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**ThePerfectHacker** We can just think of $\displaystyle \Omega = \{1,2,...,n\}$.

Without lose of generality say that $\displaystyle t=(12)$. Let $\displaystyle g$ be a non-identity element with cycle decomposition $\displaystyle g=c_1c_2...c_k$ where these cycles are non-identity cycles. The way you prove this is by exploring various cases. The first case is that $\displaystyle 1,2$ are not found as entries in any one of the cycles and so $\displaystyle tg$ has cycle decomposition $\displaystyle (12)c_1...c_k$ and now apply the definition of the signum function. The second case is that $\displaystyle 1,2$ are found as entries in the cycle but $\displaystyle 1,2$ are found both in a single cycle $\displaystyle c_1$ (by relabeling if necessary). Therefore, we can think of $\displaystyle c_1$ as $\displaystyle c=(1,a_1,...,a_j,2,b_1,...,b_i)$. Now $\displaystyle (12)c_1 = (1,a_1,...,a_j)(2,b_1,...,b_i)$ and so $\displaystyle (12)g = (1,a_1,...,a_j)(2,b_1,...,b_i)c_2...c_k$ and apply the definition of signum function. Another case is when $\displaystyle 1,2$ are found as entries in the cycle by $\displaystyle 1,2$ are found in different cycles say $\displaystyle c_1,c_2$ (by relabeling if necessary). So say that $\displaystyle c_1 = (1,a_1,...,a_j)$ and $\displaystyle c_2 = (2,b_1,...,b_k)$ now compute $\displaystyle (12)c_1c_2$ and apply the definition again. And finally there are cases when $\displaystyle 1\text{ or }2$ (but not both) appear as entries in one of the cycles. Just do all these cases and the proof ought to be complete. For the second part I imagine you need to use the result that every permutation is a product of transpositions.