So I have a set $G = \{(1, 2, 3, ... n)^a \cdot (n+1, n+2, n+3, ... 2n)^b \mid 0 \leq a,b \leq n-1\}$ which is defined for all $n \in \mathbb{N}$, $n>3$. Note, $(1, 2, 3, ... n)$ and $(n+1, n+2, n+3, ... 2n)$ are permutations expressed as k-cycles, and the operation associated with this "potential group" is composition of functions.
I know it is obviously a non-empty subset of $S_{2n}$ - the symmetric group on $2n$ points - and I'm trying to show that it is a subgroup by checking the closure and inverses axioms.
CLOSURE:
Let $\alpha = (1, 2, 3, ... n)$ and $\beta = (n+1, n+2, n+3, ... 2n)$. Let $g = \alpha^{a_1} \cdot \beta^{b_1}, h = \alpha^{a_2} \cdot \beta^{b_2} \in G$ where $0 \leq a_1, a_2, b_1, b_2 \leq n-1$. Then $g \cdot h = \alpha^{a_1+a_2} \cdot \beta^{b_1+b_2}$ since $\alpha$ and $\beta$ are disjoint cycles. I don't know how to show (formally) that this is also in $G$.
INVERSES:
The identity of $G$ is the identity permutation (surely?). I want to say that $g^{-1} = \alpha^{n-a_1} \cdot \beta^{n-b_1} \in G$ because then $g \cdot g^{-1} = g^{-1} \cdot g = e$ but I don't know how to (formally) show that $\alpha^{n-a_1} \cdot \beta^{n-b_1}$ is always in $G$.
I have been aware of the fact that this is a direct product of two groups - and therefore it is immediately a group, but I haven't learnt about direct products yet, and have been asked to prove that it is a subgroup by checking the closure and inverses axioms.
I would really appreciate any help,
Jack
The point you missed in the Closure proof (other than stating the obvious fact $\alpha^k$ commutes with $\beta^m$ for all $k,m$) is that $\alpha^n=\beta^n=I$ so $$ g\cdot h = \alpha^{(a_1+a_2)\pmod n} \beta^{(b_1+b_2)\pmod n } $$ and both those terms fit the requirements so that $g\cdot h \in G$.
The inverse property uses the same mod property since when $a_1 =0$, $n-a_1 = n = 0\pmod{n}$ which places your candidate for $g^{-1}$ in $G$.