Let $G$ be the subgroup of $S_5$ generated by the cycle $(12345)$ and the element $(15)(24)$. Prove that $G \cong D_5$. Let $G$ be the subgroup of $S_5$ generated by the cycle $(12345)$ and the element $(15)(24)$. Prove that $G \cong D_5$, where $D_5$ is the dehidral group of order $10$.
I understand intuitively that $(12345)$ corresponds to an rotation in $D_5$ and $(15)(24)$ corresponds with a reflection in $D_5$. I would say let $\rho$ be a reflection in $D_5$ and $\sigma$ a reflection in $D_5$. Then $D_5$ is generated by $\sigma$ and $\rho$. Let $r=(12345), s=(15)(24)$.
Define $f: D_5 \to G : \rho ^i \circ \sigma ^j \mapsto r^i s^j$.
The thing where I get stuck is that I can't show that $r^i s^j$ has order $2$ if $j=1$. With reflections this is quite simple, as a any reflection has order $2$, but how do I show this with permutations ?
Can anybody help me how I can construct a solid proof here ?
 A: Another hint may be: $$D_n\cong\mathbb Z_n\rtimes\mathbb Z_2$$ Form this; you can conclude what @Tobias Kildetoft commented first easily.
A: Hint: One definition of $D_n$ is $\langle s,r:o(r) = n, o(s) =2, srs = r^{-1}\rangle$. 
Using this fact, you wouldn't even need to construct an isomorphism, just show that by labelling your given permutations correctly they satisfy the above conditions.
A: You seem to want to multiply permutations. For starters 
$$
rs=(12345)(15)(24)=(1)(25)(34)
$$
(I use the rule that permutations are composed such as functions, and as I write functions to the left from their arguments, I need to compose functions from right to left).
The way to see this is to check it one number at a time. Let's start with $1$. The 2-cycle $(24)$ doesn't do anything to it. But the 2-cycle $(15)$ sends $1$ to $5$. Then the 5-cycle $(12345)$ sends $5$ back to $1$. Therefore $1$ is a fixed point (or a 1-cycle) of the composition $rs$. 
The number $2$ is up next. The 2-cycle $(24)$ sends it to $4$. The 2-cycle $(15)$ doesn't do anything to that, but the 5-cycle $(12345)$ maps $4$ to $5$. Thus the composition $rs$ maps $2$ to $5$.
Similarly you can check that the composition $rs=(12345)(15)(24)$ maps the numbers as
$$
5\mapsto5\mapsto1\mapsto2,\quad 3\mapsto3\mapsto3\mapsto4,\quad 4\mapsto2\mapsto 2\mapsto 3.
$$
Therefore $rs=(25)(34)$ as a product of disjoint cycles.
Do the other compositions $r^js$ with $j=2,3,4$ in the same way. After a little bit of practice the process becomes automatic. After doing one number I always continue with its image under the composition. That way I build up the composition one of its cycles at a time.
A: Let $V$ be a five-sided polygon. And name the different vertices as $1,2,3,4,5$. Define $f:D_5 \to S_5$ as a function that maps any element in $D_5$ to a permutation of the vertices. This a injective homomorphism.
If $x,y \in D_5$ then $f(xy)$ is the composition of the permutations $f(y) \circ f(x)$. If $x$ fixes all the vertices then $x$ is the idendity, therefore $f$ is injective. Let $r$ the standard rotation then $f(r)=(12345)$ and let $s$ the reflection in the line that connects the origin with the vertex $3$ then $f(s)=(15)(24)$. 
Therefore $f$ is an isomorphism of $D_5$ to $f(D_5)$. As $f(D_5)$ is generated by $f(r)$ and $f(s)$ we get exactly what we wanted to prove.
