Please, help
How to prove there are only 6 symmetries of an equilateral triangle?

Question

Please, help
How to prove there are only 6 symmetries of an equilateral triangle?

loser66 · Answer

This is what I get so far:
Let S ={A, B, C} where A, B, C are vertices of the triangle.
symmetric function f : {A,B,C} --> {A,B,C} has at most 6 possibilities outcome. But I don't know how to make it in logic.

loser66 · Answer

@SithsAndGiggles

loser66 · Answer

From the previous question, I had proof of all 6. To this problem, I have to prove no more than 6 of them.

anonymous · Answer

Hmm... I think just establishing that there are at most 6 functions would involve a fairly simple combinatorial proof.

If you can prove how many functions you can make that maps a set of $n$ elements to a set of $k$ elements, you need only plug in $n=k=3$ and you should be done.

loser66 · Answer

Is there any proposition relate to the number of outcomes of permutation among them?
I mean possibility outcomes of f.

anonymous · Answer

Well you can consider how many outputs you can match with each input. For example, you can let $A=\{n_1,n_2,...,n_n\}$ and $B=\{k_1,k_2,...,k_k\}$, where the the subscript denotes there are so many elements in each set.

Let $n=k$, so that $|A|=|B|$.

For $n_1$, you have $k$ possibilities ($k_1,k_2,...,k_n$).

For $n_2$, you have $k-1$ possibilities $\bigg(\left|B\backslash\{k_i\}\right|$, where $f(n_1)=k_i\bigg)$. The function must be one-to-one and onto, otherwise it's not symmetric, right?

By induction, you can establish that for $n_n$, you have only one remaining possibility.

Counting the total, you have $k(k-1)\cdots(2)(1)=k!$. Since $k=3$, you have $3!=6$ total possible functions.