Hello I am struggling with these 2 questions.

1) Prove that $\displaystyle S_3 $ and $\displaystyle D_6 $ (symmetry group of the equilateral triangle) are isomorphic.

It's quite clear that they are but is there a way to show this without having to build an isomorphism that would take a long time as I would have to show every pair of permutations f,g in $\displaystyle S_3 $ would give $\displaystyle \phi (fg) = \phi (f) \phi (g) $

I have this theorem in my notes: Every group G is isomorphic to a subgroup of $\displaystyle S_x $ for some choice of set X. [X = { bijections f: f: X -> X }]

**In particular, if $\displaystyle |G| < \infty $ then G and H are isomorphic for some $\displaystyle H \le S_n $ for some $\displaystyle n \in \mathbb{N} $.**
Am I supposed to use this? Or can I just draw a Cayley table of both groups and say look the two groups are algebraically indistinguishable and hence are isomorphic? I found a few proofs on the internet but I don't understand the notation or theorems they use.

2)Let G be a group and suppose that G has subgroups H and K for which the following conditions all hold:

(a) G = HK, (b) $\displaystyle H \cap K = {1} $, (c) $\displaystyle H \lhd G $, (d) $\displaystyle K \lhd G $.

Show also that $\displaystyle G \cong H x K $.

HINT:

Show first that any element of G can be uniquely expressed in the form hk, where $\displaystyle h \in H, k \in K $. Then, by considering elements of the form$\displaystyle h^{-1}k^{-1}hk $ where $\displaystyle h \in H, k \in K $ show that every element in H

commutes with every element of K. Finally by considering a suitable mapping from H X K to G obtain the required isomorphism.

OK I can make the isomorphism from H x K to G but in proving it is bijective I need to use commutativity and uniqueness of elements in G so how do i prove these 2 things ( in red ) ? I have pages and pages of h's and k's. I was trying to show that $\displaystyle h^{-1}k^{-1}hk $was in H and in K and therefore in $\displaystyle H \cap K = {1} $

And of course if $\displaystyle h^{-1}k^{-1}hk = 1 $ then $\displaystyle hk = kh $

Any help and discussion would be appreciated.