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
Best Answer
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 $D_8$