Thread: Isomorphism between S(4) and D(4).

The problem asks to consider the subgroup of S(4) consisting of the four permutations id,(12)(34),(13)(24),(14)(23) and the group of symmetries of a rectangle and find all the isomorphism between these two groups. They already defined one by taking id to e, (12)(34) to σ, (13)(24) to R and (14)(23) to τ (τ=σR).

The hint says "Choose any two elements of order 2; what are the restrictions on where an isomorphism can take them? Having determined where these two elements are sent by the isomorphism, is there any choice for the destinations of the other two elements?

I'm not sure what they mean by restriction. I looked at the multiplication tables of the two and it seemed like as long id was sent to e, there would be an isomorphism between the two groups. So, I came up with 5 more by taking
1. id to e, (12)(34) to σ, (14)(23) to R, and (13)(24) to τ.
2. id to e, (13)(24) to σ, (12)(34) to R, and (14)(23) to τ.
3. id to e, (13)(24) to σ, (14)(23) to R, and (12)(34) to τ.
4. id to e, (14)(23) to σ, (12)(34) to R, and (13)(24) to τ.
5. id to e, (14)(23) to σ, (13)(24) to R, and (12)(34) to τ.

Is this correct? I'm a bit lost with the topic of groups, so any help is appreciated.

Hi mi986!

The restriction is in the definition of an isomorphism.
Do you have a definition with its boundary conditions handy?

Do you mean the definition of an isomorphism?

My textbooks says,

Let G and H be groups. A function θ: G->H is an isomorphism if it is a bijection and if for all x,y in G we have θ(xy)=θ(x)θ(y).
Groups G and H are isomorphic if there exists an isomorphism from G to H.

Then there are notes.
(1) If G is any group then the identity function from G to itself is an isomorphism from G to G.
(2) if θ:G->H is an isomorphism then, the inverse map θ^-1:H->G is an isomorphism.
(3) if θ:G->H and ψ:H->K are isomorphisms, then the composition ψθ:G->K is an isomorphism.
(4) Taking (1),(2),(3) together show that the relation of being isomorphic is an equivalence relation.

For the five isomorphisms that I listed, I thought that I had θ(xy)=θ(x)θ(y). And that since both groups had 4 elements, it had to be 1-1 and onto?

Yes, the restrictions are that an isomorphism is bijective and that for all x,y in G we have θ(xy)=θ(x)θ(y).

In particular that means that identity is sent to identity.
And indeed θ(xy)=θ(x)θ(y) holds in all your cases.

So you are correct!

both groups are isomorphic to the klein 4-group V = {e,a,b,ab} a² = b² = e, ab = ba.

so let's say you have two isomorphisms from G = {e, (1 2)(3 4), (1 4)(2 3), (1 3)(2 4)} to H = {e, ρ, σ, τ}, which we will call θ and ψ.

then ψ^-1θ is an automorphism of G (isomorphism from G to itself) and ψθ^-1 is an automorphism of H.

so what we are really doing is determining Aut(V) (or G, or H).

that is: if φ:V→G is the isomorphism:

e→e
a→(1 2)(3 4)
b→(1 4)(2 3)
ab→(1 3)(2 4)

and λ:V→H is the isomorphism:

e→e
a→ρ
b→σ
ab→τ

then if f is in Aut(V), then φf is in Aut(G) and λf is in Aut(H).

note that any automorphism of V is completely determined by where we send the two generators, a and b (e has to go back to e itself, and ab has to go to the the product of the image of a and the image of b). this gives the following automorphisms:

a→a
b→b (the identity map)

a→a
b→ab (and so ab→(a)(ab) = b)

a→b
b→ab (and so ab→(b)(ab) = (ab)(b) = a)

a→b
b→a (guess what ab goes to...?)

a→ab
b→b

a→ab
b→a

note that 3 of these maps switch two of the elements of order 2 (a,b, and ab)...these are the second, fourth and fifth automorphisms on our list. thus these automorphisms are of order 2. the identity leaves all elements of V fixed, and the two remaining automorphisms permute a,b, and ab in a 3-cycle (the third and sixth ones, the third does: a→b→ab→a, and the sixth one goes in the reverse order a→ab→b→a). hopefully this is enough to convince you that Aut(V), and thus also Aut(G) and Aut(H) is a non-cyclic group of order 6 (which is thus isomorphic to S3).

so if you found six isomorphisms from G to H, you can now be confident you've found them all. here is what is time-saving about this approach:

if you pick the "a" of G to be (1 2)(3 4) and the "a" of H to be ρ, and the "b" of G to be (1 4)(2 3) and the "b" of H to be σ, and θ to be your proposed isomorphism between G and H, then all you need to do to verify that θ is a homomorphism is verify the SINGLE equation:
θ((1 2)(3 4)(1 4)(2 3)) = θ((1 3)(2 4)) = θ((1 2)(3 4))θ((1 4)(2 3)), instead of checking EVERY possible product (which would mean checking 16 possible products for EACH isomorphism θ).