq 2m$. So the point is to show that the given group $D_n$ above really has order $2n$.

It is not true that every two groups of order $2n$ are isomorphic. As you said, for example, $Q_8$ is not isomorphic to $D_4$: Question about Quaternion group $Q_8$ and Dihedral group $D_8$

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  • $\begingroup$ Thanks for answering. My question is "Are two dihedral groups of order 2n are isomorphic? if not how can i prove it" $\endgroup$
    – senapideci
    Commented May 17, 2020 at 17:49
  • $\begingroup$ @senapideci There is only one dihedral group, by definition. It is the one you have named $D_{2n}$ and wrote down the definition. There is no other one of the same order. $\endgroup$ Commented May 17, 2020 at 18:05
  • $\begingroup$ Thanks a lot :) $\endgroup$
    – senapideci
    Commented May 17, 2020 at 18:08
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The dihedral group $D_{2n}$ has presentation: $\langle r,f|r^n,f^2,(fr)^2\rangle$.

This group has order $2n$. There are $n$ reflections and $n$ rotations. Thus, if $n\ne m$, then $D_{2n}\not\cong D_{2m}$.

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  • $\begingroup$ Thanks for answering. My question is "Are two dihedral groups of order 2n are isomorphic? if not how can i prove it" $\endgroup$
    – senapideci
    Commented May 17, 2020 at 17:49

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