Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Below is the cycle graph of $\operatorname{Dih}_4$. What I don't understand is that, since $(ba)^2=a^2$, why there isn't a link between $ba$ and $a^2$, and hence of course also $a^2$ and $ba^3$? I can see that "$e$ - $ba$" is not a cycle at all, because $(ba)^2\ne e$.

    a^2
   / \
  a   a^3
   \ /
  __e_____
 / / \    \
b ba ba^2 ba^3
share|improve this question
1  
But $(ba)^2=baba=1$ surely...(as $bab=a^{-1}$) –  user1729 Nov 8 '11 at 10:40
    
@user1729: No, $(ba)^2=b^2 a^2=a^2$. ($bab=a$, $a^3=a^{-1}$) –  Voldemort Nov 8 '11 at 10:44
    
@ Voldemort: No, $(ba)^2=a^{-1}a=1$. (bab=a^{-1}, a^3=a^{-1}) –  user1729 Nov 8 '11 at 10:48
    
How is it that you are defining the Dihedral group? If you are given it in terms of a presentation (see "Equivalent Definitions" in the wikipedia article) then seeing that $(ba)^2=1$ is easy. Otherwise, I would recommend playing around with squares some more, or working out a permutation representation and playing with that (to work out a permutation representation, label the corners of your square $1$, $2$, $3$ and $4$ then work out what a rotation does to them and what a flip does to them. A flip sends $1$ to what? $2$ to what? etc.). –  user1729 Nov 8 '11 at 10:50
add comment

1 Answer

up vote 2 down vote accepted

...but $(ba)^2=baba=1$ surely...as $bab=a^{-1}$. So, $a^2\neq (ba)^2$, so they should not appear in the same cycle.

Indeed, taking an arbitrary element of $D_4$ which is not a power of $a$, $a^ib$ say, then $(a^ib)^2=a^iba^ib=a^ia^{-i}=1$ as $ba^ib=a^{-i}$. So, basically, every element has order two apart from $a$, which has order $4$, and $e$, which is trivial.

share|improve this answer
    
I'm sorry. It turns out I didn't quite think this through, the Dihedral Group is not commutative. Thank you. –  Voldemort Nov 8 '11 at 10:54
    
That's all right - for my part I should have spotted that that was your problem earlier! –  user1729 Nov 8 '11 at 11:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.