$\newcommand{\sgn}{\operatorname{sgn}}$Define $\sgn:S_{n}\to\left\{1,-1\right\}$ by
$\sgn(\sigma)=1$ if $\sigma$ is even, and $\sgn(\sigma)=-1$ if $\sigma$ is odd.
Prove that $\sgn$ is a homomorphism, where $\left\{1,-1\right\}$ is a group under multiplication.
Here is my attempt:
Proof. Let $\sgn$ be as defined and suppose $\alpha_1,\alpha_2$ are two odd permutations, while $\beta_1,\beta_2$ are two even permutations from $S_n$. It follows that $\alpha_1\alpha_2$ and $\beta_1\beta_2$ are even, while $\alpha_i\beta_i$ and $\beta_i\alpha_i$ are odd permutations for $i\in\left\{1,2\right\}$. Therefore under $\sgn$ we obtain:
\begin{align}
\sgn(\alpha_1\alpha_2) & =1=-1\cdot -1 = \sgn(\alpha_1)\sgn(\alpha_2) \\[10pt]
\sgn(\beta_1\beta_2) & =1=1\cdot 1 = \sgn(\beta_1)\sgn(\beta_2) \\[10pt]
\sgn(\alpha_i\beta_i) & =-1=-1\cdot 1 = \sgn(\alpha_i)\sgn(\beta_i) \\[10pt]
\sgn(\beta_i\alpha_i) & =-1=1\cdot -1 = \sgn(\beta_i)\sgn(\alpha_i)
\end{align}
Hence, for any $\sigma\in S_n$, $\sgn$ defines a homomorphism to $\left\{1,-1\right\}$. QED.
—
I am pretty sure this will suffice as a proof but I'd like feedback on anything that seems unusual or that could use some attention. In particular if there is a simpler way of doing this, please let me know.
Best Answer
As the proof is currently written, it could easily be interpreted as only proving (up to) six special cases of what you actually need to prove, which would not be sufficient. It would be much clearer to start with: let $\alpha, \beta \in S_n$ be any two elements. Then, you can split the proof into four cases according to whether $\alpha, \beta$ are even or odd in showing that $\sgn(\alpha\beta) = \sgn(\alpha) \sgn(\beta)$. Your proof has the essential idea needed in each of the four cases, but only restricted to one or two special cases that you fixed beforehand.
As another way of stating my objection: at most, you have proved:
$$ \forall \alpha_1, \alpha_2, \beta_1, \beta_2 \in S_n, odd(\alpha_1) \wedge odd(\alpha_2) \wedge even(\beta_1) \wedge even(\beta_2) \rightarrow \\ \sgn(\alpha_1 \alpha_2) = \sgn(\alpha_1) \sgn(\alpha_2) \wedge \cdots \wedge \sgn(\beta_2 \alpha_2) = \sgn(\beta_2) \sgn(\alpha_2). $$ So from here, you would also need to make an argument as to why this would imply: $$ \forall \alpha, \beta \in S_n, \sgn(\alpha \beta) = \sgn(\alpha) \sgn(\beta). $$ If you make this argument, you will find that it would make the overall proof be more convoluted than the proof outline I suggested.