$\newcommand{\RR}{\mathbb{R}}$Dehn twists are orientation preserving, so you can never write a reflection as a product of Dehn twists. On the other hand writing an orientation preserving, periodic element $\phi$ as a product of Dehn twists is a non-trivial problem -- that is to say, there are mechanical ways to get the desired product, but often the number of Dehn twists required is surprisingly large! For a slick (read, comprehensible) solution we need more details about how the surface and periodic element are given.

In general, the mechanical solution is as follows.

1. Draw a filling collection of curves on the surface, so that the collection is permuted by $\phi$. It will also be helpful to make the number of intersection points as small as possible.

2. Now move the curves back to their original positions using Dehn twists. This is described in Lickorish's paper. To get you started - show that if a pair of curves $a$ and $b$ meet once then you can send $a$ to $b$ using first a twist on $b$ and then a twist on $a$.

Slick solutions involve knowing about the braid relation and its generalizations to "chains" of curves. In your comment, it sounds like you are drawing the surface in $\RR^3$ and the periodic element $\phi$ is induced by a rotation of $\RR^3$, so I will just point out the following nice fact, given as an exercise in Thurston's book: not every periodic element is induced by a symmetric embedding of the surface into $\RR^3$. Consider the element of order six in the mapping class group of the torus...

$\newcommand{\RR}{\mathbb{R}}$Dehn twists are orientation preserving, so you can never write a reflection as a product of Dehn twists. On the other hand writing an orientation preserving, periodic element $\phi$ as a product of Dehn twists is a non-trivial problem -- that is to say, there are mechanical ways to get the desired product, but often the number of Dehn twists required is surprisingly large! For a slick (read, comprehensible) solution we need more details about how the surface and periodic element are given.

In general, the mechanical solution is as follows.

1. Draw a filling collection of curves on the surface, so that the collection is permuted by $\phi$. It will also be helpful to make the number of intersection points as small as possible.

2. Now move the curves back to their original positions using Dehn twists. This is described in Lickorish's paper. To get you started - show that if a pair of curves $a$ and $b$ meet once then you can send $a$ to $b$ using first a twist on $b$ and then a twist on $a$.

Sometimes a slick solution can be found using the braid relation and its generalization, the "chain relation". In your comment, it sounds like you are drawing the surface in $\RR^3$ and the periodic element $\phi$ is induced by a rotation of $\RR^3$, so I will just point out the following nice fact, given as an exercise in Thurston's book: not every periodic element is induced by a symmetric embedding of the surface into $\RR^3$. Consider the element of order six in the mapping class group of the torus...

$\newcommand{\RR}{\mathbb{R}}$Dehn twists are orientation preserving, so you can never write a reflection as a product of Dehn twists. On the other hand writing orientation preserving, periodic elements as a product of Dehn twists is a nice exercise. The method of solution depends on how the surface and periodic element are given to us.

In your comment, it sounds like you are drawing the surface in $\RR^3$ and the periodic element $\phi$ is induced by a rotation of $\RR^3$. In this situation, you should draw a filling collection of curves on the surface, so that the collection is permuted by $\phi$. It will also be helpful to make the number of intersection points as small as possible. Now you need to move the curves back to their original positions using Dehn twists. This is described in the Lickorish's paper. To get you started - show that if a pair of curves $a$ and $b$ meet once then you can send $a$ to $b$ using first a twist on $b$ and then a twist on $a$.

And a final remark - not every periodic element is induced by a symmetric embedding of the surface into $\RR^3$. Consider the element of order six in the mapping class group of the torus...

$\newcommand{\RR}{\mathbb{R}}$Dehn twists are orientation preserving, so you can never write a reflection as a product of Dehn twists. On the other hand writing an orientation preserving, periodic element $\phi$ as a product of Dehn twists is a non-trivial problem -- that is to say, there are mechanical ways to get the desired product, but often the number of Dehn twists required is surprisingly large! For a slick (read, comprehensible) solution we need more details about how the surface and periodic element are given.

In general, the mechanical solution is as follows.

1. Draw a filling collection of curves on the surface, so that the collection is permuted by $\phi$. It will also be helpful to make the number of intersection points as small as possible.

2. Now move the curves back to their original positions using Dehn twists. This is described in Lickorish's paper. To get you started - show that if a pair of curves $a$ and $b$ meet once then you can send $a$ to $b$ using first a twist on $b$ and then a twist on $a$.

Slick solutions involve knowing about the braid relation and its generalizations to "chains" of curves. In your comment, it sounds like you are drawing the surface in $\RR^3$ and the periodic element $\phi$ is induced by a rotation of $\RR^3$, so I will just point out the following nice fact, given as an exercise in Thurston's book: not every periodic element is induced by a symmetric embedding of the surface into $\RR^3$. Consider the element of order six in the mapping class group of the torus...