Why is vertical flip not in the symmetric group of isosceles triangle?

Question

Why is flip around the horizontal axis (2 times) not in the symmetric group of isosceles triangle?

Many sources say that there is only a $0^°$ rotation and a flip around the vertical axis.

Answer 1

Why is flip around the horizontal axis (2 times) not in the symmetric group of isosceles triangle?

The transformation you are describing is the identity transformation, so it is in the symmetry group.

Answer 2

Let us denote vertices of the triangle with $\{A,B,C\}$, where $\overline{AB}$ is the base. It is obvious that any symmetry of the triangle acts as permutation on its vertices. Also, since symmetries are isometries, it is easy to argue that $C$ is fixed under all the symmetries. We conclude that the symmetry group is subgroup of $S_3$ and that it leaves one element fixed, so it is a subgroup of $S_2\cong \mathbb Z/2\mathbb Z$. Obviously, the symmetry group is not trivial so it is isomorphic to $\mathbb Z/2\mathbb Z$ and is generated by reflection with respect to the line of symmetry of the base $\overline{AB}$.

The above reasoning is purely abstract argument why the group of symmetries of the isosceles triangle is what it is, but doesn't address your question. You are completely right, transformation that you describe is by all means a symmetry: it is composition of two reflections, thus an isometry, and it maps the triangle into itself. But this is not a different symmetry from what we already have, you described the identity map, which is already included.

Couple of things that might be helpful:

• any reflection $s$ is of order $2$, i.e. $s^2 =\operatorname{id}$
• composition of two reflections is a rotation; particularly composition of a reflection with itself is rotation by $0^\circ$, i.e. identity map
• any symmetry can be factored as composition of other isometries (not necessarily symmetries) in infinitely many ways (it is enough to take any vector $v$ and compose translations by vectors $v$ and $-v$ to get identity) but this doesn't mean that there are infinitely many symmetries