The following is an exercise from "Guide to Abstract Algebra, 1st ed, Carol Whitehead".
Let $A_n=\{\gamma_1, \gamma_1,\gamma_1, \ldots, \gamma_r\}$, where $r=\frac{1}{2}(n!)$, and let $\alpha$ be any permutation in $S_n$. Prove the the set $\{\alpha^{-1}\gamma_i\alpha: 1 \leq i \leq r \} = > A_n$.
Here $A_n$ is understood to be the alternating group on $n$ symbols, where elements are even permutations of $S_n$, the symmetric group of order $n$.
My solution:
We first note that $\gamma_i \alpha \in S_n$ because $\gamma_i$ and $\alpha$ are elements of the group $S_n$ and composition of its elements obeys closure.
By the same closure argument $\alpha^{-1}(\gamma_i \alpha) \in S_n$, because $a^{-1}\in S_n$. That is, every element of a group has an inverse in that group.
So we have shown $\alpha^{-1}\gamma_i\alpha \in S_n$.
To show the set $\{\alpha^{-1}\gamma_i\alpha: 1 \leq i \leq r \}$ is $A_n$ we have to show two things:
- That $\alpha^{-1}\gamma_i\alpha$ is an even permutation,
- and that it is unique over $i$.
The parity of compositions is additive, so $\alpha^{-1}\gamma_i\alpha$ has a parity that is the sum of the parities of $\alpha^{-1}$, $\gamma_i$, and $\alpha$. The parity of $\gamma_i$ is even by definition. The parity of $\alpha^{-1}$ is the same as $\alpha$. The sum of the parities of $\alpha$ and $\alpha^{-1}$ is always even (odd+odd, even+even). So $\alpha^{-1}\gamma_i\alpha$ is an even permutation and hence a member of $A_n$.
We observe that $\gamma_i\alpha$ is unique for all $i$. To see this we note that $\gamma_i \alpha = \gamma_j \alpha$ implies $\gamma_i = \gamma_j$. By the same argument $\alpha^{-1}(\gamma_i\alpha)$ is unique over $i$.
Finally, we note that there are $\frac{1}{2}(n!)$ elements $\alpha^{-1}\gamma_i\alpha$, the number we expect in an alternating subgroup $A_n$ of $S_n$.
So all $\alpha^{-1}\gamma_i\alpha$ are unique, even permutations and enumerate to $|A_n|$, completing the proof.
Question: Is it a valid / sufficient proof argument to show $\alpha^{-1}\gamma_i\alpha$ are (a) unique, (b) even permutations and (c) enumerate to $|A_n|$ ?
The textbook I am using to self-teach has not up to this point discussed proving two sets are the same in this manner (showing elements have the right property, are unique, and have the right cardinality).
I have researched potential duplicates but have found none, eg Characterization of symmetric group elements.
Yes, your argument is sufficient (and very methodical). This is more of a long comment than an answer, but I will try to provide a few comments regarding how to organize your solution a bit.
Let me give a name to your set in question, say $T=\{\alpha^{-1}\gamma_i\alpha\mid 1\le i\le r\}$. To show that the sets $T$ and $A_n$ are equal, you must show that $T\subseteq A_n$ and $A_n\subseteq T$. Note that you have shown the first of these two conditions by showing that every element of $T$ is an even permutation ($A_n$ is precisely the even permutations).
So now you have a subset $T\subseteq A_n$. Because $A_n$ is finite, to show that the subset $T$ is all of $A_n$, you just have to show that these sets have the same cardinality, which is what you do. (Note the importance of the word finite here, since e.g. $\mathbb{N}\subseteq\mathbb{Z}$ and these sets have the same cardinality, but are not equal.)
This is effectively the same as providing a function $T\to A_n$, then proving that the function is both injective and surjective. However, the argument is simplified because instead of a function, you already have one as a subset of the other.