Let $G$ denote the dihedral group $D_n$, $\rho$ a rotation of order $n$, and $\sigma$ any reflection. Prove that $\langle\rho,\sigma\rangle=G$.

Question

I can see how this is true for a given $n$, but need help in deciding how exactly this can be proved for the general case?

Also question about notation and its implications, does this mean that the exponent of $\rho$ must be equal to the exponent of $\sigma$ when multiplying them to get another element?

No, the exponents don't have to be the same. For example, you can have products such as $\rho^3\sigma$. — 2017-02-23

user id:274 112k · Accepted Answer

The subgroup generated by the rotation has n elements, and the reflection is not in it. It follows that the subgroup generated by the rotation and the reflection has more than half the number of elements in the whole group. This can only happen if the subgroup is equal to the whole group.

Why exactly is what the case? You need to specify exactly which step in the proof you don't understand. — 2017-02-23

Let $G$ denote the dihedral group $D_n$, $\rho$ a rotation of order $n$, and $\sigma$ any reflection. Prove that $\langle\rho,\sigma\rangle=G$.

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