Originally Posted by

**johng** Dan,

1 ,2 and 4 look good.

3. $\displaystyle D_8=\{1,x,x^2,x^3,y,xy,x^2y,x^3y\}$, here $\displaystyle x^4=y^2=1$ and $\displaystyle yxy=x^{-1}$. So the only elements of order 4 are $\displaystyle x$ and $\displaystyle x^3$; $\displaystyle x^2, y$ have order 2 and $\displaystyle (x^iy)(x^iy)=x^ix^{-i}=1$

$\displaystyle Q_8$ formally has the same elements as $\displaystyle D_8$, but $\displaystyle x^4=1, y^2=x^2$ and $\displaystyle y{-1}xy=x^{-1}$. I leave it to you to compute exactly 4 elements of order 4.

So you have the right idea, but your counts are off.

5. $\displaystyle D_{24}$ has 13 elements of order 2, whereas $\displaystyle S_4$ has 9 elements of order 2. This is a lot of computation, but it does show they are not isomorphic. Maybe easier: convince your self that the order of a product of disjoint cycles in a permutation group has order the least common multiple of the cycle lengths. So the maximum order of an element of $\displaystyle S_4$ is 4 (a 4 cycle), whereas $\displaystyle D_{24}$ has an element of order 12.

By the way, I think the usual symbol for a dihedral group is $\displaystyle D_n$, the dihedral group with 2n elements (at least it used to be). Also the quaternion group (your notation $\displaystyle Q_8$) is usually denoted just Q. The generalized quaternion group has 2n elements; I don't remember any standard notation for this generalized quaternion group.

Generally, a good job.