# Proving/Disproving (via formal proof) that the Alternating group $A_n$ is a subgroup of the Symmetric group $S_n$

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

• You have to show that for every element $\sigma\in A_n$ also $\sigma^{-1}\in A_n$ and for all $\sigma,\tau\in A_n$ it follows that $\sigma\circ\tau\in A_n$; not a few you have chosen at random. – mrtaurho Apr 28 '20 at 21:45
• $A_n$ is finite, so nonemptiness and closure are enough. – user750041 Apr 28 '20 at 21:47
• Your definition of $A_n$ is wrong. The permutation $(1,2)(3,4,5)$ has even order (6) but it is NOT an even permutation. Also $(1,2,3)$ is even permutation but has odd order. – N. S. Apr 28 '20 at 21:47
• @N.S. hmmm, i see your point. would defining it as { f in $S_n$ | f is even} be more clear?? also, could you define what you mean by 'order', from my enjoyment readings i thought order was like cardinality where |(1 2)(3 4 5)| would be order 5 since it contains 5 elements? – user1618033988749895 Apr 28 '20 at 22:43

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 $$S_n$$.

• Let me know if you have any further problem in this. – Pritam Apr 28 '20 at 22:22
• Thanks! i tried to upvote this answer but since im a newbie and only have a reputation of 11 it doesn't show.... But this definitely clarifies how the inverse relation works with these groups! – user1618033988749895 Apr 28 '20 at 22:45
• Yes, this is how it works. No problem and welcome to MSE..! I've given your question an upvote.😉 – Pritam Apr 28 '20 at 22:56
• By the way you can verify the answer I guess. I'll appreciate if you do that. – Pritam Apr 28 '20 at 23:08
• after the recent upvotes/answers i have enough reputation! haha thanks again for the help on thinking this through – user1618033988749895 Apr 29 '20 at 5:42