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I have some difficulties finding the elements of Dihedral Group $D_8$ (note that the order $|D_8|=8$). I know the geometric approach for defining $D_8$, but I prefer the algebraic way. In this case, I'd like to work with the group presentation.

Namely, $D_8=\langle x,a|a^n=x^2=e, xax^{-1}=a^{-1} \rangle$ with $n=4$. My question is about what the elements are in $D_8$. I really want to know why $a^3$ is not equal to $1$. I tried this work with this way: firstly, I consider the Dihedral Group $D_8 = X / S$, where $X$ is the free group generated by $\{x,a\}$, and $S$ is the smallest normal subgroup of $\langle x,a\rangle$ that contains $\{a^4,x^2,axax^{-1}\}$. Thus I want to figure out whether $a^3$ is in $S$ or not.

I agree $a^3$ is most likely not in $S$, but I want to make sure that it is true in very rigorous way. And I want to generalize this way for the general case, $D_{2n}$.

If I violated some rules of this awesome forum, Please let me know so that I can fix it.

Thanks.

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  • $\begingroup$ I can't use the 'hi' in the first place in my question. Is it some rule? $\endgroup$
    – nicksohn
    May 6, 2015 at 6:53
  • $\begingroup$ Shouldn't $S$ be the subgroup generated by $\{a^4, x^2, xax^{-1}a\}$? $\endgroup$ May 6, 2015 at 7:03
  • $\begingroup$ Here if $a^3=1=a^4$, then $o(a) \mid (3,4)=1$, so $a=1$, which is clearly not true. In general, $o(a) = n$ in $D_{2n}$ $\endgroup$ May 6, 2015 at 7:05
  • $\begingroup$ @PrahladVaidyanathan I guess they are equivalent. and.. Oh! My bad. Sorry S should be the smallest 'normal' subgroup of <{x,a}> that contains {x4,a2,axax}. Namely, It is spanned by conjugacy class of S. $\endgroup$
    – nicksohn
    May 6, 2015 at 10:29
  • $\begingroup$ Why a = 1 is not true? and I think a^3 is just one example. I can't generalize by your way. How about a^2? Anyway, Thanks for your answer!:) $\endgroup$
    – nicksohn
    May 6, 2015 at 10:31

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