intuitively this makes sense and i can conceptualize how this would work, but i struggle to formalize a proof to express what my ideas are…
Henceforth,
Let n be an element of $ \mathbb Z^+ $ and let the alternating group $A_n$ = all the permutations of even order (which is what i believe the definition to be)
so now when trying to prove that this is a subgroup of $S_n$ i know that i would have to find a way to show that $A_n$ is non-empty, closed under operation, and closed under inverses
My idea:
non-empty: there exists an even permutation that defines the identity (by definition on identity being even), so with there existing an identity in $S_n$ then we know it's even definition will also be in $A_n$
operation: (1 2 3)(1 2 3) = an element in $S_n$ and $A_n$ = (1 2 3)(1 2 3) and this is even because it can be written as an even number of transpositions (1 2)(1 3)(1 3)(1 2) thus being an element of the Alternating group
inverses: take arbitrary element that is in $A_n$ and to prove that it's inverse is also in $A_n$ I would show that (3 8 1) is in $A_n$ since it's cycle can be written as even number of transpositions, then by the definition of cycle notation we know that it's inverse is the same cycle and is thus also even and an element of $A_n$
any and all input is appreciated and will help me think through this problem
Best Answer
Let $ σ = γ_1γ_2 . . . γ_{2k} $ and $ τ = δ_1δ_2 . . . δ_{2l}$
Then $στ^{−1} = γ_1γ_2 . . . γ_{2k}δ_{2l}^{-1}δ_{2l−1}^{-1}. . . δ_{1}^1 = γ_1γ_2 . . . γ_{2k}δ_{2l}δ_{2l−1} . . . δ_1$ $\in A_n$
Therefore $σ,τ\in A_n \implies στ^{-1} \in A_n$. So $A_n$ is a subgroup of $S_n$.