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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 ?

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First, show that $G$ has order $10$. Then show that $G$ is not abelian. Are there any other non-abelian groups of order $10$? – Tobias Kildetoft Mar 25 '13 at 18:59
@TobiasKildetoft: Would you suggest the OP to satisfy the relations in $D_5$ with above permutations and then using the Von dyck theorem? – Babak S. Mar 25 '13 at 19:01
We usually use the prefix "sub" for inclusion, as in subset, subgroup, subring, subfield, subspace, &c. – Pedro Tamaroff Mar 25 '13 at 19:13
up vote 2 down vote accepted

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.

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+1 cause it is ready to be taught at the class easily. – Babak S. Mar 30 '13 at 20:24

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.

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+1 Good point about the comment of Tobias, though this may be slightly more advanced stuff for the OP... – DonAntonio Mar 25 '13 at 23:49

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.

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This is the way I would have tried to do it. I think it is the simpler, more direct one – DonAntonio Mar 25 '13 at 23:36

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.

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