-1
$\begingroup$

Let G= $ D_8$ be dihedral group of symmetries of square. Find the minimal number of generators for G.

My book directly writes thar answer is 2. In order to do this do we have to remember the group elements and start finding out which pair of elements form generating set of this group. Since it is not cyclic there shoudn't be one element that is generator.

$\endgroup$
  • 1
    $\begingroup$ What is the question? Is it, how do we know that the minimum is $2$? $\endgroup$ – Travis Jan 23 '15 at 5:56
6
$\begingroup$

If you know Lagranges theorem then it doesn't take much calculation to know that dihedral groups can be generated by two elements.

I presume you know that $D_8$ has order $8$. There is a very obvious element which has order $4$, call it $\rho$. Take $\tau$ to be any element outside of $\{1, \rho, \rho^2, \rho^3\}$. Then the subgroup generated by $\rho$ and $\tau$ has more than $4$ elements in it (all the powers of $\rho$ plus the element $\tau$) and by Lagranges theorem the order of this subgroup must divide $8$. Well the only number bigger than $4$ that divides $8$ is $8$, so the subgroup generated by $\rho$ and $\tau$ is all of $D_8$.

$\endgroup$
1
$\begingroup$

The symmetries are of two kinds: rotations by angles that are multiples of $\pi/4$ and reflections about 4 axes. It is clear that rotation by $pi/4$ generates a subgroup of order 4. Use Lagrange's theorem to determine the order of subgroup generated by two elements one which is this rotation and the other is any reflection. This works for all dihedral groups $D_n$, the group of symmetries of a general regular polygon on $n$-sides.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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