Dihedral group and cyclic group question Let $D_n=\langle r,s\rangle$ such that $r^n=s^2=1$ be the dihedral group on the $n$-gon.
If $n$ is odd, then $r^2$ generates all the rotations, so $s$ with $r^2$ generates the whole group. 
A friend told me that when $n$ is odd then $⟨s,r^2⟩$ and $⟨sr,r^2⟩$ generate the whole group $D_n$ and when $n$ is even $⟨s,r^2⟩$ and $⟨sr,r^2⟩$ only generate part of $D_n$, but he can't explain to me why. 
I wonder would like to explain this to me without using isomorphism, since I haven't learn that yet.
 A: If I read this, I get the feeling that you'd prefer a more geometrical approach. Before doing anything else, recall that $D_n$ is the group of rotations and reflection of regular $n$-gons. In order to understand what is going on, it may be useful to actually draw a regular pentagon and hexagon to understand the cases where $n$ is odd or even, respectively. Using the standard coordinate system of $\mathbb{R}^2$, you can always draw the regular $n$-gon on the unit circle with $(1,0)$ as one of its vertices. If $n$ is even, that means $(-1,0)$ is a vertex as well, while in the case $n$ is odd, you don't get any other vertex on the $x$-axis.
We can now consider the group $D_n$ as generated by the reflection $s$ in the $x$-axis, together with the smallest possible rotation $r$. Since $s\in K=\langle s,r^2\rangle$, we need only ask whether $r\in K$. If it is, we'll see that $G=D_n$, while it is a proper subgroup otherwise.
First look at the case where $n$ is odd and try to see why $r\in K$. First of all, since $n$ is odd, $n-1$ is even and therefore the rotation $r^{n-1}$ can be found in $K$ by using $r^2$ a couple of times: $r^{n-1}=(r^2)^{(n-1)/2}$. Therefore, $sr^{n-1}s$ is an element of $K$ as well and by using the relations for reflections and rotations, this equals $r$ and hence $K=D_n$. To understand this, it may help to actually use the previously mentioned drawing of a pentagon.
For even $n$ you want to show that $K\neq D_n$, so you'll have to show that no odd power of $r$ is an element of $K$. There are many ways to do this, but I think that doing this by using the concept of a group action is the most intuitive. This is a formal way to describe, for instance, how $D_n$ gives the symmetries of the $n$-gons. More precise, let $G$ be a group and $X$ a set. A group action is a map $\lambda:G\times X\to X$ where we also write $gx$ for $\lambda(g,x)$, such that
(i) $ex=x$ and (ii) $g(hx)=(gh)x$ for all $g,h\in G$, $x\in X$.
In the case of $D_6$, we may take $X=\{1,2,3,4,5,6\}$ by drawing a hexagon and numbering the vertices (counterclockwise). If you do not know group actions yet, just think about this for a moment.
Given a group action on a set $X$, it is possible to define orbits to be subsets of $X$ as follows: $x,y\in X$ are in the same orbit iff there exists $g\in G$ such that $gx=y$ (this is actually an equivalence relation). The set of orbits is written as $X/G$. In the case where $G=D_n$, there is just one orbit, i.e. $X/G=\{X\}$.
Now notice that if $G$ is a group acting on a set $X$, any subgroup $H\leq G$ acts on the same set by restricting as well. If $H=G$, it will of course have the same orbits (even though the converse is not necessarily true). If you have a subgroup $H\leq G$ and you can show that $X/H\neq X/G$, it then follows that $H\neq G$.
Now return to $K\leq D_n$. We already saw that $X=\{1,2,\ldots,n\}$ has just one orbit under $D_n$. In the case where $n$ is even, it is possible to consider the effect of the elements of $K$ under this action. You will notice that the even numbers are always mapped to even numbers and that odd numbers are always mapped to odd numbers. I'm afraid you'll really have to do this yourself. As a result, it follows that $X/K=\{\{1,3,5,\ldots,n-1\},\{2,4,\ldots,n\}\}$. We have now found that $X/K\neq X/D_n$ and thus $K\neq D_n$.
A: Let us look at the cyclic subgroup generated by $r^2$, where $r$ denotes the rotation.  Then since $n$ is even, write $m=n/2$. $(r^2)^m=r^n=1$, so $\langle r^2\rangle$ does not meet any of the odd-degree rotations. Can you see how that would be a problem once we add the reflection $s$ into the mix? 
